Chapter 3 Pair of Linear Equations in Two Variables (Additional Questions)

The correct option is (D) $a \neq 0$ and $b \neq 0$.

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A linear equation in two variables, say $x$ and $y$, is an equation that can be written in the form $ax + by + c = 0$, where $a$, $b$, and $c$ are real numbers.

For the equation to be considered a linear equation in two variables, both variables must be present with non-zero coefficients.

Consider the condition $a \neq 0$ and $b \neq 0$. In this case, both the $x$ term and the $y$ term are present in the equation $ax + by + c = 0$, making it a linear equation in two variables.

Let's examine other options:

If $a = 0$ and $b \neq 0$, the equation becomes $0 \cdot x + by + c = 0$, which simplifies to $by + c = 0$. This is a linear equation in only one variable ($y$).

If $a \neq 0$ and $b = 0$, the equation becomes $ax + 0 \cdot y + c = 0$, which simplifies to $ax + c = 0$. This is a linear equation in only one variable ($x$).

If $a = 0$ and $b = 0$, the equation becomes $0 \cdot x + 0 \cdot y + c = 0$, which simplifies to $c = 0$. This is not a linear equation involving any variable; it is just a condition on the constant $c$.

Therefore, for the equation $ax + by + c = 0$ to be a linear equation specifically in two variables $x$ and $y$, it is essential that the coefficients of both $x$ and $y$ are non-zero.

Hence, the condition is $a \neq 0$ and $b \neq 0$.

The correct option is (C) Straight line.

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A linear equation in two variables, such as $ax + by + c = 0$ (where $a$ and $b$ are not both zero), represents a relationship between the variables $x$ and $y$ such that plotting all pairs $(x, y)$ that satisfy the equation results in a specific geometric shape.

By definition, the graph of a linear equation in two variables is always a straight line in the Cartesian coordinate system.

Options (A) Curve, (B) Parabola, and (D) Circle represent graphs of other types of equations (non-linear equations in general). A parabola is typically the graph of a quadratic equation, and a circle is represented by an equation like $(x-h)^2 + (y-k)^2 = r^2$, which involves squared terms.

Since the equation $ax + by + c = 0$ involves only variables raised to the power of 1, its graph is linear, i.e., a straight line.

The correct option is (C) Infinitely many solutions.

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A linear equation in two variables, such as $ax + by + c = 0$ (where $a$ and $b$ are not both zero), represents a relationship between the variables $x$ and $y$.

A solution to this equation is any ordered pair $(x, y)$ that makes the equation true when substituted into it.

Geometrically, the set of all solutions to a linear equation in two variables forms a straight line in the Cartesian coordinate system.

A straight line consists of an infinite number of points.

Each point on the line corresponds to a unique solution $(x, y)$ of the equation, and every solution $(x, y)$ corresponds to a point on the line.

Since there are infinitely many points on a straight line, a linear equation in two variables typically has infinitely many solutions.

The correct option is (A) Intersecting lines.

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For a pair of linear equations in two variables:

$a_1 x + b_1 y + c_1 = 0$

$a_2 x + b_2 y + c_2 = 0$

The nature of their graphical representation (and the number of solutions) is determined by the ratios of their coefficients.

There are three main cases:

1. If $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, the lines are intersecting at exactly one point. This system has a unique solution.

2. If $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the lines are parallel and distinct. They do not intersect. This system has no solution.

3. If $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, the lines are coincident (one lies exactly on top of the other). They intersect at infinitely many points. This system has infinitely many solutions.

The given condition is $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$. According to the rules, this corresponds to the case where the lines are intersecting.

Option (D) Perpendicular lines is a specific type of intersecting lines, but the general condition $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$ does not guarantee perpendicularity; it only guarantees intersection.

The correct option is (C) Coincident.

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For a pair of linear equations in two variables, the number of solutions is directly related to how the lines representing the equations intersect on a graph.

If the lines are intersecting at a single point, the system has exactly one unique solution.

If the lines are parallel and distinct, they never intersect, so the system has no solution.

If the lines are coincident, it means one line lies exactly on top of the other. They share all their points in common.

Since every point on a line represents a solution to its equation, and coincident lines share all their points, a system whose lines are coincident has infinitely many solutions.

The condition for coincident lines for the equations $a_1 x + b_1 y + c_1 = 0$ and $a_2 x + b_2 y + c_2 = 0$ is $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$. When this condition holds, the equations are essentially multiples of each other, representing the same line.

Therefore, if a pair of linear equations has infinitely many solutions, the lines are coincident.

The correct option is (C) It has at least one solution.

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In the context of a pair of linear equations in two variables, a system of equations is classified based on the number of solutions it has.

A system is called consistent if it has one or more solutions.

This includes two cases:

1. The system has exactly one unique solution (corresponding to intersecting lines).

2. The system has infinitely many solutions (corresponding to coincident lines).

A system is called inconsistent if it has no solution (corresponding to parallel and distinct lines).

Therefore, a consistent system is one that has at least one solution, covering both the unique solution and infinitely many solutions cases.

The correct option is (C) Inconsistent.

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For a pair of linear equations in two variables, the graphical representation provides a clear understanding of the number of solutions.

If the lines are parallel and distinct, they never meet or intersect at any point in the coordinate plane.

The solution(s) to a system of linear equations correspond to the point(s) where the graphs of the equations intersect.

Since parallel lines have no intersection point, a pair of linear equations represented by parallel lines has no solution.

A system of linear equations is classified as inconsistent if it has no solution.

A system is classified as consistent if it has at least one solution (either a unique solution or infinitely many solutions).

Therefore, if a pair of linear equations is represented by parallel lines, it has no solution and is classified as inconsistent.

Note: Option (D) is incorrect because a consistent system *by definition* must have at least one solution.

The correct option is (C) $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$.

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For a pair of linear equations $a_1 x + b_1 y + c_1 = 0$ and $a_2 x + b_2 y + c_2 = 0$, the number of solutions depends on the ratios of the coefficients $a_1, b_1, c_1$ and $a_2, b_2, c_2$.

Here are the conditions:

1. Unique Solution: The system has exactly one solution if the lines are intersecting. This occurs when the ratio of the $x$ coefficients is not equal to the ratio of the $y$ coefficients.

The condition is: $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$.

2. Infinitely Many Solutions: The system has infinitely many solutions if the lines are coincident. This occurs when the ratios of all corresponding coefficients are equal.

The condition is: $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.

3. No Solution: The system has no solution if the lines are parallel and distinct. This occurs when the ratio of the $x$ coefficients is equal to the ratio of the $y$ coefficients, but this is not equal to the ratio of the constant terms.

The condition is: $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$.

The question asks for the condition when the system has no solution, which matches the third case: $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$.

The correct option is (B) Intersection.

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The graphical method for solving a pair of linear equations involves plotting the graph of each equation on the same coordinate plane.

As established earlier, the graph of a linear equation in two variables is a straight line.

A solution to a system of two linear equations is an ordered pair $(x, y)$ that satisfies both equations simultaneously.

On a graph, this corresponds to a point that lies on both lines.

The point(s) that lie on both lines are the point(s) where the lines intersect.

If the lines are intersecting, they meet at exactly one point, giving a unique solution.

If the lines are parallel, they do not intersect, indicating no solution.

If the lines are coincident, they overlap completely, meaning they intersect at infinitely many points, indicating infinitely many solutions.

In all cases where solutions exist (intersecting or coincident lines), the graphical method finds these solutions by identifying the point(s) of intersection of the lines.

The correct option is (B) Substitution method.

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Algebraic methods for solving a pair of linear equations are techniques that involve manipulating the equations themselves using algebraic operations (like addition, subtraction, multiplication, substitution) to find the values of the variables.

Common algebraic methods include:

• Substitution method
• Elimination method (or Addition method)
• Cross-multiplication method

Let's look at the given options:

(A) Graphical method: This involves plotting the lines representing the equations and finding their intersection point. This is a geometric or graphical method, not algebraic.

(B) Substitution method: This method involves solving one equation for one variable and substituting that expression into the other equation. This is a purely algebraic process.

(C) Factoring method: This method is used to solve polynomial equations by expressing a polynomial as a product of its factors. It is not a standard method for solving systems of linear equations.

(D) Remainder theorem: This theorem is related to polynomial division and is used to find the remainder when a polynomial is divided by a linear factor. It is not a method for solving systems of linear equations.

Therefore, the Substitution method is an algebraic method for solving a pair of linear equations.

The correct option is (A) $x=3, y=2$.

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We are given the system of linear equations:

$x + y = 5 \quad$ ... (1)

$x - y = 1 \quad$ ... (2)

We will use the substitution method to solve this system.

From equation (1), we can express $y$ in terms of $x$:

$y = 5 - x \quad$ ... (3)

Now, substitute the expression for $y$ from equation (3) into equation (2):

$x - (5 - x) = 1$

Simplify and solve for $x$:

$x - 5 + x = 1$

$2x - 5 = 1$

$2x = 1 + 5$

$2x = 6$

$x = \frac{6}{2}$

$x = 3$

Now substitute the value of $x=3$ back into equation (3) to find the value of $y$:

$y = 5 - x$

$y = 5 - 3$

$y = 2$

Thus, the solution to the system of equations is $x=3$ and $y=2$.

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Verification:

Substitute $x=3$ and $y=2$ into equation (1):

$x + y = 3 + 2 = 5$. (Correct)

Substitute $x=3$ and $y=2$ into equation (2):

$x - y = 3 - 2 = 1$. (Correct)

The solution is verified.

The correct option is (B) $x=2, y=1$.

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We are given the system of linear equations:

$2x + 3y = 7 \quad$ ... (1)

$3x - 2y = 4 \quad$ ... (2)

We will use the elimination method to solve this system.

To eliminate $y$, we can multiply equation (1) by 2 and equation (2) by 3. The coefficient of $y$ in equation (1) will become $3y \times 2 = 6y$, and in equation (2), it will become $-2y \times 3 = -6y$. Then, we can add the resulting equations.

Multiply equation (1) by 2:

$2 \times (2x + 3y) = 2 \times 7$

$4x + 6y = 14 \quad$ ... (3)

Multiply equation (2) by 3:

$3 \times (3x - 2y) = 3 \times 4$

$9x - 6y = 12 \quad$ ... (4)

Now, add equation (3) and equation (4):

$(4x + 6y) + (9x - 6y) = 14 + 12$

$4x + 9x + 6y - 6y = 26$

$13x + 0y = 26$

$13x = 26$

$x = \frac{26}{13}$

$x = 2$

Now substitute the value of $x=2$ back into equation (1) to find the value of $y$:

$2x + 3y = 7$

$2(2) + 3y = 7$

$4 + 3y = 7$

$3y = 7 - 4$

$3y = 3$

$y = \frac{3}{3}$

$y = 1$

Thus, the solution to the system of equations is $x=2$ and $y=1$.

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Verification:

Substitute $x=2$ and $y=1$ into equation (1):

$2x + 3y = 2(2) + 3(1) = 4 + 3 = 7$. (Correct)

Substitute $x=2$ and $y=1$ into equation (2):

$3x - 2y = 3(2) - 2(1) = 6 - 2 = 4$. (Correct)

The solution is verified.

The correct option is (A) $x = \frac{b_1 c_2 - b_2 c_1}{a_1 b_2 - a_2 b_1}$.

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For a system of linear equations in two variables:

$a_1 x + b_1 y + c_1 = 0$

$a_2 x + b_2 y + c_2 = 0$

The cross-multiplication method provides a formula to directly find the values of $x$ and $y$. The setup for the cross-multiplication method is based on the following relationship:

$\frac{x}{b_1 c_2 - b_2 c_1} = \frac{y}{c_1 a_2 - c_2 a_1} = \frac{1}{a_1 b_2 - a_2 b_1}$

Provided that the denominator $(a_1 b_2 - a_2 b_1)$ is not equal to zero (which indicates a unique solution).

To find the formula for $x$, we equate the first fraction with the third fraction:

$\frac{x}{b_1 c_2 - b_2 c_1} = \frac{1}{a_1 b_2 - a_2 b_1}$

Multiplying both sides by $(b_1 c_2 - b_2 c_1)$, we get:

$x = \frac{b_1 c_2 - b_2 c_1}{a_1 b_2 - a_2 b_1}$

This matches the formula given in option (A).

Similarly, to find the formula for $y$, we equate the second fraction with the third fraction:

$\frac{y}{c_1 a_2 - c_2 a_1} = \frac{1}{a_1 b_2 - a_2 b_1}$

Multiplying both sides by $(c_1 a_2 - c_2 a_1)$, we get:

$y = \frac{c_1 a_2 - c_2 a_1}{a_1 b_2 - a_2 b_1}$

The question specifically asks for the formula for $x$, which is $\frac{b_1 c_2 - b_2 c_1}{a_1 b_2 - a_2 b_1}$.

The correct option is (A) Let $\frac{1}{x} = p$, $\frac{1}{y} = q$.

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We are given the pair of equations:

$\frac{2}{x} + \frac{3}{y} = 13 \quad$ ... (1)

$\frac{5}{x} - \frac{4}{y} = -2 \quad$ ... (2)

These equations are not linear in $x$ and $y$ because the variables appear in the denominators.

However, if we look at the terms involving the variables, they are of the form $\frac{1}{x}$ and $\frac{1}{y}$.

Consider the substitution given in option (A):

Let $\frac{1}{x} = p$ and $\frac{1}{y} = q$.

Substitute these new variables into equation (1):

$2\left(\frac{1}{x}\right) + 3\left(\frac{1}{y}\right) = 13$

$2p + 3q = 13 \quad$ ... (3)

Substitute the new variables into equation (2):

$5\left(\frac{1}{x}\right) - 4\left(\frac{1}{y}\right) = -2$

$5p - 4q = -2 \quad$ ... (4)

The resulting system of equations (3) and (4):

$2p + 3q = 13$

$5p - 4q = -2$

is a pair of linear equations in the variables $p$ and $q$. We can solve this system for $p$ and $q$ using methods like substitution or elimination, and then use the original substitutions ($\frac{1}{x} = p$, $\frac{1}{y} = q$) to find the values of $x$ and $y$.

The other options (B), (C), and (D) would not transform the original equations into a linear form.

Therefore, the suitable substitution to reduce the given equations to linear form is letting $\frac{1}{x} = p$ and $\frac{1}{y} = q$.

The correct option is (A) Both A and R are true and R is the correct explanation of A.

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Let's analyze the given pair of linear equations:

Equation 1: $x + y = 4$, which can be written as $x + y - 4 = 0$.

Here, $a_1 = 1$, $b_1 = 1$, $c_1 = -4$.

Equation 2: $2x + 2y = 8$, which can be written as $2x + 2y - 8 = 0$.

Here, $a_2 = 2$, $b_2 = 2$, $c_2 = -8$.

Let's find the ratios of the corresponding coefficients:

$\frac{a_1}{a_2} = \frac{1}{2}$

$\frac{b_1}{b_2} = \frac{1}{2}$

$\frac{c_1}{c_2} = \frac{-4}{-8} = \frac{4}{8} = \frac{1}{2}$

We observe that $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$. Specifically, $\frac{1}{2} = \frac{1}{2} = \frac{1}{2}$ (as $\frac{4}{8}$ simplifies to $\frac{1}{2}$). The Reason states $\frac{1}{2} = \frac{1}{2} = \frac{4}{8}$, which is also true before simplification.

Assertion (A): The condition $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$ means that the lines representing the pair of equations are coincident. When lines are coincident, they share all points, resulting in infinitely many solutions.

Since the given equations satisfy this condition ($\frac{1}{2} = \frac{1}{2} = \frac{1}{2}$), the pair of equations has infinitely many solutions. Thus, Assertion (A) is true.

Reason (R): The reason correctly states that the lines are coincident and provides the condition $\frac{1}{2} = \frac{1}{2} = \frac{4}{8}$, which is the correct ratio of coefficients for the given equations ($a_1/a_2=1/2, b_1/b_2=1/2, c_1/c_2 = -4/-8=4/8$). The condition $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$ is the indeed the condition for coincident lines.

Thus, Reason (R) is true.

Furthermore, the fact that the lines are coincident is precisely why the system has infinitely many solutions. Therefore, Reason (R) correctly explains Assertion (A).

Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation for Assertion (A).

The correct option is (A) Both A and R are true and R is the correct explanation of A.

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We are given the pair of linear equations:

$x - 2y = 4$, which can be written as $x - 2y - 4 = 0$.

Comparing this with $a_1 x + b_1 y + c_1 = 0$, we have $a_1 = 1$, $b_1 = -2$, $c_1 = -4$.

$2x - 4y = 12$, which can be written as $2x - 4y - 12 = 0$.

Comparing this with $a_2 x + b_2 y + c_2 = 0$, we have $a_2 = 2$, $b_2 = -4$, $c_2 = -12$.

Let's calculate the ratios of the corresponding coefficients:

$\frac{a_1}{a_2} = \frac{1}{2}$

$\frac{b_1}{b_2} = \frac{-2}{-4} = \frac{2}{4} = \frac{1}{2}$

$\frac{c_1}{c_2} = \frac{-4}{-12} = \frac{4}{12} = \frac{1}{3}$

We observe that $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$. Specifically, $\frac{1}{2} = \frac{1}{2} \neq \frac{1}{3}$.

Assertion (A): The pair of equations is inconsistent.

A pair of linear equations is inconsistent if it has no solution. A system has no solution if the lines representing the equations are parallel and distinct. The condition $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$ corresponds to parallel and distinct lines, which means there is no intersection point and thus no solution. Therefore, the system is inconsistent. Assertion (A) is true.

Reason (R): The lines representing the equations are parallel because $\frac{1}{2} = \frac{-2}{-4} \neq \frac{4}{12}$.

The condition for a pair of linear equations to be represented by parallel lines is $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$. The reason states this condition using the coefficients of the given equations, which evaluates to $\frac{1}{2} = \frac{1}{2} \neq \frac{1}{3}$. This condition is indeed satisfied by the given equations, and it is the correct condition for parallel lines. Thus, Reason (R) is true.

Since an inconsistent system is defined as a system with no solution, and the lack of solution arises precisely because the lines are parallel and distinct (as stated and justified in the Reason), the Reason correctly explains why the Assertion is true.

Therefore, both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

The correct option is (A) (i)-(c), (ii)-(b), (iii)-(a), (iv)-(d).

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For a pair of linear equations $a_1 x + b_1 y + c_1 = 0$ and $a_2 x + b_2 y + c_2 = 0$, the relationship between the coefficients and the nature of the solutions and graph is as follows:

1. If $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$:

The lines are intersecting.

The system has a unique solution.

This matches option (c) in Column B. So, (i) - (c).

2. If $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$:

The lines are coincident.

The system has infinitely many solutions.

This matches option (b) in Column B. So, (ii) - (b).

3. If $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$:

The lines are parallel and distinct.

The system has no solution.

This matches option (a) in Column B. So, (iii) - (a).

4. The condition $\frac{a_1}{a_2} \neq \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$:

If $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, the system already falls into case (i) and has a unique solution and intersecting lines, regardless of the relation with $\frac{c_1}{c_2}$. The way condition (iv) is written does not represent a distinct standard case beyond the first three for classifying linear systems. Therefore, based on the provided options, this condition is likely intended to correspond to the case where the standard classifications using ratios don't apply in the format given by (iv), or it's a poorly phrased condition. Matching it with (d) "Cannot determine nature of solution and graph" within the context of this specific question and options seems the most appropriate choice.

So, (iv) - (d).

Combining the matches, we get: (i)-(c), (ii)-(b), (iii)-(a), (iv)-(d).

This corresponds to option (A).

The correct option is (A) $y+10 = 2(x-10)$, $x+10 = 3(y-10)$.

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Let Rohit's initial amount of money be $\textsf{₹} x$.

Let Virat's initial amount of money be $\textsf{₹} y$.

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Condition 1: If Rohit gives $\textsf{₹} 10$ to Virat.

Money left with Rohit = $\textsf{₹} (x - 10)$

Money Virat will have = $\textsf{₹} (y + 10)$

According to the condition, Virat will have twice the money left with Rohit:

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Condition 2: If Virat gives $\textsf{₹} 10$ to Rohit.

Money Rohit will have = $\textsf{₹} (x + 10)$

Money left with Virat = $\textsf{₹} (y - 10)$

According to the condition, Rohit will have thrice the money left with Virat:

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The pair of linear equations representing the situation is:

$y + 10 = 2(x - 10)$

$x + 10 = 3(y - 10)$

This matches the equations given in option (A).

We are given the pair of linear equations derived from the case study in Question 18:

If Rohit gives $\textsf{₹} 10$ to Virat: $y+10 = 2(x-10)$

If Virat gives $\textsf{₹} 10$ to Rohit: $x+10 = 3(y-10)$

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Let's simplify these equations.

From the first equation:

$y + 10 = 2x - 20$

$2x - y = 10 + 20$

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From the second equation:

$x + 10 = 3y - 30$

$x - 3y = -30 - 10$

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We now have the system of linear equations:

$2x - y = 30$

$x - 3y = -40$

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We can solve this system using the elimination method. Let's eliminate $y$. Multiply equation (1) by 3:

$3 \times (2x - y) = 3 \times 30$

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Now subtract equation (2) from equation (3):

$(6x - 3y) - (x - 3y) = 90 - (-40)$

$6x - 3y - x + 3y = 90 + 40$

$5x = 130$

$x = \frac{130}{5}$

$x = 26$

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Substitute the value of $x=26$ back into equation (1):

$2x - y = 30$

$2(26) - y = 30$

$52 - y = 30$

$-y = 30 - 52$

$-y = -22$

$y = 22$

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The solution to the system of equations is $x=26$ and $y=22$.

Therefore, Rohit's initial money is $\textsf{₹} 26$ and Virat's initial money is $\textsf{₹} 22$.

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Note: The calculated solution ($x=26, y=22$) based on the problem statement and the derived equations does not match any of the options provided in the question.

The correct option is (B) $\frac{1}{x-1} + \frac{1}{y-2} = 3$.

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An equation is said to be reducible to a linear equation in two variables if, by making a suitable substitution for expressions involving the original variables, it can be transformed into an equation that is linear in the new variables.

A linear equation in two variables (say, $p$ and $q$) is of the form $Ap + Bq + C = 0$ or $Ap + Bq = C$, where $A, B, C$ are constants and $A, B$ are not both zero.

Let's examine the given options:

(A) $x^2 + y^2 = 10$. If we let $p = x^2$ and $q = y^2$, the equation becomes $p + q = 10$. This is a linear equation in the new variables $p$ and $q$.

(B) $\frac{1}{x-1} + \frac{1}{y-2} = 3$. Let $p = \frac{1}{x-1}$ and $q = \frac{1}{y-2}$. Substituting these into the equation gives $p + q = 3$. This is a linear equation in the new variables $p$ and $q$.

(C) $\sqrt{x} + \sqrt{y} = 5$. Let $p = \sqrt{x}$ and $q = \sqrt{y}$. Substituting these into the equation gives $p + q = 5$. This is a linear equation in the new variables $p$ and $q.

(D) $xy = 7$. If we take logarithms on both sides (assuming $x, y > 0$), $\log(xy) = \log 7$, which is $\log x + \log y = \log 7$. Let $p = \log x$ and $q = \log y$. Then $p + q = \log 7$. This is a linear equation in the new variables $p$ and $q$.

Based on a strict interpretation, options (A), (B), (C), and (D) (using logs) can all be reduced to a linear equation in two new variables. However, in the context of the chapter "Pairs of Linear Equations in Two Variables" (especially at the secondary school level), "reducible to linear form" typically refers to equations where the variables (or linear expressions involving variables) are in the denominator, allowing substitution of reciprocals to obtain a linear system.

Among the given options, the form shown in (B) is the most standard example of an equation that is part of a system commonly solved by reducing it to a linear system through substitution of reciprocal expressions.

While (A) and (C) can also be made linear in other variables, the form in (B) is the primary type discussed as "reducible" in this specific topic alongside reciprocal forms involving just $x$ and $y$ (like $\frac{1}{x} + \frac{1}{y} = k$). Option (D) requires logarithms, which isn't the typical method covered in this specific topic of reducing equations to linear form.

Therefore, option (B) represents the most conventional example of an equation type considered "reducible to a linear equation in two variables" in this context.

The correct option is (C) intersect.

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The graphical method for solving a pair of linear equations involves plotting the lines represented by each equation.

A solution to a system of equations is an ordered pair $(x, y)$ that satisfies both equations simultaneously.

Graphically, this means the point $(x, y)$ lies on both lines.

The point(s) that lie on both lines are the point(s) where the lines meet.

If the lines are intersecting, they meet at exactly one point, which corresponds to a unique solution. The coordinates of this single point of intersection are the solution.

If the lines are coincident (overlap completely), they "intersect" at every point along their length. This means there are infinitely many points of intersection, and each of these points is a solution. The coordinates of any point on the line represent a solution.

If the lines are parallel and distinct, they never meet, meaning there are no points of intersection, and thus no solution.

In all cases where solutions exist (intersecting or coincident lines), the solutions are the coordinates of the point(s) where the lines intersect.

The correct option is (C) Coincident.

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We are given the system of linear equations:

$2x + y = 6$

$4x + 2y = 12$

To determine the nature of the lines represented by these equations, we compare the ratios of their coefficients in the standard form $a_1 x + b_1 y + c_1 = 0$ and $a_2 x + b_2 y + c_2 = 0$.

Rewrite the equations in standard form:

$2x + y - 6 = 0 \quad$ ($a_1 = 2, b_1 = 1, c_1 = -6$)

$4x + 2y - 12 = 0 \quad$ ($a_2 = 4, b_2 = 2, c_2 = -12$)

Calculate the ratios of the corresponding coefficients:

$\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}$

$\frac{b_1}{b_2} = \frac{1}{2}$

$\frac{c_1}{c_2} = \frac{-6}{-12} = \frac{6}{12} = \frac{1}{2}$

We observe that all three ratios are equal:

$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$

Specifically, $\frac{1}{2} = \frac{1}{2} = \frac{1}{2}$.

This condition indicates that the two linear equations are equivalent (one is a multiple of the other). Graphically, this means the lines representing the equations occupy the exact same position in the coordinate plane.

Lines that lie exactly on top of each other are called coincident lines.

Coincident lines have infinitely many points in common, which corresponds to a system with infinitely many solutions.

Therefore, the lines for the given system of equations are coincident.

The correct option is (D) 10.

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We are given the pair of linear equations:

$x + 2y = 3$

$5x + ky = 15$

We rewrite these equations in the standard form $ax + by + c = 0$:

$x + 2y - 3 = 0 \quad$ ... (1)

$5x + ky - 15 = 0 \quad$ ... (2)

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Comparing equation (1) with $a_1 x + b_1 y + c_1 = 0$, we get $a_1 = 1$, $b_1 = 2$, $c_1 = -3$.

Comparing equation (2) with $a_2 x + b_2 y + c_2 = 0$, we get $a_2 = 5$, $b_2 = k$, $c_2 = -15$.

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For a system of linear equations to have infinitely many solutions, the lines representing the equations must be coincident. The condition for coincident lines is:

$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$

---

Substitute the coefficients from our equations into this condition:

$\frac{1}{5} = \frac{2}{k} = \frac{-3}{-15}$

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Let's simplify the ratio of the constant terms:

$\frac{-3}{-15} = \frac{3}{15} = \frac{1}{5}$

So the condition becomes:

$\frac{1}{5} = \frac{2}{k} = \frac{1}{5}$

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To find the value of $k$, we can use the first equality:

$\frac{1}{5} = \frac{2}{k}$

Cross-multiply:

$1 \times k = 5 \times 2$

$k = 10$

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Thus, the value of $k$ for which the system of equations has infinitely many solutions is 10.

The correct option is (A) $x+y=20, x-y=4$.

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Let the two numbers be $x$ and $y$.

The first condition given is that the sum of the two numbers is 20.

This can be written as a linear equation:

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The second condition given is that their difference is 4.

This can be written as a linear equation. Assuming $x$ is the larger number, the difference is $x - y$.

If we assumed $y$ was the larger number, the difference would be $y-x=4$. However, option (A) uses $x-y=4$. We can proceed with option (A) and check if a valid solution exists.

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The pair of equations representing this situation is:

$x + y = 20$

$x - y = 4$

This matches the equations in option (A).

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Solving the system (Optional, as the question only asks for the pair of equations):

We can add the two equations:

$(x + y) + (x - y) = 20 + 4$

$2x = 24$

$x = \frac{24}{2}$

$x = 12$

Substitute $x=12$ into the first equation:

$12 + y = 20$

$y = 20 - 12$

$y = 8$

The numbers are 12 and 8. Their sum is $12+8=20$ and their difference is $12-8=4$. This confirms that the equations in option (A) correctly represent the problem.

The point $(2, 1)$ is a solution to a pair of equations if substituting $x=2$ and $y=1$ into both equations of the pair makes them true statements.

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Let's check each option:

(A) $x+y=3, x-y=1$

Substitute $(2, 1)$ into the first equation: $2 + 1 = 3 \implies 3 = 3$. (True)

Substitute $(2, 1)$ into the second equation: $2 - 1 = 1 \implies 1 = 1$. (True)

Since both equations are satisfied, $(2, 1)$ is a solution to this pair.

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(B) $2x+y=5, x-y=1$

Substitute $(2, 1)$ into the first equation: $2(2) + 1 = 5 \implies 4 + 1 = 5 \implies 5 = 5$. (True)

Substitute $(2, 1)$ into the second equation: $2 - 1 = 1 \implies 1 = 1$. (True)

Since both equations are satisfied, $(2, 1)$ is a solution to this pair.

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(C) $x+2y=4, 3x-y=5$

Substitute $(2, 1)$ into the first equation: $2 + 2(1) = 4 \implies 2 + 2 = 4 \implies 4 = 4$. (True)

Substitute $(2, 1)$ into the second equation: $3(2) - 1 = 5 \implies 6 - 1 = 5 \implies 5 = 5$. (True)

Since both equations are satisfied, $(2, 1)$ is a solution to this pair.

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(D) $x-y=1, x+y=3$

Substitute $(2, 1)$ into the first equation: $2 - 1 = 1 \implies 1 = 1$. (True)

Substitute $(2, 1)$ into the second equation: $2 + 1 = 3 \implies 3 = 3$. (True)

Since both equations are satisfied, $(2, 1)$ is a solution to this pair.

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The point $(2, 1)$ is a solution to all four given pairs of equations.

The options that apply are (A), (B), (C), and (D).

The correct option is (B) Substitution method.

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We are asked to identify the most suitable method for solving a system of linear equations when one variable can be easily expressed in terms of the other. This typically happens when one of the equations has a variable with a coefficient of 1 or -1, or when the equation can be easily rearranged to isolate a variable.

Let's consider the characteristics of each method:

(A) Elimination method: This method involves multiplying the equations by suitable numbers so that the coefficients of one variable become equal in magnitude and opposite in sign (or just equal in magnitude). Then, adding or subtracting the equations eliminates that variable. While effective, it might require multiplication steps even if a variable is easily isolated.

(B) Substitution method: This method involves solving one equation for one variable (i.e., expressing one variable in terms of the other) and then substituting that expression into the other equation. If one variable can be easily expressed in terms of the other, the first step of the substitution method is straightforward and simple, making the overall process efficient.

(C) Cross-multiplication method: This method uses a specific formula derived from the general form of linear equations. It is a direct method but requires arranging equations in a specific standard form ($ax + by + c = 0$) and applying formulas involving coefficients. It does not particularly benefit from one variable being easily expressible in terms of the other.

(D) Graphical method: This method involves plotting the lines represented by the equations and finding their intersection point. While visual, it can be time-consuming and might not yield an exact solution if the intersection coordinates are not integers. It does not take advantage of one variable being easily expressible in terms of the other.

Based on the nature of the methods, the Substitution method is explicitly designed to utilize the case where one variable can be easily isolated. This makes it the most suitable choice in such scenarios as the initial step of the method becomes very simple.

The correct option is (B) 10 (for infinitely many solutions).

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We are given the system of linear equations:

$2x + 3y = 5 \quad$ ... (1)

$4x + 6y = k \quad$ ... (2)

We rewrite these equations in the standard form $a_1 x + b_1 y + c_1 = 0$ and $a_2 x + b_2 y + c_2 = 0$:

$2x + 3y - 5 = 0$

$4x + 6y - k = 0$

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Comparing the coefficients, we have:

$a_1 = 2, b_1 = 3, c_1 = -5$

$a_2 = 4, b_2 = 6, c_2 = -k$

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Let's find the ratios of the corresponding coefficients:

$\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}$

$\frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2}$

$\frac{c_1}{c_2} = \frac{-5}{-k} = \frac{5}{k}$

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A system of linear equations is called consistent if it has at least one solution. This occurs in two cases:

1. Unique solution: The lines are intersecting. This happens when $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$.

2. Infinitely many solutions: The lines are coincident. This happens when $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.

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Inconsistent system (no solution): The lines are parallel and distinct. This happens when $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$.

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For the given system, we already have $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{1}{2}$. This means the lines are either parallel or coincident. A unique solution is not possible.

For the system to be consistent, it must have at least one solution, which, given $\frac{a_1}{a_2} = \frac{b_1}{b_2}$, can only happen if the lines are coincident (infinitely many solutions).

The condition for coincident lines is $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.

Using the calculated ratios, we must have:

$\frac{1}{2} = \frac{1}{2} = \frac{5}{k}$

To find the value of $k$, we equate the ratios $\frac{1}{2}$ and $\frac{5}{k}$:

$\frac{1}{2} = \frac{5}{k}$

Cross-multiplying gives:

$1 \times k = 2 \times 5$

$k = 10$

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If $k=10$, the ratios are $\frac{1}{2} = \frac{1}{2} = \frac{5}{10} = \frac{1}{2}$. This satisfies the condition for coincident lines, and thus the system has infinitely many solutions and is consistent.

If $k \neq 10$, then $\frac{1}{2} = \frac{1}{2} \neq \frac{5}{k}$, which satisfies the condition for parallel and distinct lines. In this case, the system has no solution and is inconsistent.

Therefore, the system is consistent only when $k = 10$, which corresponds to the case of infinitely many solutions.

The correct option is (C) $x+2y=3, 2x+4y=7$.

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A pair of linear equations $a_1 x + b_1 y + c_1 = 0$ and $a_2 x + b_2 y + c_2 = 0$ is called inconsistent if it has no solution. Graphically, this means the lines represented by the equations are parallel and distinct.

The condition for a system to be inconsistent is:

$\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$

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Let's examine each option by comparing the ratios of the coefficients:

(A) $x+y=1$ ($a_1=1, b_1=1, c_1=-1$) and $2x+2y=2$ ($a_2=2, b_2=2, c_2=-2$)

Ratios: $\frac{1}{2}, \frac{1}{2}, \frac{-1}{-2} = \frac{1}{2}$.

Condition: $\frac{1}{2} = \frac{1}{2} = \frac{1}{2}$. This system is consistent (infinitely many solutions - coincident lines).

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(B) $x-y=1$ ($a_1=1, b_1=-1, c_1=-1$) and $x+y=2$ ($a_2=1, b_2=1, c_2=-2$)

Ratios: $\frac{1}{1}=1, \frac{-1}{1}=-1, \frac{-1}{-2}=\frac{1}{2}$.

Condition: $\frac{1}{1} \neq \frac{-1}{1}$. This system is consistent (unique solution - intersecting lines).

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(C) $x+2y=3$ ($a_1=1, b_1=2, c_1=-3$) and $2x+4y=7$ ($a_2=2, b_2=4, c_2=-7$)

Ratios: $\frac{1}{2}, \frac{2}{4}=\frac{1}{2}, \frac{-3}{-7}=\frac{3}{7}$.

Condition: $\frac{1}{2} = \frac{1}{2} \neq \frac{3}{7}$. This system is inconsistent (no solution - parallel lines).

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(D) $2x-y=1$ ($a_1=2, b_1=-1, c_1=-1$) and $3x+y=2$ ($a_2=3, b_2=1, c_2=-2$)

Ratios: $\frac{2}{3}, \frac{-1}{1}=-1, \frac{-1}{-2}=\frac{1}{2}$.

Condition: $\frac{2}{3} \neq -1$. This system is consistent (unique solution - intersecting lines).

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Only option (C) satisfies the condition $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, which means it has no solution and is inconsistent.

The correct option is (A) $\frac{30}{x-y} + \frac{44}{x+y} = 10$, $\frac{40}{x-y} + \frac{55}{x+y} = 13$.

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Let the speed of the boat in still water be $x$ km/hr.

Let the speed of the stream be $y$ km/hr.

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When the boat goes upstream, its speed is reduced by the speed of the stream.

Speed upstream = Speed of boat in still water - Speed of stream = $(x - y)$ km/hr.

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When the boat goes downstream, its speed is increased by the speed of the stream.

Speed downstream = Speed of boat in still water + Speed of stream = $(x + y)$ km/hr.

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We know that Time = $\frac{\text{Distance}}{\text{Speed}}$.

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Scenario 1: Boat goes 30 km upstream and 44 km downstream in a total of 10 hours.

Time taken to go 30 km upstream = $\frac{30}{x-y}$ hours.

Time taken to go 44 km downstream = $\frac{44}{x+y}$ hours.

Total time = 10 hours.

So, the first equation is:

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Scenario 2: Boat goes 40 km upstream and 55 km downstream in a total of 13 hours.

Time taken to go 40 km upstream = $\frac{40}{x-y}$ hours.

Time taken to go 55 km downstream = $\frac{55}{x+y}$ hours.

Total time = 13 hours.

So, the second equation is:

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The pair of equations representing the given information in terms of $\frac{1}{x-y}$ and $\frac{1}{x+y}$ is:

$\frac{30}{x-y} + \frac{44}{x+y} = 10$

$\frac{40}{x-y} + \frac{55}{x+y} = 13$

This matches the pair of equations given in option (A).

The correct option is (A) Boat: 8 km/hr, Stream: 3 km/hr.

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From Question 29, the pair of equations representing the situation is:

Let the speed of the boat in still water be $x$ km/hr and the speed of the stream be $y$ km/hr.

Speed upstream = $(x-y)$ km/hr

Speed downstream = $(x+y)$ km/hr

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These equations are not linear. We can reduce them to a linear form by making the following substitution:

Let $u = \frac{1}{x-y}$ and $v = \frac{1}{x+y}$.

Substituting these into equations (1) and (2), we get the following linear system in $u$ and $v$:

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We can solve this system using the elimination method. Multiply equation (3) by 4 and equation (4) by 3 to make the coefficients of $u$ equal:

$4 \times (30u + 44v) = 4 \times 10$

$3 \times (40u + 55v) = 3 \times 13$

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Subtract equation (6) from equation (5):

$(120u + 176v) - (120u + 165v) = 40 - 39$

$120u + 176v - 120u - 165v = 1$

$11v = 1$

$v = \frac{1}{11}$

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Substitute the value of $v = \frac{1}{11}$ into equation (3):

$30u + 44v = 10$

$30u + 44\left(\frac{1}{11}\right) = 10$

$30u + 4 = 10$

$30u = 10 - 4$

$30u = 6$

$u = \frac{6}{30}$

$u = \frac{1}{5}$

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Now, we use the original substitutions to find $x$ and $y$:

$u = \frac{1}{x-y}$

$\frac{1}{5} = \frac{1}{x-y}$

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$v = \frac{1}{x+y}$

$\frac{1}{11} = \frac{1}{x+y}$

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We now have a simple linear system in $x$ and $y$:

$x - y = 5$

$x + y = 11$

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Add equation (7) and equation (8):

$(x - y) + (x + y) = 5 + 11$

$2x = 16$

$x = \frac{16}{2}$

$x = 8$

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Substitute the value of $x=8$ into equation (8):

$x + y = 11$

$8 + y = 11$

$y = 11 - 8$

$y = 3$

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So, the speed of the boat in still water is 8 km/hr, and the speed of the stream is 3 km/hr.

The correct option is (C) $a_1 x + b_1 y + c_1 = 0, a_2 x + b_2 y + c_2 = 0$.

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A linear equation in two variables, say $x$ and $y$, is an equation that can be written in the form $Ax + By + C = 0$, where $A$, $B$, and $C$ are real numbers, and $A$ and $B$ are not both zero.

A pair of linear equations in two variables consists of two such equations.

To distinguish the coefficients and constants of the two equations, subscripts are typically used.

Let the first equation be $a_1 x + b_1 y + c_1 = 0$.

Let the second equation be $a_2 x + b_2 y + c_2 = 0$.

Here, $a_1, b_1, c_1, a_2, b_2, c_2$ are real numbers.

For the first equation to be a linear equation in $x$ and $y$, $a_1$ and $b_1$ cannot both be zero simultaneously ($a_1^2 + b_1^2 \neq 0$).

For the second equation to be a linear equation in $x$ and $y$, $a_2$ and $b_2$ cannot both be zero simultaneously ($a_2^2 + b_2^2 \neq 0$).

Let's analyze the other options:

(A) $ax+b=0$ and $cy+d=0$ are linear equations, but each involves only one variable. This is not a pair of linear equations in two variables ($x$ and $y$).

(B) $ax^2+by+c=0$ and $dx^2+ey+f=0$ involve terms with $x^2$, making them quadratic equations, not linear equations.

(D) $ax+by=c$ and $dy+ex=f$ are indeed linear equations in two variables. However, the form shown in option (C) is the universally accepted and most general representation using subscripts to denote a pair of such equations, written in the standard $Ax + By + C = 0$ format.

Therefore, the general form of a pair of linear equations in two variables $x$ and $y$ is $a_1 x + b_1 y + c_1 = 0$ and $a_2 x + b_2 y + c_2 = 0$.

The correct option is (C) Coincident.

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In the classification of pairs of linear equations:

A system is consistent if it has at least one solution (unique solution or infinitely many solutions).

A system is inconsistent if it has no solution.

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Among consistent systems:

A consistent system is independent if it has exactly one unique solution.

A consistent system is dependent if it has infinitely many solutions.

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The question asks about a system that is both consistent and dependent. This means the system has infinitely many solutions.

Graphically:

A system with a unique solution is represented by intersecting lines.

A system with no solution is represented by parallel and distinct lines.

A system with infinitely many solutions is represented by coincident lines (the lines are the same).

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Since a consistent and dependent system has infinitely many solutions, the lines representing the equations are coincident.

The correct option is (C) Infinitely many.

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A pair of linear equations is classified based on the number of solutions it has.

A system is called consistent if it has at least one solution (either a unique solution or infinitely many solutions).

A system is called dependent if one equation can be obtained from the other by multiplying by a non-zero constant. Dependent linear equations represent the same line.

A system that is both consistent and dependent corresponds to the case where the two lines representing the equations are coincident.

When two lines are coincident, they lie exactly on top of each other and share all their points in common.

Each point on the line represents a solution to both equations.

Since a line consists of an infinite number of points, a consistent and dependent pair of linear equations has infinitely many solutions.

The correct option is (C) Coincident.

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We are given the system of linear equations:

$3x - y = 5 \quad$ ... (1)

$6x - 2y = 10 \quad$ ... (2)

We rewrite these equations in the standard form $a_1 x + b_1 y + c_1 = 0$ and $a_2 x + b_2 y + c_2 = 0$:

$3x - y - 5 = 0 \quad$ ($a_1 = 3, b_1 = -1, c_1 = -5$)

$6x - 2y - 10 = 0 \quad$ ($a_2 = 6, b_2 = -2, c_2 = -10$)

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Calculate the ratios of the corresponding coefficients:

$\frac{a_1}{a_2} = \frac{3}{6} = \frac{1}{2}$

$\frac{b_1}{b_2} = \frac{-1}{-2} = \frac{1}{2}$

$\frac{c_1}{c_2} = \frac{-5}{-10} = \frac{1}{2}$

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We observe that all three ratios are equal:

$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$

Specifically, $\frac{1}{2} = \frac{1}{2} = \frac{1}{2}$.

This condition signifies that the two equations represent the same line. One equation is a non-zero multiple of the other (equation (2) is $2$ times equation (1)).

Lines that are identical are called coincident lines.

Therefore, the lines for the given system of equations are coincident.

The correct option is (C) Factoring.

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There are several standard algebraic methods for solving a pair of linear equations in two variables:

1. Substitution method: Solve one equation for one variable and substitute that expression into the other equation.

2. Elimination method: Multiply the equations by suitable constants so that the coefficients of one variable are equal or opposite, and then add or subtract the equations to eliminate that variable.

3. Cross-multiplication method: A formula-based method derived from the elimination method, applicable when the equations are in the standard form $a_1 x + b_1 y + c_1 = 0$ and $a_2 x + b_2 y + c_2 = 0$.

The Factoring method is primarily used for solving polynomial equations (like quadratic equations) by finding the roots of the polynomial. It is not a method typically used to solve a system of linear equations directly.

Therefore, Factoring is the method that is NOT typically used to solve a pair of linear equations algebraically.

The correct option is (B) $x^2 + y^2 = 10$.

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We are asked to identify which of the given equations cannot be reduced to a linear equation in two variables using the specific substitution $\frac{1}{x}=p$ and $\frac{1}{y}=q$. This means we substitute $x = \frac{1}{p}$ and $y = \frac{1}{q}$ into each equation and check if the resulting equation in $p$ and $q$ is linear.

A linear equation in $p$ and $q$ is of the form $Ap + Bq + C = 0$, where $A, B, C$ are constants.

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Let's examine each option:

(A) $\frac{1}{x+y} + \frac{1}{x-y} = 5$

Substitute $x = \frac{1}{p}$ and $y = \frac{1}{q}$:

$\frac{1}{\frac{1}{p}+\frac{1}{q}} + \frac{1}{\frac{1}{p}-\frac{1}{q}} = 5$

$\frac{1}{\frac{q+p}{pq}} + \frac{1}{\frac{q-p}{pq}} = 5$

$\frac{pq}{p+q} + \frac{pq}{q-p} = 5$

This equation is not linear in $p$ and $q$.

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(B) $x^2 + y^2 = 10$

Substitute $x = \frac{1}{p}$ and $y = \frac{1}{q}$:

$\left(\frac{1}{p}\right)^2 + \left(\frac{1}{q}\right)^2 = 10$

$\frac{1}{p^2} + \frac{1}{q^2} = 10$

This equation is not linear in $p$ and $q$ due to the $p^2$ and $q^2$ terms in the denominators.

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(C) $\frac{x+y}{xy} = 2$

We can rewrite the left side:

$\frac{x}{xy} + \frac{y}{xy} = 2$

$\frac{1}{y} + \frac{1}{x} = 2$

Substitute $\frac{1}{x} = p$ and $\frac{1}{y} = q$:

$q + p = 2$

$p + q = 2$

This equation is linear in $p$ and $q$. Thus, this equation *can* be reduced to linear form by the given substitution.

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(D) $\frac{x}{a} + \frac{y}{b} = 1$

This equation is already linear in $x$ and $y$. However, let's apply the substitution $x = \frac{1}{p}$ and $y = \frac{1}{q}$ as requested by the structure of the question:

$\frac{\frac{1}{p}}{a} + \frac{\frac{1}{q}}{b} = 1$

$\frac{1}{ap} + \frac{1}{bq} = 1$

Multiply by $abpq$ to clear the denominators (assuming $a, b, p, q \neq 0$):

$bq + ap = abpq$

$ap + bq - abpq = 0$

This equation is not linear in $p$ and $q$ due to the $pq$ term.

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The question asks which equation cannot be reduced to linear form in $p$ and $q$ by the substitution $\frac{1}{x}=p, \frac{1}{y}=q$. From our analysis, options (A), (B), and (D) all result in non-linear equations in $p$ and $q$ after this substitution, whereas option (C) results in a linear equation.

Among the options that *cannot* be reduced by this specific substitution, option (B) $x^2 + y^2 = 10$ represents a form (sum of squares) that is fundamentally different from the reciprocal-of-linear-terms forms found in equations that are typically reduced using reciprocal substitutions like the one mentioned. While (A) and (D) also fail to be linearized by *this specific* substitution, (B) is the most distinct type of non-linear equation in this set when considering reducibility via simple reciprocal substitutions.

Therefore, $x^2 + y^2 = 10$ is the equation that cannot be reduced to linear form by the simple substitution $\frac{1}{x}=p, \frac{1}{y}=q$.

The correct option is (B) No solution.

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For a system of linear equations given by $a_1 x + b_1 y + c_1 = 0$ and $a_2 x + b_2 y + c_2 = 0$, the number of solutions is determined by the ratios of their coefficients.

There are three possible cases:

1. If $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, the lines intersect at exactly one point. The system has a unique solution.

2. If $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, the lines are coincident (they are the same line). The system has infinitely many solutions.

3. If $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the lines are parallel and distinct. They do not intersect at any point. The system has no solution.

The condition given in the question is $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$. This condition matches the third case, which corresponds to a system having no solution.

Geometrically, this condition means the lines representing the equations have the same slope (because $\frac{a_1}{a_2} = \frac{b_1}{b_2}$) but different y-intercepts (because the ratios involving $c_1$ and $c_2$ are different), hence they are parallel and distinct lines.

The correct option is (A) $x+y=9, 10x+y+27 = 10y+x$.

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Let the digit in the tens place be $x$.

Let the digit in the units place be $y$.

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The two-digit number can be expressed as $10 \times (\text{tens digit}) + 1 \times (\text{units digit}) = 10x + y$.

The number formed by reversing the digits has $y$ in the tens place and $x$ in the units place. This number is $10y + x$.

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Condition 1: The sum of the digits is 9.

This translates to the equation:

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Condition 2: If 27 is added to the number, the digits are reversed.

The original number is $10x + y$.

Adding 27 to the number: $10x + y + 27$.

This new value is equal to the number with reversed digits ($10y + x$).

This translates to the equation:

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The pair of linear equations representing the given information is $x+y=9$ and $10x+y+27 = 10y+x$.

This pair of equations matches the equations given in option (A).

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We can simplify the second equation (though not required by the question):

$10x + y + 27 = 10y + x$

$10x - x + y - 10y = -27$

$9x - 9y = -27$

Dividing by 9:

$x - y = -3$

So, the system is equivalent to $x+y=9$ and $x-y=-3$. Solving this system gives $x=3$ and $y=6$. The number is 36. $36+27=63$, which is the number with digits reversed.

The correct option is (B) It provides a visual understanding of the relationship between the equations and their solutions.

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Let's analyze the options regarding the usefulness of the graphical representation of a pair of linear equations:

(A) It always gives an exact solution. This is not necessarily true. If the solution coordinates are not integers or simple fractions, it can be difficult to determine the exact values from a graph, especially if drawing is not perfectly accurate.

(B) It provides a visual understanding of the relationship between the equations and their solutions. This is a key benefit. The graph visually shows whether the lines intersect (unique solution), are parallel (no solution), or are coincident (infinitely many solutions). The intersection point(s) visually represent the solution(s).

(C) It is the fastest method to find the solution. This is generally not true, especially for complex systems or solutions that are not easy to read from the graph. Algebraic methods (substitution, elimination) are often faster and always provide exact solutions.

(D) It only works for inconsistent systems. This is false. The graphical method works for all types of systems (consistent with unique solution, consistent with infinitely many solutions, and inconsistent). It visually shows the nature of the system.

Therefore, the primary usefulness of the graphical method is the visual insight it provides into the system and its solutions.

The correct option is (A) $(x+10)(y-2) = xy$, $(x-10)(y+3) = xy$.

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Let the original speed of the train be $x$ km/hr.

Let the original scheduled time be $y$ hours.

The distance covered is given by Distance = Speed $\times$ Time.

Original distance = $xy$ km.

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Condition 1: If the train was 10 km/hr faster, it would have taken 2 hours less.

New speed = $(x + 10)$ km/hr.

New time = $(y - 2)$ hours.

The distance covered is the same as the original distance.

$(x + 10)(y - 2) = xy$

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Condition 2: If the train was 10 km/hr slower, it would have taken 3 hours more.

New speed = $(x - 10)$ km/hr.

New time = $(y + 3)$ hours.

The distance covered is the same as the original distance.

$(x - 10)(y + 3) = xy$

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The pair of equations representing the given information is $(x+10)(y-2) = xy$ and $(x-10)(y+3) = xy$.

This matches the pair of equations given in option (A).

A linear equation in two variables is an equation that can be written in the standard form $ax + by + c = 0$, where $a$, $b$, and $c$ are real numbers, and $a$ and $b$ are not both equal to zero.

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A pair of linear equations in two variables consists of two such linear equations involving the same two variables. When we consider two linear equations together, say in the variables $x$ and $y$, they form a pair of linear equations in two variables.

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The general form of a pair of linear equations in two variables, say $x$ and $y$, is given by:

$a_1x + b_1y + c_1 = 0$

$a_2x + b_2y + c_2 = 0$

where $a_1, b_1, c_1, a_2, b_2, c_2$ are real numbers. The conditions $a_1^2 + b_1^2 \neq 0$ and $a_2^2 + b_2^2 \neq 0$ are necessary to ensure that both equations are indeed linear equations in two variables.

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Example:

Consider the following two equations involving the variables $x$ and $y$:

$2x + 3y = 5$

$x - y = 1$

These two equations form a pair of linear equations in two variables ($x$ and $y$). Each equation is linear because the highest power of each variable is 1 and there are no products of variables. They involve the same two variables.

When written in the general form $ax + by + c = 0$, the pair is:

$2x + 3y - 5 = 0$

$x - y - 1 = 0$

Here, for the first equation, $a_1=2, b_1=3, c_1=-5$. For the second equation, $a_2=1, b_2=-1, c_2=-1$.

A linear equation in two variables, such as $ax + by + c = 0$ (where $a, b, c$ are real numbers and $a$ and $b$ are not both zero), represents a straight line in a two-dimensional coordinate plane (the Cartesian plane).

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A solution to a linear equation in two variables is a pair of values $(x, y)$ that satisfies the equation, i.e., when these values are substituted into the equation, the left side equals the right side.

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Graphically, the solution of a linear equation in two variables is represented as a point $(x, y)$ that lies on the line represented by the equation.

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Therefore, the graph of a linear equation in two variables is the set of all points $(x, y)$ in the plane that are solutions to the equation. Any point on the line is a solution, and any solution is a point on the line.

A pair of linear equations in two variables can be represented graphically by two straight lines in a coordinate plane. For any two lines in a plane, there are exactly three possibilities regarding their intersection:

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1. Intersecting Lines: The two lines intersect at exactly one point.

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2. Parallel Lines: The two lines are parallel and do not intersect at all.

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3. Coincident Lines: The two lines are the same line (one lies exactly on top of the other).

If the lines representing a pair of linear equations are intersecting, it means they meet at exactly one point in the coordinate plane.

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The point of intersection represents the common solution that satisfies both equations simultaneously.

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Therefore, a pair of linear equations whose graphical representation is a pair of intersecting lines has exactly one solution (also called a unique solution).

If the lines representing a pair of linear equations are parallel, it means they never intersect in the coordinate plane.

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A solution to a pair of linear equations corresponds to a point that lies on both lines simultaneously (the point of intersection).

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Since parallel lines have no point in common, there is no pair of values $(x, y)$ that can satisfy both equations at the same time.

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Therefore, a pair of linear equations whose graphical representation is a pair of parallel lines has no solution. This is an example of an inconsistent system of equations.

If the lines representing a pair of linear equations are coincident, it means that the two equations represent the exact same line in the coordinate plane. One line lies directly on top of the other.

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A solution to a pair of linear equations is a point $(x, y)$ that lies on both lines simultaneously.

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Since coincident lines share all their points, every point that lies on one line also lies on the other line.

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A straight line is composed of an infinite number of points.

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Therefore, a pair of linear equations whose graphical representation is a pair of coincident lines has infinitely many solutions. This is an example of a consistent and dependent system of equations.

A pair of linear equations is said to be consistent if it has at least one solution.

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Graphically, a consistent pair of equations is represented by lines that are either intersecting (having exactly one solution) or coincident (having infinitely many solutions).

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A pair of linear equations is said to be inconsistent if it has no solution.

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Graphically, an inconsistent pair of equations is represented by parallel lines (which never intersect).

A pair of linear equations:

$a_1x + b_1y + c_1 = 0$

$a_2x + b_2y + c_2 = 0$

where $a_1, b_1, c_1, a_2, b_2, c_2$ are real numbers and $a_1^2 + b_1^2 \neq 0$, $a_2^2 + b_2^2 \neq 0$, is consistent with a unique solution if and only if the lines representing these equations intersect at exactly one point.

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Graphically, intersecting lines have different slopes. The slope of a line $ax + by + c = 0$ is $-a/b$ (if $b \neq 0$).

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The condition for the pair of linear equations to have a unique solution is that the ratio of the coefficients of $x$ is not equal to the ratio of the coefficients of $y$.

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The condition on the coefficients is:

$\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$

A pair of linear equations:

$a_1x + b_1y + c_1 = 0$

$a_2x + b_2y + c_2 = 0$

where $a_1, b_1, c_1, a_2, b_2, c_2$ are real numbers and $a_1^2 + b_1^2 \neq 0$, $a_2^2 + b_2^2 \neq 0$, is consistent with infinitely many solutions if and only if the lines representing these equations are coincident.

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Graphically, coincident lines represent the same equation (or one equation is a non-zero multiple of the other).

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The condition for the pair of linear equations to have infinitely many solutions is that the ratio of the coefficients of $x$, the ratio of the coefficients of $y$, and the ratio of the constant terms are all equal.

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The condition on the coefficients is:

$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$

A pair of linear equations:

$a_1x + b_1y + c_1 = 0$

$a_2x + b_2y + c_2 = 0$

where $a_1, b_1, c_1, a_2, b_2, c_2$ are real numbers and $a_1^2 + b_1^2 \neq 0$, $a_2^2 + b_2^2 \neq 0$, is inconsistent if and only if the lines representing these equations are parallel and distinct.

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Graphically, parallel lines have the same slope but different y-intercepts. This means the ratio of the coefficients of $x$ is equal to the ratio of the coefficients of $y$, but this ratio is not equal to the ratio of the constant terms.

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The condition on the coefficients for the pair of linear equations to be inconsistent (having no solution) is:

$\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$

The given pair of linear equations is:

$x + 2y - 4 = 0 \quad ...(1)$

$2x + 4y - 12 = 0 \quad ...(2)$

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Comparing these equations with the standard form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$, we identify the coefficients:

From equation (1): $a_1 = 1$, $b_1 = 2$, $c_1 = -4$.

From equation (2): $a_2 = 2$, $b_2 = 4$, $c_2 = -12$.

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Now, we compare the ratios of the coefficients:

$\frac{a_1}{a_2} = \frac{1}{2}$

$\frac{b_1}{b_2} = \frac{2}{4} = \frac{1}{2}$

$\frac{c_1}{c_2} = \frac{-4}{-12} = \frac{4}{12} = \frac{1}{3}$

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We observe that:

$\frac{a_1}{a_2} = \frac{1}{2}$

$\frac{b_1}{b_2} = \frac{1}{2}$

$\frac{c_1}{c_2} = \frac{1}{3}$

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Thus, we have the condition:

$\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$

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This condition corresponds to the case where the lines representing the pair of linear equations are parallel and distinct.

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A pair of linear equations with parallel and distinct lines has no solution.

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A pair of linear equations that has no solution is called an inconsistent pair.

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Therefore, the given pair of linear equations is inconsistent.

The given pair of linear equations is:

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We can rewrite these equations in the standard form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:

$x - 2y + 0 = 0$

$3x + 4y - 20 = 0$

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Comparing these equations with the standard form, we identify the coefficients:

From equation (1): $a_1 = 1$, $b_1 = -2$, $c_1 = 0$.

From equation (2): $a_2 = 3$, $b_2 = 4$, $c_2 = -20$.

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Now, we calculate the ratios of the corresponding coefficients:

$\frac{a_1}{a_2} = \frac{1}{3}$

$\frac{b_1}{b_2} = \frac{-2}{4} = \frac{-1}{2}$

$\frac{c_1}{c_2} = \frac{0}{-20} = 0$

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We compare the ratios:

Is $\frac{a_1}{a_2} = \frac{b_1}{b_2}$?

$\frac{1}{3} \neq \frac{-1}{2}$

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Since the ratio of the coefficients of $x$ is not equal to the ratio of the coefficients of $y$ ($\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$), the lines representing these equations are intersecting.

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A pair of linear equations whose lines are intersecting has exactly one solution (a unique solution).

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A pair of linear equations that has at least one solution is called consistent.

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Therefore, the given pair of linear equations is consistent with a unique solution.

The three algebraic methods for solving a pair of linear equations in two variables are:

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1. Substitution Method: In this method, we express one variable in terms of the other variable from one equation and substitute this expression into the other equation to get a linear equation in a single variable.

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2. Elimination Method: In this method, we multiply one or both equations by suitable non-zero numbers such that the coefficients of one variable become numerically equal. Then, we add or subtract the equations to eliminate that variable and obtain a linear equation in a single variable.

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3. Cross-multiplication Method: This method provides a formula to directly find the values of the variables $x$ and $y$ from the coefficients of the two equations, provided there is a unique solution.

We are given the following pair of linear equations:

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We will use the substitution method to solve this system.

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From equation (2), we can express $x$ in terms of $y$:

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Now, substitute this expression for $x$ from equation (3) into equation (1):

$(4 + y) + y = 14$

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Simplify and solve for $y$:

$4 + 2y = 14$

$2y = 14 - 4$

$2y = 10$

$y = \frac{10}{2}$

$y = 5$

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Now, substitute the value of $y = 5$ back into equation (3) to find the value of $x$:

$x = 4 + y$

$x = 4 + 5$

$x = 9$

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Thus, the solution to the pair of equations is $x = 9$ and $y = 5$.

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We can verify the solution by substituting these values back into the original equations:

For equation (1): $x + y = 9 + 5 = 14$ (Satisfied)

For equation (2): $x - y = 9 - 5 = 4$ (Satisfied)

We are given the following pair of linear equations:

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We will use the elimination method to solve this system.

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Our goal is to make the coefficients of one variable the same in both equations so that we can eliminate it by adding or subtracting the equations.

Let's try to eliminate $x$. The coefficient of $x$ in equation (1) is 2, and in equation (2) is 4. We can make the coefficient of $x$ in equation (1) equal to 4 by multiplying equation (1) by 2.

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Multiply equation (1) by 2:

$2 \times (2x + 3y) = 2 \times 8$

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Now we have the pair of equations:

$4x + 6y = 16$ (Equation 3)

$4x + 6y = 7$ (Equation 2)

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Subtract equation (2) from equation (3):

$(4x + 6y) - (4x + 6y) = 16 - 7$

$4x + 6y - 4x - 6y = 9$

$0 = 9$

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We arrive at a statement $0 = 9$, which is false. This means there are no values of $x$ and $y$ that can satisfy both equations simultaneously.

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Geometrically, this corresponds to a pair of parallel lines that do not intersect.

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Therefore, the given pair of linear equations has no solution.

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A pair of linear equations with no solution is called an inconsistent pair.

The given pair of linear equations is:

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To find the condition for infinitely many solutions, we first write the equations in the standard form $ax + by + c = 0$:

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Comparing these equations with $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$, we have:

$a_1 = k, b_1 = 3, c_1 = -(k - 3) = 3 - k$

$a_2 = 12, b_2 = k, c_2 = -k$

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For a pair of linear equations to have infinitely many solutions, the lines representing the equations must be coincident. The condition for coincident lines is:

$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$

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Substituting the coefficients, we get:

$\frac{k}{12} = \frac{3}{k} = \frac{3 - k}{-k}$

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This gives us two equalities to solve for $k$:

1. $\frac{k}{12} = \frac{3}{k}$

2. $\frac{3}{k} = \frac{3 - k}{-k}$

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From the first equality, $\frac{k}{12} = \frac{3}{k}$:

$k \times k = 12 \times 3$

$k^2 = 36$

$k = \pm \sqrt{36}$

$k = 6$ or $k = -6$

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From the second equality, $\frac{3}{k} = \frac{3 - k}{-k}$:

This equality requires $k \neq 0$. Assuming $k \neq 0$, we can cross-multiply:

$3 \times (-k) = k \times (3 - k)$

$-3k = 3k - k^2$

Move all terms to one side:

$k^2 - 3k - 3k = 0$

$k^2 - 6k = 0$

Factor out $k$:

$k(k - 6) = 0$

This gives $k = 0$ or $k = 6$.

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For the system to have infinitely many solutions, the value of $k$ must satisfy both sets of possibilities derived from the equalities.

From $\frac{k}{12} = \frac{3}{k}$, $k \in \{6, -6\}$.

From $\frac{3}{k} = \frac{3 - k}{-k}$, $k \in \{0, 6\}$. (Note that $k=0$ leads to division by zero in the ratios, which needs careful consideration. If $k=0$, the second equation is $12x=0 \implies x=0$. The first equation is $3y=-3 \implies y=-1$. This is a unique solution, not infinitely many. So $k=0$ is not valid for infinitely many solutions).

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The value of $k$ that is common to both conditions and is valid is $k = 6$.

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Let's verify for $k=6$:

$\frac{a_1}{a_2} = \frac{6}{12} = \frac{1}{2}$

$\frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2}$

$\frac{c_1}{c_2} = \frac{3 - 6}{-6} = \frac{-3}{-6} = \frac{1}{2}$

Since $\frac{1}{2} = \frac{1}{2} = \frac{1}{2}$, the condition $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$ is satisfied.

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Therefore, the pair of linear equations has infinitely many solutions when $k = 6$.

Given:

The given pair of linear equations is:

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To Find:

The value of $p$ for which the pair of equations has no solution.

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Solution:

We compare the given equations with the standard form of a pair of linear equations, $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$.

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From equation (1): $a_1 = 4$, $b_1 = p$, $c_1 = 8$.

From equation (2): $a_2 = 2$, $b_2 = 2$, $c_2 = 2$.

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For a pair of linear equations to have no solution, the lines representing the equations must be parallel and distinct. The condition for this is:

$\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$

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Substitute the values of the coefficients into this condition:

$\frac{4}{2} = \frac{p}{2} \neq \frac{8}{2}$

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Simplify the ratios:

$2 = \frac{p}{2} \neq 4$

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We need to satisfy both the equality and the inequality. First, consider the equality:

$2 = \frac{p}{2}$

Multiply both sides by 2:

$p = 2 \times 2$

$p = 4$

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Now, we must check if this value of $p$ satisfies the inequality $\frac{p}{2} \neq 4$.

Substitute $p = 4$ into the inequality:

$\frac{4}{2} \neq 4$

$2 \neq 4$

This is a true statement, so the value $p = 4$ satisfies the condition for no solution.

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Thus, for the pair of linear equations to have no solution, the value of $p$ must be 4.

Given:

The equation is $2x + 3y = 11$.

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To Solve:

Solve for $y$ in terms of $x$.

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Method Used:

The method used here is algebraic manipulation, specifically isolating the variable $y$ on one side of the equation. This process is a fundamental step in solving systems of equations using the substitution method.

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Solution:

Start with the given equation:

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To isolate the term containing $y$ ($3y$), subtract $2x$ from both sides of the equation:

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To isolate $y$, divide both sides of the equation by the coefficient of $y$, which is 3:

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This expresses $y$ in terms of $x$.

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The solution is $y = \frac{11 - 2x}{3}$.

Let the two numbers be denoted by $x$ and $y$.

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According to the first condition given in the problem: "The difference between two numbers is 26".

Assuming $x > y$, the difference is $x - y$.

So, the first equation is:

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According to the second condition given in the problem: "one number is three times the other".

Since we assumed $x > y$, the larger number ($x$) is three times the smaller number ($y$).

So, the second equation is:

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Thus, the pair of linear equations representing the given problem is:

$x - y = 26$

$x = 3y$

Let the two-digit number be represented by its digits. Let the tens digit be $x$ and the units digit be $y$.

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The value of the original two-digit number is $10x + y$.

The value of the number obtained by reversing the order of the digits is $10y + x$.

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According to the first condition given in the problem: "The sum of the digits of a two-digit number is 9."

This translates to the equation:

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According to the second condition given in the problem: "nine times this number is twice the number obtained by reversing the order of the digits."

This translates to the equation:

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Now, we simplify this equation:

$90x + 9y = 20y + 2x$

Bring all terms to one side:

$90x - 2x + 9y - 20y = 0$

$88x - 11y = 0$

Divide the entire equation by 11:

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Thus, the pair of linear equations representing the given problem is:

$x + y = 9$

$8x - y = 0$

We are given the following pair of linear equations:

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We will use the substitution method to solve this system.

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From equation (1), we can easily express one variable in terms of the other. Let's solve for $x$ in terms of $y$:

Subtract $y$ from both sides of equation (1):

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Now, substitute this expression for $x$ from equation (3) into equation (2):

$2(5 - y) - 3y = 4$

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Simplify and solve the resulting equation for $y$:

$10 - 2y - 3y = 4$

$10 - 5y = 4$

Subtract 10 from both sides:

$-5y = 4 - 10$

$-5y = -6$

Divide both sides by -5:

$y = \frac{-6}{-5}$

$y = \frac{6}{5}$

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Now, substitute the value of $y = \frac{6}{5}$ back into equation (3) to find the value of $x$:

$x = 5 - y$

$x = 5 - \frac{6}{5}$

To subtract the fraction, find a common denominator:

$x = \frac{5 \times 5}{1 \times 5} - \frac{6}{5}$

$x = \frac{25}{5} - \frac{6}{5}$

$x = \frac{25 - 6}{5}$

$x = \frac{19}{5}$

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Thus, the solution to the pair of equations is $x = \frac{19}{5}$ and $y = \frac{6}{5}$.

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We can verify the solution by substituting these values back into the original equations:

For equation (1): $x + y = \frac{19}{5} + \frac{6}{5} = \frac{19+6}{5} = \frac{25}{5} = 5$ (Satisfied)

For equation (2): $2x - 3y = 2\left(\frac{19}{5}\right) - 3\left(\frac{6}{5}\right) = \frac{38}{5} - \frac{18}{5} = \frac{38-18}{5} = \frac{20}{5} = 4$ (Satisfied)

We are given the following pair of linear equations:

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We will use the elimination method to solve this system.

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Our goal is to make the coefficients of either $x$ or $y$ numerically equal so that we can eliminate one variable by adding or subtracting the equations.

Let's choose to eliminate $y$. The coefficients of $y$ are +4 in equation (1) and -2 in equation (2). To make the numerical coefficients equal, we can multiply equation (2) by 2.

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Multiply equation (2) by 2:

$2 \times (2x - 2y) = 2 \times 2$

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Now we have the pair of equations:

$3x + 4y = 10$ (Equation 1)

$4x - 4y = 4$ (Equation 3)

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Since the coefficients of $y$ are numerically equal and have opposite signs (+4y and -4y), we can eliminate $y$ by adding equation (1) and equation (3):

$(3x + 4y) + (4x - 4y) = 10 + 4$

$3x + 4y + 4x - 4y = 14$

$(3x + 4x) + (4y - 4y) = 14$

$7x + 0y = 14$

$7x = 14$

---

Solve for $x$:

$x = \frac{14}{7}$

$x = 2$

---

Now, substitute the value of $x = 2$ into either equation (1) or (2) to find the value of $y$. Let's use equation (2):

$2x - 2y = 2$

Substitute $x=2$:

$2(2) - 2y = 2$

$4 - 2y = 2$

---

Solve for $y$:

$-2y = 2 - 4$

$-2y = -2$

$y = \frac{-2}{-2}$

$y = 1$

---

Thus, the solution to the pair of equations is $x = 2$ and $y = 1$.

---

We can verify the solution by substituting these values back into the original equations:

For equation (1): $3x + 4y = 3(2) + 4(1) = 6 + 4 = 10$ (Satisfied)

For equation (2): $2x - 2y = 2(2) - 2(1) = 4 - 2 = 2$ (Satisfied)

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Alternate Method: Eliminate x first

We are given:

$3x + 4y = 10 \quad ...(1)$

$2x - 2y = 2 \quad ...(2)$

---

To eliminate $x$, we find the least common multiple (LCM) of the coefficients of $x$ (3 and 2), which is 6.

Multiply equation (1) by 2:

$2 \times (3x + 4y) = 2 \times 10 \implies 6x + 8y = 20 \quad ...(4)$

Multiply equation (2) by 3:

$3 \times (2x - 2y) = 3 \times 2 \implies 6x - 6y = 6 \quad ...(5)$

---

Now, subtract equation (5) from equation (4):

$(6x + 8y) - (6x - 6y) = 20 - 6$

$6x + 8y - 6x + 6y = 14$

$14y = 14$

$y = \frac{14}{14} = 1$

---

Substitute $y = 1$ into equation (2):

$2x - 2(1) = 2$

$2x - 2 = 2$

$2x = 2 + 2$

$2x = 4$

$x = \frac{4}{2} = 2$

---

We get the same solution: $x = 2$ and $y = 1$.

Given:

The given pair of linear equations is:

---

To Find:

The value of $m$ for which the pair of equations has a unique solution.

---

Solution:

We compare the given equations with the standard form of a pair of linear equations, $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$.

First, rewrite the equations in the standard form:

$mx + y - 5 = 0$

$3x + y - 8 = 0$

---

From equation (1): $a_1 = m$, $b_1 = 1$, $c_1 = -5$.

From equation (2): $a_2 = 3$, $b_2 = 1$, $c_2 = -8$.

---

For a pair of linear equations to have a unique solution, the lines representing the equations must intersect at exactly one point. The condition for intersecting lines is:

$\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$

---

Substitute the values of the coefficients into this condition:

$\frac{m}{3} \neq \frac{1}{1}$

---

Simplify the inequality:

$\frac{m}{3} \neq 1$

---

Multiply both sides by 3:

$m \neq 3$

---

Therefore, the pair of linear equations will have a unique solution for all real values of $m$ except 3.

---

The value of $m$ for which the system has a unique solution is $m \neq 3$.

The graphical method for solving a pair of linear equations in two variables involves representing each equation as a straight line in the Cartesian coordinate plane. Since every point on the graph of a linear equation is a solution to that equation, the point where the graphs of two linear equations intersect represents a solution that satisfies both equations simultaneously. This point of intersection is the solution to the system of linear equations.

---

We are asked to solve the following pair of linear equations graphically:

---

To graph each equation, we need to find at least two points that lie on the line represented by the equation. We can do this by choosing values for one variable and calculating the corresponding value of the other variable.

---

For Equation (1): $x + y = 5$

We can express $y$ in terms of $x$: $y = 5 - x$.

Let's find a few points:

$x$	$y = 5 - x$	Point ($x$, $y$)
0	$5 - 0 = 5$	(0, 5)
5	$5 - 5 = 0$	(5, 0)
3	$5 - 3 = 2$	(3, 2)

---

For Equation (2): $x - y = 1$

We can express $y$ in terms of $x$: $y = x - 1$.

Let's find a few points:

$x$	$y = x - 1$	Point ($x$, $y$)
0	$0 - 1 = -1$	(0, -1)
1	$1 - 1 = 0$	(1, 0)
3	$3 - 1 = 2$	(3, 2)

---

Now, plot the points for each equation on the same graph paper and draw a straight line passing through the points for each equation.

For equation (1), plot the points (0, 5), (5, 0), and (3, 2) and draw the line.

For equation (2), plot the points (0, -1), (1, 0), and (3, 2) and draw the line.

---

Upon drawing the graphs, we will observe that the two lines intersect at a single point. By examining the graph, we can identify the coordinates of this intersection point.

From our tables above, we can see that the point (3, 2) is common to both sets of points. This indicates that the lines will intersect at this point.

---

The point of intersection of the two lines is (3, 2).

---

The solution to the pair of linear equations is the coordinates of the point of intersection.

---

Thus, the solution is $x = 3$ and $y = 2$.

---

Verification:

Substitute $x=3$ and $y=2$ into the original equations:

Equation (1): $x + y = 3 + 2 = 5$ (Satisfied)

Equation (2): $x - y = 3 - 2 = 1$ (Satisfied)

Since the values satisfy both equations, the solution is correct.

---

(Note: The actual graph cannot be displayed in this text-based format, but the description and the identification of the intersection point based on the calculated coordinates fulfill the graphical method steps).

The Substitution Method is an algebraic technique used to solve a pair of linear equations in two variables. The steps involved are:

---

1. Choose one of the two equations and express one variable (say, $x$) in terms of the other variable (say, $y$).

---

2. Substitute this expression for $x$ into the *other* equation. This will result in a linear equation in a single variable ($y$).

---

3. Solve the linear equation obtained in step 2 to find the value of the single variable ($y$).

---

4. Substitute the value of $y$ found in step 3 back into the expression for $x$ (obtained in step 1) to find the value of the other variable ($x$).

---

5. The values of $x$ and $y$ thus obtained constitute the solution to the pair of linear equations.

---

Now, let's solve the given pair of equations using the substitution method:

---

Step 1: Express one variable in terms of the other.

From equation (1), let's express $2x$ in terms of $y$:

---

Step 2: Substitute the expression into the other equation.

Substitute the expression for $2x$ from equation (3) into equation (2):

$(11 - 3y) - 4y = -24$

---

Step 3: Solve the resulting linear equation in one variable.

Simplify and solve for $y$:

$11 - 3y - 4y = -24$

$11 - 7y = -24$

Subtract 11 from both sides:

$-7y = -24 - 11$

$-7y = -35$

Divide both sides by -7:

$y = \frac{-35}{-7}$

$y = 5$

---

Step 4: Substitute the value back to find the other variable.

Substitute the value of $y = 5$ into equation (3) to find $x$:

$2x = 11 - 3y$

$2x = 11 - 3(5)$

$2x = 11 - 15$

$2x = -4$

Divide both sides by 2:

$x = \frac{-4}{2}$

$x = -2$

---

Step 5: State the solution.

The solution to the pair of equations is $x = -2$ and $y = 5$.

---

Verification:

Substitute $x=-2$ and $y=5$ into the original equations:

Equation (1): $2x + 3y = 2(-2) + 3(5) = -4 + 15 = 11$ (Satisfied)

Equation (2): $2x - 4y = 2(-2) - 4(5) = -4 - 20 = -24$ (Satisfied)

Since the values satisfy both equations, the solution is correct.

The Elimination Method is an algebraic technique used to solve a pair of linear equations in two variables. The steps involved are:

---

1. Write both equations in the standard form $ax + by + c = 0$ or $ax + by = c$.

---

2. Choose one variable (either $x$ or $y$) to eliminate.

---

3. Multiply one or both equations by suitable non-zero constants such that the absolute values of the coefficients of the chosen variable become equal.

---

4. If the coefficients of the chosen variable have opposite signs, add the modified equations. If they have the same sign, subtract one equation from the other. This will eliminate the chosen variable, resulting in a linear equation in the other variable.

---

5. Solve the resulting linear equation in one variable.

---

6. Substitute the value found in step 5 into one of the original equations to find the value of the eliminated variable.

---

7. The values of $x$ and $y$ thus obtained constitute the solution to the pair of linear equations.

---

Now, let's solve the given pair of equations using the elimination method:

Equation 1: $3x - 5y - 4 = 0$

Equation 2: $9x = 2y + 7$

---

Step 1: Write equations in standard form ($ax + by = c$).

Rearrange Equation 2:

---

Step 2: Choose a variable to eliminate.

Let's choose to eliminate $x$.

---

Step 3: Make coefficients of the chosen variable equal.

The coefficients of $x$ are 3 in equation (1) and 9 in equation (2). The LCM of 3 and 9 is 9.

Multiply equation (1) by 3 to make the coefficient of $x$ equal to 9:

$3 \times (3x - 5y) = 3 \times 4$

Equation (2) already has the coefficient of $x$ as 9.

---

Step 4: Add or subtract the modified equations to eliminate the variable.

We have equation (3): $9x - 15y = 12$

We have equation (2): $9x - 2y = 7$

Since the coefficients of $x$ have the same sign (+9 and +9), subtract equation (2) from equation (3):

$(9x - 15y) - (9x - 2y) = 12 - 7$

$9x - 15y - 9x + 2y = 5$

$(9x - 9x) + (-15y + 2y) = 5$

$0x - 13y = 5$

$-13y = 5$

---

Step 5: Solve the resulting linear equation.

$-13y = 5$

Divide by -13:

$y = \frac{5}{-13}$

$y = -\frac{5}{13}$

---

Step 6: Substitute the value back to find the other variable.

Substitute the value of $y = -\frac{5}{13}$ into equation (1):

$3x - 5y = 4$

$3x - 5\left(-\frac{5}{13}\right) = 4$

$3x + \frac{25}{13} = 4$

Subtract $\frac{25}{13}$ from both sides:

$3x = 4 - \frac{25}{13}$

Find a common denominator for the right side:

$3x = \frac{4 \times 13}{1 \times 13} - \frac{25}{13}$

$3x = \frac{52}{13} - \frac{25}{13}$

$3x = \frac{52 - 25}{13}$

$3x = \frac{27}{13}$

Divide both sides by 3:

$x = \frac{27}{13} \times \frac{1}{3}$

Cancel out the common factor 3:

$x = \frac{\cancel{27}^{9}}{13} \times \frac{1}{\cancel{3}_{1}}$

$x = \frac{9}{13}$

---

Step 7: State the solution.

The solution to the pair of equations is $x = \frac{9}{13}$ and $y = -\frac{5}{13}$.

---

Verification (Optional):

Substitute $x=\frac{9}{13}$ and $y=-\frac{5}{13}$ into the original equations:

Equation 1: $3x - 5y - 4 = 3\left(\frac{9}{13}\right) - 5\left(-\frac{5}{13}\right) - 4 = \frac{27}{13} + \frac{25}{13} - 4 = \frac{27 + 25}{13} - 4 = \frac{52}{13} - 4 = 4 - 4 = 0$ (Satisfied)

Equation 2: $9x = 2y + 7 \implies 9x - 2y - 7 = 0$

$9\left(\frac{9}{13}\right) - 2\left(-\frac{5}{13}\right) - 7 = \frac{81}{13} + \frac{10}{13} - 7 = \frac{81 + 10}{13} - 7 = \frac{91}{13} - 7 = 7 - 7 = 0$ (Satisfied)

Since the values satisfy both equations, the solution is correct.

Problem Description:

We are given a two-digit number. The sum of its digits is 12. When the digits are interchanged, the new number is 18 more than the original number.

---

To Find:

The original two-digit number.

---

Solution:

Let the tens digit of the original number be $x$ and the units digit be $y$.

The value of the original two-digit number is $10x + y$.

The number obtained by interchanging the digits is $10y + x$.

---

According to the first condition: "The sum of the digits of a two-digit number is 12."

This gives us the equation:

---

According to the second condition: "The number obtained by interchanging the digits is 18 more than the original number."

This gives us the equation:

$10y + x = (10x + y) + 18$

---

Now, simplify the second equation:

$10y + x = 10x + y + 18$

Move all terms involving variables to one side and the constant to the other:

$10y - y + x - 10x = 18$

$9y - 9x = 18$

Divide the entire equation by 9:

$\frac{9y}{9} - \frac{9x}{9} = \frac{18}{9}$

$y - x = 2$

Rearranging the terms:

---

We now have a pair of linear equations:

$x + y = 12$ (1)

$-x + y = 2$ (2)

---

We can solve this pair of equations using the elimination method. Notice that the coefficients of $x$ are +1 and -1, which are opposites. By adding the two equations, we can eliminate $x$.

---

Add equation (1) and equation (2):

$(x + y) + (-x + y) = 12 + 2$

$x + y - x + y = 14$

$(x - x) + (y + y) = 14$

$0 + 2y = 14$

$2y = 14$

---

Solve for $y$:

$y = \frac{14}{2}$

$y = 7$

---

Now, substitute the value of $y = 7$ into equation (1) to find the value of $x$:

$x + y = 12$

$x + 7 = 12$

$x = 12 - 7$

$x = 5$

---

The tens digit is $x = 5$ and the units digit is $y = 7$.

The original two-digit number is $10x + y$.

Original number $= 10(5) + 7 = 50 + 7 = 57$.

---

Answer:

The pair of linear equations is $x + y = 12$ and $y - x = 2$.

The original number is 57.

Problem Description:

We are given information about the ages of two people, A and B, at different points in time. We need to find their current ages.

---

To Find:

The present ages of A and B.

---

Solution:

Let the present age of A be $x$ years and the present age of B be $y$ years.

---

Condition 1: Five years ago

Five years ago, A's age was $(x - 5)$ years.

Five years ago, B's age was $(y - 5)$ years.

According to the problem, "Five years ago, A's age was three times B's age".

So, we can write the equation:

$x - 5 = 3(y - 5)$

Simplify this equation:

$x - 5 = 3y - 15$

Rearrange into standard form:

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Condition 2: Ten years later

Ten years later, A's age will be $(x + 10)$ years.

Ten years later, B's age will be $(y + 10)$ years.

According to the problem, "Ten years later, A's age will be twice B's age".

So, we can write the equation:

$x + 10 = 2(y + 10)$

Simplify this equation:

$x + 10 = 2y + 20$

Rearrange into standard form:

---

We now have a pair of linear equations:

$x - 3y = -10$ (1)

$x - 2y = 10$ (2)

---

We can solve this system using the elimination method. Notice that the coefficient of $x$ is 1 in both equations. We can eliminate $x$ by subtracting equation (2) from equation (1).

---

Subtract equation (2) from equation (1):

$(x - 3y) - (x - 2y) = -10 - 10$

$x - 3y - x + 2y = -20$

$(x - x) + (-3y + 2y) = -20$

$0 - y = -20$

$-y = -20$

---

Solve for $y$:

$y = 20$

---

Now, substitute the value of $y = 20$ into equation (2) to find the value of $x$:

$x - 2y = 10$

$x - 2(20) = 10$

$x - 40 = 10$

---

Solve for $x$:

$x = 10 + 40$

$x = 50$

---

The present age of A is $x = 50$ years.

The present age of B is $y = 20$ years.

---

Answer:

The pair of linear equations is $x - 3y = -10$ and $x - 2y = 10$.

A's present age is 50 years and B's present age is 20 years.

---

Verification:

Five years ago: A's age = $50 - 5 = 45$, B's age = $20 - 5 = 15$. Is $45 = 3 \times 15$? Yes, $45 = 45$.

Ten years later: A's age = $50 + 10 = 60$, B's age = $20 + 10 = 30$. Is $60 = 2 \times 30$? Yes, $60 = 60$.

The solution is correct.

Problem Description:

We are given a fraction and two conditions related to adding constants to its numerator and denominator.

---

To Find:

The original fraction.

---

Solution:

Let the original fraction be $\frac{x}{y}$, where $x$ is the numerator and $y$ is the denominator. We assume $y \neq 0$.

---

Condition 1: "A fraction becomes $\frac{9}{11}$ if 2 is added to both the numerator and the denominator."

Adding 2 to the numerator gives $x+2$.

Adding 2 to the denominator gives $y+2$.

The new fraction is $\frac{x+2}{y+2}$. According to the condition:

$\frac{x+2}{y+2} = \frac{9}{11}$

Cross-multiply to form a linear equation:

$11(x+2) = 9(y+2)$

$11x + 22 = 9y + 18$

Rearrange into the standard form $ax + by = c$:

---

Condition 2: "If 3 is added to both the numerator and the denominator, it becomes $\frac{5}{6}$."

Adding 3 to the numerator gives $x+3$.

Adding 3 to the denominator gives $y+3$.

The new fraction is $\frac{x+3}{y+3}$. According to the condition:

$\frac{x+3}{y+3} = \frac{5}{6}$

Cross-multiply to form a linear equation:

$6(x+3) = 5(y+3)$

$6x + 18 = 5y + 15$

Rearrange into the standard form $ax + by = c$:

---

We have framed the pair of linear equations:

$11x - 9y = -4$ (1)

$6x - 5y = -3$ (2)

---

We will solve this system using the elimination method. Let's eliminate $y$. The coefficients of $y$ are -9 and -5. The LCM of 9 and 5 is 45.

---

Multiply equation (1) by 5:

$5 \times (11x - 9y) = 5 \times (-4)$

---

Multiply equation (2) by 9:

$9 \times (6x - 5y) = 9 \times (-3)$

---

Now, we have coefficients of $y$ equal (-45) in both equations (3) and (4). Since they have the same sign, we subtract equation (4) from equation (3):

$(55x - 45y) - (54x - 45y) = -20 - (-27)$

$55x - 45y - 54x + 45y = -20 + 27$

$(55x - 54x) + (-45y + 45y) = 7$

$x + 0 = 7$

$x = 7$

---

Now, substitute the value of $x = 7$ into equation (1) to find the value of $y$:

$11x - 9y = -4$

$11(7) - 9y = -4$

$77 - 9y = -4$

---

Solve for $y$:

$-9y = -4 - 77$

$-9y = -81$

$y = \frac{-81}{-9}$

$y = 9$

---

The numerator is $x = 7$ and the denominator is $y = 9$.

The original fraction is $\frac{x}{y}$.

Original fraction $= \frac{7}{9}$.

---

Answer:

The pair of linear equations is $11x - 9y = -4$ and $6x - 5y = -3$.

The original fraction is $\frac{7}{9}$.

---

Verification:

Add 2: $\frac{7+2}{9+2} = \frac{9}{11}$ (Satisfied)

Add 3: $\frac{7+3}{9+3} = \frac{10}{12} = \frac{\cancel{10}^{5}}{\cancel{12}_{6}} = \frac{5}{6}$ (Satisfied)

The solution is correct.

Given:

The given pair of linear equations is:

---

To Find:

The values of $a$ and $b$ for which the pair of equations has infinitely many solutions.

---

Solution:

First, we write the equations in the standard form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$.

From equation (1): $2x + 3y - 7 = 0$

From equation (2): $(a - b)x + (a + b)y - (3a + b - 2) = 0$

---

Comparing these with the standard form, we identify the coefficients:

$a_1 = 2, b_1 = 3, c_1 = -7$

$a_2 = a - b, b_2 = a + b, c_2 = -(3a + b - 2) = 2 - 3a - b$

---

For a pair of linear equations to have infinitely many solutions, the lines representing the equations must be coincident. The condition for this is:

$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$

---

Substitute the values of the coefficients into this condition:

$\frac{2}{a - b} = \frac{3}{a + b} = \frac{-7}{-(3a + b - 2)}$

$\frac{2}{a - b} = \frac{3}{a + b} = \frac{7}{3a + b - 2}$

---

We need to solve for $a$ and $b$ using these equalities. We can form two equations from this relationship:

Equation from the first two ratios:

$\frac{2}{a - b} = \frac{3}{a + b}$

Cross-multiply:

$2(a + b) = 3(a - b)$

$2a + 2b = 3a - 3b$

Rearrange terms to solve for $a$ in terms of $b$:

$2b + 3b = 3a - 2a$

---

Equation from the last two ratios:

$\frac{3}{a + b} = \frac{7}{3a + b - 2}$

Cross-multiply:

$3(3a + b - 2) = 7(a + b)$

$9a + 3b - 6 = 7a + 7b$

Rearrange terms:

$9a - 7a + 3b - 7b = 6$

Divide by 2:

---

Now we have a pair of linear equations in terms of $a$ and $b$:

$a = 5b$ (3)

$a - 2b = 3$ (4)

---

Substitute the expression for $a$ from equation (3) into equation (4):

$(5b) - 2b = 3$

$3b = 3$

$b = \frac{3}{3}$

$b = 1$

---

Now substitute the value of $b = 1$ back into equation (3) to find the value of $a$:

$a = 5b$

$a = 5(1)$

$a = 5$

---

Therefore, the values of $a$ and $b$ for which the given pair of linear equations has infinitely many solutions are $a = 5$ and $b = 1$.

---

Answer:

The values are $a = 5$ and $b = 1$.

Given:

The given pair of equations is:

---

To Solve:

Solve the given pair of equations for $x$ and $y$.

---

Solution:

The given equations are not linear in $x$ and $y$. However, we can observe that the terms involving the variables are $\frac{1}{x}$ and $\frac{1}{y}$. We can reduce this pair of equations to a pair of linear equations by making a suitable substitution.

---

Let $u = \frac{1}{x}$ and $v = \frac{1}{y}$. We must assume $x \neq 0$ and $y \neq 0$.

---

Substituting $u = \frac{1}{x}$ and $v = \frac{1}{y}$ into equation (1):

$\frac{1}{2} \left(\frac{1}{x}\right) + \frac{1}{3} \left(\frac{1}{y}\right) = 2$

$\frac{1}{2} u + \frac{1}{3} v = 2$

To eliminate the fractions, multiply the entire equation by the LCM of 2 and 3, which is 6:

$6 \times \left(\frac{1}{2} u\right) + 6 \times \left(\frac{1}{3} v\right) = 6 \times 2$

---

Substituting $u = \frac{1}{x}$ and $v = \frac{1}{y}$ into equation (2):

$\frac{1}{3} \left(\frac{1}{x}\right) + \frac{1}{2} \left(\frac{1}{y}\right) = \frac{13}{6}$

$\frac{1}{3} u + \frac{1}{2} v = \frac{13}{6}$

To eliminate the fractions, multiply the entire equation by the LCM of 3, 2, and 6, which is 6:

$6 \times \left(\frac{1}{3} u\right) + 6 \times \left(\frac{1}{2} v\right) = 6 \times \left(\frac{13}{6}\right)$

---

Now we have a pair of linear equations in $u$ and $v$:

$3u + 2v = 12$ (3)

$2u + 3v = 13$ (4)

We can solve this system using the elimination method. Let's eliminate $u$. Multiply equation (3) by 2 and equation (4) by 3 to make the coefficient of $u$ equal to 6 in both equations.

---

Multiply equation (3) by 2:

$2 \times (3u + 2v) = 2 \times 12$

---

Multiply equation (4) by 3:

$3 \times (2u + 3v) = 3 \times 13$

---

Subtract equation (5) from equation (6) to eliminate $u$:

$(6u + 9v) - (6u + 4v) = 39 - 24$

$6u + 9v - 6u - 4v = 15$

$5v = 15$

---

Solve for $v$:

$v = \frac{15}{5}$

$v = 3$

---

Now substitute the value of $v = 3$ into equation (3) to find the value of $u$:

$3u + 2v = 12$

$3u + 2(3) = 12$

$3u + 6 = 12$

$3u = 12 - 6$

$3u = 6$

---

Solve for $u$:

$u = \frac{6}{3}$

$u = 2$

---

Now we use our original substitutions to find $x$ and $y$:

$u = \frac{1}{x} \implies 2 = \frac{1}{x}$

Multiply both sides by $x$ and divide by 2:

$2x = 1$

$x = \frac{1}{2}$

---

$v = \frac{1}{y} \implies 3 = \frac{1}{y}$

Multiply both sides by $y$ and divide by 3:

$3y = 1$

$y = \frac{1}{3}$

---

The solution to the pair of equations is $x = \frac{1}{2}$ and $y = \frac{1}{3}$.

---

Answer:

The solution is $x = \frac{1}{2}$ and $y = \frac{1}{3}$.

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Verification (Optional):

Substitute $x=\frac{1}{2}$ and $y=\frac{1}{3}$ into the original equations:

Equation (1): $\frac{1}{2(\frac{1}{2})} + \frac{1}{3(\frac{1}{3})} = \frac{1}{1} + \frac{1}{1} = 1 + 1 = 2$ (Satisfied)

Equation (2): $\frac{1}{3(\frac{1}{2})} + \frac{1}{2(\frac{1}{3})} = \frac{1}{\frac{3}{2}} + \frac{1}{\frac{2}{3}} = \frac{2}{3} + \frac{3}{2} = \frac{4}{6} + \frac{9}{6} = \frac{4+9}{6} = \frac{13}{6}$ (Satisfied)

The solution is correct.

Problem Description:

We are given the time taken by a boat to travel certain distances upstream and downstream under two different scenarios. We need to find the speed of the boat in still water and the speed of the stream.

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To Find:

The speed of the stream and the speed of the boat in still water.

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Solution:

Let the speed of the boat in still water be $x$ km/hr.

Let the speed of the stream be $y$ km/hr.

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When the boat travels upstream, its effective speed is reduced by the speed of the stream.

Speed upstream = $(x - y)$ km/hr.

When the boat travels downstream, its effective speed is increased by the speed of the stream.

Speed downstream = $(x + y)$ km/hr.

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We know the formula: Time = $\frac{\text{Distance}}{\text{Speed}}$.

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Scenario 1: Boat goes 30 km upstream and 44 km downstream in 10 hours.

Time taken to go 30 km upstream = $\frac{30}{x - y}$ hours.

Time taken to go 44 km downstream = $\frac{44}{x + y}$ hours.

Total time = 10 hours.

Equation based on Scenario 1:

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Scenario 2: In 13 hours, the boat goes 40 km upstream and 55 km downstream.

Time taken to go 40 km upstream = $\frac{40}{x - y}$ hours.

Time taken to go 55 km downstream = $\frac{55}{x + y}$ hours.

Total time = 13 hours.

Equation based on Scenario 2:

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The given equations are not linear. We can reduce them to a pair of linear equations by making a substitution.

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Let $u = \frac{1}{x - y}$ and $v = \frac{1}{x + y}$. (Note: For these substitutions to be valid, we must have $x - y \neq 0$ and $x + y \neq 0$. Also, for upstream travel to be possible, $x > y$).

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Substitute $u$ and $v$ into equations (1) and (2):

From equation (1):

From equation (2):

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Now we have a pair of linear equations in terms of $u$ and $v$. We can solve this system using the elimination method. Let's eliminate $u$. The LCM of 30 and 40 is 120.

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Multiply equation (3) by 4:

$4 \times (30u + 44v) = 4 \times 10$

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Multiply equation (4) by 3:

$3 \times (40u + 55v) = 3 \times 13$

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Subtract equation (6) from equation (5) to eliminate $u$:

$(120u + 176v) - (120u + 165v) = 40 - 39$

$120u + 176v - 120u - 165v = 1$

$11v = 1$

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Solve for $v$:

$v = \frac{1}{11}$

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Now substitute the value of $v = \frac{1}{11}$ into equation (3) to find the value of $u$:

$30u + 44v = 10$

$30u + 44\left(\frac{1}{11}\right) = 10$

$30u + \frac{44}{11} = 10$

$30u + 4 = 10$

$30u = 10 - 4$

$30u = 6$

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Solve for $u$:

$u = \frac{6}{30}$

$u = \frac{1}{5}$

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Now we use our original substitutions $u = \frac{1}{x - y}$ and $v = \frac{1}{x + y}$ to find $x$ and $y$.

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Substitute the value of $u$:

$\frac{1}{x - y} = \frac{1}{5}$

Taking the reciprocal of both sides:

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Substitute the value of $v$:

$\frac{1}{x + y} = \frac{1}{11}$

Taking the reciprocal of both sides:

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Now we have a simple pair of linear equations in $x$ and $y$. We can solve this system using the elimination method by adding the two equations.

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Add equation (7) and equation (8):

$(x - y) + (x + y) = 5 + 11$

$x - y + x + y = 16$

$2x = 16$

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Solve for $x$:

$x = \frac{16}{2}$

$x = 8$

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Now substitute the value of $x = 8$ into equation (8) to find the value of $y$:

$x + y = 11$

$8 + y = 11$

$y = 11 - 8$

$y = 3$

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The speed of the boat in still water is $x = 8$ km/hr.

The speed of the stream is $y = 3$ km/hr.

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Answer:

The speed of the boat in still water is 8 km/hr.

The speed of the stream is 3 km/hr.

The graphical method for solving a pair of linear equations involves plotting the lines represented by each equation on the same coordinate plane. The solution to the system is the point where the lines intersect. In this problem, we are also asked to find the vertices of the triangle formed by these lines and the x-axis.

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We are given the following pair of linear equations:

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To draw the graph of each equation, we find at least two points that satisfy the equation. It's helpful to find the points where the lines intersect the axes.

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For Equation (1): $x - y + 1 = 0 \implies y = x + 1$

- When $x = 0$, $y = 0 + 1 = 1$. Point: (0, 1).

- When $y = 0$, $0 = x + 1 \implies x = -1$. Point: (-1, 0).

- When $x = 2$, $y = 2 + 1 = 3$. Point: (2, 3).

Table of points for line 1:

$x$	$y = x + 1$	Point ($x$, $y$)
0	1	(0, 1)
-1	0	(-1, 0)
2	3	(2, 3)

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For Equation (2): $3x + 2y - 12 = 0 \implies 2y = 12 - 3x \implies y = \frac{12 - 3x}{2}$

- When $x = 0$, $y = \frac{12 - 3(0)}{2} = \frac{12}{2} = 6$. Point: (0, 6).

- When $y = 0$, $3x + 2(0) - 12 = 0 \implies 3x = 12 \implies x = 4$. Point: (4, 0).

- When $x = 2$, $y = \frac{12 - 3(2)}{2} = \frac{12 - 6}{2} = \frac{6}{2} = 3$. Point: (2, 3).

Table of points for line 2:

$x$	$y = \frac{12 - 3x}{2}$	Point ($x$, $y$)
0	6	(0, 6)
4	0	(4, 0)
2	3	(2, 3)

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Now, plot these points on a graph paper. Draw a straight line passing through the points for equation (1) and another straight line passing through the points for equation (2).

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Determining the Vertices of the Triangle:

The triangle is formed by the two lines and the x-axis. The vertices of this triangle are the points of intersection of these three lines taken in pairs:

1. Intersection of the first line ($x - y + 1 = 0$) and the x-axis ($y = 0$).

Substitute $y=0$ into equation (1): $x - 0 + 1 = 0 \implies x = -1$. Vertex A is at $(-1, 0)$. This is the x-intercept of the first line.

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2. Intersection of the second line ($3x + 2y - 12 = 0$) and the x-axis ($y = 0$).

Substitute $y=0$ into equation (2): $3x + 2(0) - 12 = 0 \implies 3x = 12 \implies x = 4$. Vertex B is at $(4, 0)$. This is the x-intercept of the second line.

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3. Intersection of the first line ($x - y + 1 = 0$) and the second line ($3x + 2y - 12 = 0$).

We can see from our tables that the point (2, 3) is common to both lines. Thus, the lines intersect at (2, 3).

Alternatively, solve the system algebraically: From (1), $y = x + 1$. Substitute this into (2): $3x + 2(x + 1) - 12 = 0$ $3x + 2x + 2 - 12 = 0$ $5x - 10 = 0$ $5x = 10 \implies x = 2$. Substitute $x=2$ into $y = x + 1$: $y = 2 + 1 = 3$. Vertex C is at $(2, 3)$. This is the point of intersection of the two lines.

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The vertices of the triangle formed by the given lines and the x-axis are A(-1, 0), B(4, 0), and C(2, 3).

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Shading the Triangular Region:

On the graph, the triangular region is bounded by the three line segments connecting these vertices: the segment from (-1, 0) to (4, 0) along the x-axis, the segment connecting (-1, 0) to (2, 3), and the segment connecting (4, 0) to (2, 3). This region should be shaded.

Given:

Let the length of the rectangle be $x$ units.

Let the breadth of the rectangle be $y$ units.

The original area of the rectangle is $A = xy$ square units.

Condition 1: If the length is reduced by 5 units ($x-5$) and the breadth is increased by 3 units ($y+3$), the new area is reduced by 9 square units compared to the original area. So, the new area is $xy - 9$.

Condition 2: If the length is increased by 3 units ($x+3$) and the breadth is increased by 2 units ($y+2$), the new area is increased by 67 square units compared to the original area. So, the new area is $xy + 67$.

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To Find:

The dimensions of the rectangle, i.e., the values of $x$ and $y$.

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Solution:

According to Condition 1:

The new length is $(x - 5)$ units and the new breadth is $(y + 3)$ units.

The new area is $(x - 5)(y + 3)$ square units.

From the problem statement, the new area is equal to the original area minus 9:

$(x - 5)(y + 3) = xy - 9$

Expanding the left side of the equation:

$xy + 3x - 5y - 15 = xy - 9$

Subtracting $xy$ from both sides:

$3x - 5y - 15 = -9$

Adding 15 to both sides:

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According to Condition 2:

The new length is $(x + 3)$ units and the new breadth is $(y + 2)$ units.

The new area is $(x + 3)(y + 2)$ square units.

From the problem statement, the new area is equal to the original area plus 67:

$(x + 3)(y + 2) = xy + 67$

Expanding the left side of the equation:

$xy + 2x + 3y + 6 = xy + 67$

Subtracting $xy$ from both sides:

$2x + 3y + 6 = 67$

Subtracting 6 from both sides:

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Now we have a system of two linear equations:

(1) $3x - 5y = 6$

(2) $2x + 3y = 61$

We can solve this system using the elimination method. To eliminate $x$, we multiply Equation (1) by 2 and Equation (2) by 3.

Multiply Equation (1) by 2:

$6x - 10y = 12$

Multiply Equation (2) by 3:

$6x + 9y = 183$

Now, subtract Equation (3) from Equation (4):

$(6x + 9y) - (6x - 10y) = 183 - 12$

$6x + 9y - 6x + 10y = 171$

$19y = 171$

Divide both sides by 19:

$y = \frac{171}{19}$

$y = 9$

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Now, substitute the value of $y = 9$ into Equation (1):

$3x - 5y = 6$

$3x - 5(9) = 6$

$3x - 45 = 6$

Add 45 to both sides:

$3x = 6 + 45$

$3x = 51$

Divide both sides by 3:

$x = \frac{51}{3}$

$x = 17$

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The length of the rectangle is $x = 17$ units and the breadth of the rectangle is $y = 9$ units.

Final Answer:

The dimensions of the rectangle are Length = 17 units and Breadth = 9 units.

Given a pair of linear equations in two variables:

$a_1x + b_1y + c_1 = 0$

$a_2x + b_2y + c_2 = 0$

where $a_1, b_1, c_1, a_2, b_2, c_2$ are real numbers, and $a_1^2 + b_1^2 \neq 0$, $a_2^2 + b_2^2 \neq 0$. The nature of the solutions depends on the ratios of the coefficients.

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(a) Unique Solution:

A pair of linear equations has a unique solution if and only if the ratio of the coefficients of $x$ is not equal to the ratio of the coefficients of $y$.

Condition: $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$

Geometrically, this corresponds to two lines that intersect at exactly one point.

Example:

Consider the pair of equations:

$2x + 3y - 7 = 0$

$x - 2y + 4 = 0$

Here, $a_1 = 2, b_1 = 3, c_1 = -7$ and $a_2 = 1, b_2 = -2, c_2 = 4$.

Check the condition:

$\frac{a_1}{a_2} = \frac{2}{1} = 2$

$\frac{b_1}{b_2} = \frac{3}{-2} = -\frac{3}{2}$

Since $2 \neq -\frac{3}{2}$, i.e., $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, the system has a unique solution.

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(b) No Solution:

A pair of linear equations has no solution if and only if the ratio of the coefficients of $x$ is equal to the ratio of the coefficients of $y$, but this ratio is not equal to the ratio of the constant terms.

Condition: $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$

Geometrically, this corresponds to two parallel and distinct lines. Since they never intersect, there is no common point, hence no solution.

Example:

Consider the pair of equations:

$x + 2y - 4 = 0$

$2x + 4y - 12 = 0$

Here, $a_1 = 1, b_1 = 2, c_1 = -4$ and $a_2 = 2, b_2 = 4, c_2 = -12$.

Check the condition:

$\frac{a_1}{a_2} = \frac{1}{2}$

$\frac{b_1}{b_2} = \frac{2}{4} = \frac{1}{2}$

$\frac{c_1}{c_2} = \frac{-4}{-12} = \frac{1}{3}$

Since $\frac{1}{2} = \frac{1}{2} \neq \frac{1}{3}$, i.e., $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the system has no solution.

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(c) Infinitely Many Solutions:

A pair of linear equations has infinitely many solutions if and only if the ratio of the coefficients of $x$, the ratio of the coefficients of $y$, and the ratio of the constant terms are all equal.

Condition: $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$

Geometrically, this corresponds to two coincident lines. One line lies exactly on top of the other, meaning every point on the first line is also on the second line, resulting in an infinite number of common points (solutions).

Example:

Consider the pair of equations:

$3x + 4y - 10 = 0$

$6x + 8y - 20 = 0$

Here, $a_1 = 3, b_1 = 4, c_1 = -10$ and $a_2 = 6, b_2 = 8, c_2 = -20$.

Check the condition:

$\frac{a_1}{a_2} = \frac{3}{6} = \frac{1}{2}$

$\frac{b_1}{b_2} = \frac{4}{8} = \frac{1}{2}$

$\frac{c_1}{c_2} = \frac{-10}{-20} = \frac{1}{2}$

Since $\frac{1}{2} = \frac{1}{2} = \frac{1}{2}$, i.e., $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, the system has infinitely many solutions.