Chapter 3 Pair Of Linear Equations In Two Variables

Given Equations:

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Solution using Substitution Method:

Step 1: Choose one equation and express one variable in terms of the other.

From equation (2), it is easy to express $x$ in terms of $y$.

Step 2: Substitute the expression for $x$ from equation (3) into the other equation (equation (1)).

Substituting $x = 3 - 2y$ in equation (1), we get:

$7(3 - 2y) - 15y = 2$

Step 3: Solve the resulting linear equation in one variable ($y$).

$21 - 14y - 15y = 2$

$21 - 29y = 2$

$-29y = 2 - 21$

$-29y = -19$

$y = \frac{-19}{-29}$

$y = \frac{19}{29}$

Step 4: Substitute the value of $y$ back into equation (3) to find the value of $x$.

Substituting $y = \frac{19}{29}$ in equation (3):

$x = 3 - 2 \left( \frac{19}{29} \right)$

$x = 3 - \frac{38}{29}$

$x = \frac{3 \times 29}{29} - \frac{38}{29}$

$x = \frac{87 - 38}{29}$

$x = \frac{49}{29}$

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Result:

The solution to the pair of linear equations is:

$x = \frac{49}{29}$ and $y = \frac{19}{29}$.

Algebraic Representation (from Q.1 of Ex 3.1):

Let the present age of Aftab be $x$ years.

Let the present age of his daughter be $y$ years.

The pair of linear equations formed in Q.1 of Exercise 3.1 is:

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Solution using Substitution Method:

We need to solve the above pair of equations for $x$ and $y$.

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

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

Step 2: Substitute the expression for $x$ from equation (iii) into equation (i).

Substituting $x = 3y + 6$ in equation (i), we get:

$(3y + 6) - 7y = -42$

Step 3: Solve the resulting linear equation in one variable ($y$).

$3y + 6 - 7y = -42$

$-4y + 6 = -42$

$-4y = -42 - 6$

$-4y = -48$

$y = \frac{-48}{-4}$

$y = 12$

Step 4: Substitute the value of $y$ back into equation (iii) to find the value of $x$.

Substituting $y = 12$ in equation (iii):

$x = 3(12) + 6$

$x = 36 + 6$

$x = 42$

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Result:

The solution to the pair of linear equations is $x = 42$ and $y = 12$.

Therefore, Aftab's present age is 42 years and his daughter's present age is 12 years.

Algebraic Representation (from Example 2):

Let the cost of one pencil be $\textsf{₹ } x$.

Let the cost of one eraser be $\textsf{₹ } y$.

The pair of linear equations representing the situation is:

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Solution using Substitution Method:

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

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

$2x = 9 - 3y$

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

$4 \left( \frac{9 - 3y}{2} \right) + 6y = 18$

Step 3: Simplify and solve the resulting equation.

$2 (9 - 3y) + 6y = 18$

$18 - 6y + 6y = 18$

$18 = 18$

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Conclusion:

The resulting statement, $18 = 18$, is a true statement, but the variable $y$ has been eliminated.

This indicates that the pair of linear equations is dependent and has infinitely many solutions.

The second equation ($4x + 6y = 18$) is essentially a multiple of the first equation ($2x + 3y = 9$), as $2(2x + 3y) = 2(9)$ gives $4x + 6y = 18$.

Therefore, we cannot find a unique cost for one pencil ($x$) and one eraser ($y$) based on the given information. Any pair of values ($x, y$) that satisfies the equation $2x + 3y = 9$ (where $x$ and $y$ are non-negative, representing costs) is a possible solution.

Given Equations (from Example 3):

The equations representing the two rails are:

The question asks whether the rails will cross each other, which means we need to determine if the system of equations has a solution (i.e., if the lines intersect).

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Solution using Substitution Method:

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

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

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

Substituting $x = 4 - 2y$ into equation (2):

$2(4 - 2y) + 4y - 12 = 0$

Step 3: Simplify and solve the resulting equation.

$8 - 4y + 4y - 12 = 0$

$8 - 12 = 0$

$-4 = 0$

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Conclusion:

The resulting statement, $-4 = 0$, is a false statement.

This indicates that the pair of linear equations has no solution.

Geometrically, this means the lines representing the two equations are parallel.

Therefore, the rails will not cross each other.

(i) $x + y = 14$ and $x - y = 4$

The given equations are:

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

Substitute the value of $x$ from equation (3) into equation (1):

$(4 + y) + y = 14$

$4 + 2y = 14$

$2y = 14 - 4$

$2y = 10$

$y = 5$

Substitute the value of $y = 5$ back into equation (3):

$x = 4 + 5$

$x = 9$

Therefore, the solution is $x = 9$ and $y = 5$.

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(ii) $s - t = 3$ and $\frac{s}{3} + \frac{t}{2} = 6$

The given equations are:

From equation (1), we express $s$ in terms of $t$:

Substitute the value of $s$ from equation (3) into equation (2):

$\frac{3 + t}{3} + \frac{t}{2} = 6$

To eliminate fractions, multiply the entire equation by the LCM of 3 and 2, which is 6:

$6 \left( \frac{3 + t}{3} \right) + 6 \left( \frac{t}{2} \right) = 6(6)$

$2(3 + t) + 3(t) = 36$

$6 + 2t + 3t = 36$

$6 + 5t = 36$

$5t = 36 - 6$

$5t = 30$

$t = 6$

Substitute the value of $t = 6$ back into equation (3):

$s = 3 + 6$

$s = 9$

Therefore, the solution is $s = 9$ and $t = 6$.

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(iii) $3x - y = 3$ and $9x - 3y = 9$

The given equations are:

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

Substitute the value of $y$ from equation (3) into equation (2):

$9x - 3(3x - 3) = 9$

$9x - 9x + 9 = 9$

$9 = 9$

This statement is true for all values of $x$. This means the equations are dependent (represent the same line).

Therefore, the pair of linear equations has infinitely many solutions. Any pair $(x, y)$ satisfying $y = 3x - 3$ is a solution.

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(iv) $0.2x + 0.3y = 1.3$ and $0.4x + 0.5y = 2.3$

The given equations are:

$0.2x + 0.3y = 1.3$

$0.4x + 0.5y = 2.3$

Multiply both equations by 10 to remove decimals:

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

$2x = 13 - 3y$

Substitute the value of $x$ from equation (3) into equation (2):

$4 \left( \frac{13 - 3y}{2} \right) + 5y = 23$

$2(13 - 3y) + 5y = 23$

$26 - 6y + 5y = 23$

$26 - y = 23$

$-y = 23 - 26$

$-y = -3$

$y = 3$

Substitute the value of $y = 3$ back into equation (3):

$x = \frac{13 - 3(3)}{2}$

$x = \frac{13 - 9}{2}$

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

$x = 2$

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

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(v) $\sqrt{2}x + \sqrt{3}y = 0$ and $\sqrt{3}x - \sqrt{8}y = 0$

The given equations are:

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

$\sqrt{2}x = -\sqrt{3}y$

Substitute the value of $x$ from equation (3) into equation (2):

$\sqrt{3} \left( -\frac{\sqrt{3}}{\sqrt{2}} y \right) - \sqrt{8}y = 0$

$-\frac{3}{\sqrt{2}} y - \sqrt{8}y = 0$

Noting that $\sqrt{8} = 2\sqrt{2}$:

$-\frac{3}{\sqrt{2}} y - 2\sqrt{2}y = 0$

Multiply by $\sqrt{2}$ to clear the denominator:

$-3y - (2\sqrt{2})(\sqrt{2})y = 0$

$-3y - 4y = 0$

$-7y = 0$

$y = 0$

Substitute the value of $y = 0$ back into equation (3):

$x = -\frac{\sqrt{3}}{\sqrt{2}} (0)$

$x = 0$

Therefore, the solution is $x = 0$ and $y = 0$.

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(vi) $\frac{3x}{2} - \frac{5y}{3} = -2$ and $\frac{x}{3} + \frac{y}{2} = \frac{13}{6}$

The given equations are:

$\frac{3x}{2} - \frac{5y}{3} = -2$

$\frac{x}{3} + \frac{y}{2} = \frac{13}{6}$

Clear the fractions by multiplying by the LCM of the denominators.

For the first equation, LCM(2, 3) = 6:

$6 \left( \frac{3x}{2} \right) - 6 \left( \frac{5y}{3} \right) = 6(-2)$

For the second equation, LCM(3, 2, 6) = 6:

$6 \left( \frac{x}{3} \right) + 6 \left( \frac{y}{2} \right) = 6 \left( \frac{13}{6} \right)$

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

$2x = 13 - 3y$

Substitute the value of $x$ from equation (3) into equation (1):

$9 \left( \frac{13 - 3y}{2} \right) - 10y = -12$

Multiply the entire equation by 2 to clear the fraction:

$9(13 - 3y) - 2(10y) = 2(-12)$

$117 - 27y - 20y = -24$

$117 - 47y = -24$

$-47y = -24 - 117$

$-47y = -141$

$y = \frac{-141}{-47}$

$y = 3$

Substitute the value of $y = 3$ back into equation (3):

$x = \frac{13 - 3(3)}{2}$

$x = \frac{13 - 9}{2}$

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

$x = 2$

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

Given Equations:

And the relation:

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Solving the pair of linear equations:

We can use the substitution method or elimination method. Let's use the substitution method.

From equation (i), express $x$ in terms of $y$:

$2x = 11 - 3y$

Substitute this expression for $x$ into equation (ii):

$2 \left( \frac{11 - 3y}{2} \right) - 4y = -24$

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

$11 - 7y = -24$

$-7y = -24 - 11$

$-7y = -35$

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

$y = 5$

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

$x = \frac{11 - 3(5)}{2}$

$x = \frac{11 - 15}{2}$

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

$x = -2$

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

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Finding the value of 'm':

Now substitute the values $x = -2$ and $y = 5$ into equation (iii):

$y = mx + 3$

$5 = m(-2) + 3$

$5 = -2m + 3$

$5 - 3 = -2m$

$2 = -2m$

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

$m = -1$

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Result:

The solution to the system of equations is $x = -2$, $y = 5$.

The value of $m$ for which $y = mx + 3$ is $m = -1$.

(i) Two Numbers

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

According to the problem:

Substitute equation (2) into equation (1):

$(3y) - y = 26$

$2y = 26$

$y = \frac{26}{2}$

$y = 13$

Substitute the value of $y = 13$ back into equation (2):

$x = 3(13)$

$x = 39$

The two numbers are 39 and 13.

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(ii) Supplementary Angles

Let the larger angle be $x$ degrees and the smaller angle be $y$ degrees.

Since the angles are supplementary:

The larger angle exceeds the smaller by 18 degrees:

Substitute equation (2) into equation (1):

$(y + 18) + y = 180$

$2y + 18 = 180$

$2y = 180 - 18$

$2y = 162$

$y = \frac{162}{2}$

$y = 81$

Substitute the value of $y = 81$ back into equation (2):

$x = 81 + 18$

$x = 99$

The two angles are 99 degrees and 81 degrees.

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(iii) Bats and Balls

Let the cost of one bat be $\textsf{₹ } x$ and the cost of one ball be $\textsf{₹ } y$.

According to the problem:

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

$3x = 1750 - 5y$

Substitute the value of $x$ from equation (3) into equation (1):

$7 \left( \frac{1750 - 5y}{3} \right) + 6y = 3800$

Multiply by 3 to clear the fraction:

$7(1750 - 5y) + 3(6y) = 3(3800)$

$12250 - 35y + 18y = 11400$

$12250 - 17y = 11400$

$-17y = 11400 - 12250$

$-17y = -850$

$y = \frac{-850}{-17}$

$y = 50$

Substitute the value of $y = 50$ back into equation (3):

$x = \frac{1750 - 5(50)}{3}$

$x = \frac{1750 - 250}{3}$

$x = \frac{1500}{3}$

$x = 500$

The cost of one bat is $\textsf{₹ } 500$ and the cost of one ball is $\textsf{₹ } 50$.

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(iv) Taxi Charges

Let the fixed charge be $\textsf{₹ } x$ and the charge per km be $\textsf{₹ } y$.

According to the problem:

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

Substitute the value of $x$ from equation (3) into equation (2):

$(105 - 10y) + 15y = 155$

$105 + 5y = 155$

$5y = 155 - 105$

$5y = 50$

$y = 10$

Substitute the value of $y = 10$ back into equation (3):

$x = 105 - 10(10)$

$x = 105 - 100$

$x = 5$

The fixed charge is $\textsf{₹ } 5$ and the charge per km is $\textsf{₹ } 10$.

Now, find the charge for travelling 25 km:

Total charge = Fixed charge + (Charge per km $\times$ Distance)

Total charge = $x + 25y$

Total charge = $5 + 25(10)$

Total charge = $5 + 250$

Total charge = $255$

The charge for travelling 25 km is $\textsf{₹ } 255$.

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(v) Fraction

Let the fraction be $\frac{x}{y}$.

According to the problem:

If 2 is added to both numerator and denominator, the fraction becomes $\frac{9}{11}$:

$\frac{x+2}{y+2} = \frac{9}{11}$

$11(x+2) = 9(y+2)$

$11x + 22 = 9y + 18$

If 3 is added to both numerator and denominator, the fraction becomes $\frac{5}{6}$:

$\frac{x+3}{y+3} = \frac{5}{6}$

$6(x+3) = 5(y+3)$

$6x + 18 = 5y + 15$

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

$6x = 5y - 3$

Substitute the value of $x$ from equation (3) into equation (1):

$11 \left( \frac{5y - 3}{6} \right) - 9y = -4$

Multiply by 6 to clear the fraction:

$11(5y - 3) - 6(9y) = 6(-4)$

$55y - 33 - 54y = -24$

$y - 33 = -24$

$y = -24 + 33$

$y = 9$

Substitute the value of $y = 9$ back into equation (3):

$x = \frac{5(9) - 3}{6}$

$x = \frac{45 - 3}{6}$

$x = \frac{42}{6}$

$x = 7$

The fraction is $\frac{x}{y}$. Therefore, the fraction is $\frac{7}{9}$.

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(vi) Ages

Let Jacob's present age be $x$ years and his son's present age be $y$ years.

Five years hence:

Jacob's age = $x+5$

Son's age = $y+5$

According to the problem: $x+5 = 3(y+5)$

$x+5 = 3y + 15$

Five years ago:

Jacob's age = $x-5$

Son's age = $y-5$

According to the problem: $x-5 = 7(y-5)$

$x-5 = 7y - 35$

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

Substitute the value of $x$ from equation (3) into equation (2):

$(3y + 10) - 7y = -30$

$-4y + 10 = -30$

$-4y = -30 - 10$

$-4y = -40$

$y = 10$

Substitute the value of $y = 10$ back into equation (3):

$x = 3(10) + 10$

$x = 30 + 10$

$x = 40$

Jacob's present age is 40 years and his son's present age is 10 years.

Given:

Ratio of monthly incomes of two persons = 9 : 7.

Ratio of their monthly expenditures = 4 : 3.

Monthly savings for each person = $\textsf{₹ } 2000$.

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To Find:

The monthly incomes of the two persons.

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Solution:

Let the monthly incomes of the two persons be $\textsf{₹ } 9x$ and $\textsf{₹ } 7x$ respectively.

Let their monthly expenditures be $\textsf{₹ } 4y$ and $\textsf{₹ } 3y$ respectively.

We know that: Income - Expenditure = Savings

For the first person:

For the second person:

Now, we need to solve this pair of linear equations for $x$. We can use the elimination method.

Multiply equation (i) by 3 and equation (ii) by 4 to make the coefficients of $y$ equal:

$3 \times (9x - 4y = 2000) \implies 27x - 12y = 6000$

$4 \times (7x - 3y = 2000) \implies 28x - 12y = 8000$

Subtract equation (iii) from equation (iv):

$(28x - 12y) - (27x - 12y) = 8000 - 6000$

$28x - 12y - 27x + 12y = 2000$

$x = 2000$

Now, calculate the monthly incomes:

Income of the first person = $9x = 9 \times 2000 = 18000$.

Income of the second person = $7x = 7 \times 2000 = 14000$.

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Result:

The monthly incomes of the two persons are $\textsf{₹ } 18000$ and $\textsf{₹ } 14000$ respectively.

Given Equations:

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Solution using Elimination Method:

We want to make the coefficients of either $x$ or $y$ the same in both equations.

Let's make the coefficient of $x$ the same. The coefficient of $x$ in equation (1) is 2 and in equation (2) is 4. The least common multiple (LCM) of 2 and 4 is 4.

Multiply equation (1) by 2:

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

Now we have the following system of equations:

Subtract equation (2) from equation (3) to eliminate both $x$ and $y$:

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

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

$0 = 9$

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Conclusion:

The resulting equation $0 = 9$ is a false statement.

This means that the given pair of linear equations is inconsistent and has no possible solution.

Geometrically, the lines represented by these equations are parallel and never intersect.

Algebraic Representation:

Let the two-digit number be $10x + y$, where $x$ is the digit in the tens place and $y$ is the digit in the units place.

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

According to the first condition, the sum of the number and the reversed number is 66:

$(10x + y) + (10y + x) = 66$

$11x + 11y = 66$

Divide the equation by 11:

According to the second condition, the digits differ by 2. This gives two possibilities:

Possibility 1: $x - y = 2$

Possibility 2: $y - x = 2$

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Solving Case 1 (Equations (1) and (2)):

We have the system:

Using the elimination method, add equations (1) and (2):

$(x + y) + (x - y) = 6 + 2$

$2x = 8$

$x = 4$

Substitute $x = 4$ into equation (1):

$4 + y = 6$

$y = 2$

In this case, the number is $10x + y = 10(4) + 2 = 42$.

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Solving Case 2 (Equations (1) and (3)):

We have the system:

Using the elimination method, add equations (1) and (3):

$(x + y) + (-x + y) = 6 + 2$

$2y = 8$

$y = 4$

Substitute $y = 4$ into equation (1):

$x + 4 = 6$

$x = 2$

In this case, the number is $10x + y = 10(2) + 4 = 24$.

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Conclusion:

The numbers satisfying the given conditions are 42 and 24.

There are two such numbers.

(i) $x + y = 5$ and $2x - 3y = 4$

The given equations are:

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Elimination Method:

Multiply equation (1) by 3:

Add equation (2) and equation (3):

$(2x - 3y) + (3x + 3y) = 4 + 15$

$5x = 19$

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

Substitute $x = \frac{19}{5}$ into equation (1):

$\frac{19}{5} + y = 5$

$y = 5 - \frac{19}{5}$

$y = \frac{25 - 19}{5}$

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

Solution by elimination: $x = \frac{19}{5}, y = \frac{6}{5}$.

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Substitution Method:

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

Substitute this value of $y$ into equation (2):

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

$2x - 15 + 3x = 4$

$5x = 4 + 15$

$5x = 19$

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

Substitute $x = \frac{19}{5}$ back into equation (4):

$y = 5 - \frac{19}{5}$

$y = \frac{25 - 19}{5}$

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

Solution by substitution: $x = \frac{19}{5}, y = \frac{6}{5}$.

The solution is $x = \frac{19}{5}$ and $y = \frac{6}{5}$.

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(ii) $3x + 4y = 10$ and $2x - 2y = 2$

The given equations are:

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Elimination Method:

Multiply equation (2) by 2:

Add equation (1) and equation (3):

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

$7x = 14$

$x = 2$

Substitute $x = 2$ into equation (2):

$2(2) - 2y = 2$

$4 - 2y = 2$

$-2y = 2 - 4$

$-2y = -2$

$y = 1$

Solution by elimination: $x = 2, y = 1$.

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Substitution Method:

From equation (2), divide by 2: $x - y = 1$.

Express $x$ in terms of $y$:

Substitute this value of $x$ into equation (1):

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

$3y + 3 + 4y = 10$

$7y + 3 = 10$

$7y = 7$

$y = 1$

Substitute $y = 1$ back into equation (4):

$x = 1 + 1$

$x = 2$

Solution by substitution: $x = 2, y = 1$.

The solution is $x = 2$ and $y = 1$.

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(iii) $3x - 5y - 4 = 0$ and $9x = 2y + 7$

Rewrite the equations in standard form:

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Elimination Method:

Multiply equation (1) by 3:

Subtract equation (3) from equation (2):

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

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

$13y = -5$

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

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

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

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

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

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

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

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

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

Solution by elimination: $x = \frac{9}{13}, y = -\frac{5}{13}$.

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Substitution Method:

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

$3x = 5y + 4$

Substitute this value of $x$ into equation (2):

$9 \left( \frac{5y + 4}{3} \right) - 2y = 7$

$3(5y + 4) - 2y = 7$

$15y + 12 - 2y = 7$

$13y + 12 = 7$

$13y = 7 - 12$

$13y = -5$

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

Substitute $y = -\frac{5}{13}$ back into equation (4):

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

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

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

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

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

Solution by substitution: $x = \frac{9}{13}, y = -\frac{5}{13}$.

The solution is $x = \frac{9}{13}$ and $y = -\frac{5}{13}$.

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(iv) $\frac{x}{2} + \frac{2y}{3} = -1$ and $x - \frac{y}{3} = 3$

Clear the fractions first.

Multiply the first equation by 6 (LCM of 2 and 3):

$6 \left( \frac{x}{2} \right) + 6 \left( \frac{2y}{3} \right) = 6(-1)$

Multiply the second equation by 3:

$3(x) - 3 \left( \frac{y}{3} \right) = 3(3)$

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Elimination Method:

Subtract equation (2) from equation (1):

$(3x + 4y) - (3x - y) = -6 - 9$

$3x + 4y - 3x + y = -15$

$5y = -15$

$y = -3$

Substitute $y = -3$ into equation (2):

$3x - (-3) = 9$

$3x + 3 = 9$

$3x = 6$

$x = 2$

Solution by elimination: $x = 2, y = -3$.

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Substitution Method:

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

Substitute this value of $y$ into equation (1):

$3x + 4(3x - 9) = -6$

$3x + 12x - 36 = -6$

$15x = -6 + 36$

$15x = 30$

$x = 2$

Substitute $x = 2$ back into equation (3):

$y = 3(2) - 9$

$y = 6 - 9$

$y = -3$

Solution by substitution: $x = 2, y = -3$.

The solution is $x = 2$ and $y = -3$.

(i) Fraction Problem

Let the numerator of the fraction be $x$ and the denominator be $y$. The fraction is $\frac{x}{y}$.

According to the first condition:

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

$x+1 = y-1$

According to the second condition:

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

$2x = y+1$

Now, solve equations (1) and (2) using the elimination method.

Subtract equation (1) from equation (2):

$(2x - y) - (x - y) = 1 - (-2)$

$2x - y - x + y = 3$

$x = 3$

Substitute $x = 3$ into equation (1):

$3 - y = -2$

$-y = -2 - 3$

$-y = -5$

$y = 5$

Therefore, the fraction is $\frac{3}{5}$.

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(ii) Age Problem

Let Nuri's present age be $n$ years and Sonu's present age be $s$ years.

Five years ago:

Nuri's age = $n-5$

Sonu's age = $s-5$

Condition: $n-5 = 3(s-5)$

$n-5 = 3s - 15$

Ten years later:

Nuri's age = $n+10$

Sonu's age = $s+10$

Condition: $n+10 = 2(s+10)$

$n+10 = 2s + 20$

Now, solve equations (1) and (2) using the elimination method.

Subtract equation (1) from equation (2):

$(n - 2s) - (n - 3s) = 10 - (-10)$

$n - 2s - n + 3s = 20$

$s = 20$

Substitute $s = 20$ into equation (2):

$n - 2(20) = 10$

$n - 40 = 10$

$n = 50$

Therefore, Nuri's present age is 50 years and Sonu's present age is 20 years.

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(iii) Two-Digit Number Problem

Let the digit in the tens place be $x$ and the digit in the units place be $y$.

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

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

According to the first condition (sum of digits is 9):

According to the second condition (nine times the number is twice the reversed number):

$9(10x + y) = 2(10y + x)$

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

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

$88x = 11y$

$8x = y$

Now, solve equations (1) and (2) using the elimination method.

Add equation (1) and equation (2):

$(x + y) + (8x - y) = 9 + 0$

$9x = 9$

$x = 1$

Substitute $x = 1$ into equation (1):

$1 + y = 9$

$y = 8$

The number is $10x + y = 10(1) + 8 = 18$.

Therefore, the number is 18.

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(iv) Bank Notes Problem

Let the number of $\textsf{₹ } 50$ notes be $f$ and the number of $\textsf{₹ } 100$ notes be $h$.

According to the first condition (total number of notes is 25):

According to the second condition (total value is $\textsf{₹ } 2000$):

$50f + 100h = 2000$

Divide the equation by 50:

Now, solve equations (1) and (2) using the elimination method.

Subtract equation (1) from equation (2):

$(f + 2h) - (f + h) = 40 - 25$

$f + 2h - f - h = 15$

$h = 15$

Substitute $h = 15$ into equation (1):

$f + 15 = 25$

$f = 10$

Therefore, Meena received 10 notes of $\textsf{₹ } 50$ and 15 notes of $\textsf{₹ } 100$.

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(v) Library Charges Problem

Let the fixed charge for the first three days be $\textsf{₹ } x$.

Let the additional charge for each day thereafter be $\textsf{₹ } y$.

For Saritha (book kept for 7 days):

Number of extra days = $7 - 3 = 4$

Total charge = Fixed charge + (Charge per extra day $\times$ Number of extra days)

For Susy (book kept for 5 days):

Number of extra days = $5 - 3 = 2$

Now, solve equations (1) and (2) using the elimination method.

Subtract equation (2) from equation (1):

$(x + 4y) - (x + 2y) = 27 - 21$

$x + 4y - x - 2y = 6$

$2y = 6$

$y = 3$

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

$x + 2(3) = 21$

$x + 6 = 21$

$x = 15$

Therefore, the fixed charge is $\textsf{₹ } 15$ and the charge for each extra day is $\textsf{₹ } 3$.

Algebraic Representation:

Let the fare from the bus stand in Bangalore to Malleswaram be $\textsf{₹ } x$.

Let the fare from the bus stand in Bangalore to Yeshwanthpur be $\textsf{₹ } y$.

According to the first condition:

According to the second condition:

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Solution using Cross-Multiplication Method:

Rewrite the equations in the standard form $ax + by + c = 0$:

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

$a_1 = 2, b_1 = 3, c_1 = -46$

$a_2 = 3, b_2 = 5, c_2 = -74$

Using the cross-multiplication formula:

$\frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1}$

Substitute the values:

$\frac{x}{(3)(-74) - (5)(-46)} = \frac{y}{(-46)(3) - (-74)(2)} = \frac{1}{(2)(5) - (3)(3)}$

$\frac{x}{-222 - (-230)} = \frac{y}{-138 - (-148)} = \frac{1}{10 - 9}$

$\frac{x}{-222 + 230} = \frac{y}{-138 + 148} = \frac{1}{1}$

$\frac{x}{8} = \frac{y}{10} = \frac{1}{1}$

Now, solve for $x$ and $y$:

$\frac{x}{8} = \frac{1}{1} \implies x = 8$

$\frac{y}{10} = \frac{1}{1} \implies y = 10$

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Alternate Solution using Elimination Method:

Multiply equation (1) by 3 and equation (2) by 2:

Subtract equation (3) from equation (4):

$(6x + 10y) - (6x + 9y) = 148 - 138$

$y = 10$

Substitute $y = 10$ into equation (1):

$2x + 3(10) = 46$

$2x + 30 = 46$

$2x = 16$

$x = 8$

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Result:

The fare from the bus stand to Malleswaram is $\textsf{₹ } 8$.

The fare from the bus stand to Yeshwanthpur is $\textsf{₹ } 10$.

Given Equations:

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Condition for Unique Solution:

A pair of linear equations $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ has a unique solution if and only if:

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

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Identifying Coefficients:

Comparing equation (1) with $a_1x + b_1y + c_1 = 0$, we get:

$a_1 = 4$, $b_1 = p$, $c_1 = 8$

Comparing equation (2) with $a_2x + b_2y + c_2 = 0$, we get:

$a_2 = 2$, $b_2 = 2$, $c_2 = 2$

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Applying the Condition:

Substitute the values of the coefficients into the condition for a unique solution:

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

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

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

Multiply both sides by 2:

$2 \times 2 \neq p$

$4 \neq p$

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Conclusion:

The pair of linear equations will have a unique solution for all real values of $p$ except $p = 4$.

Thus, the condition is $p \neq 4$.

Given Equations:

The pair of linear equations is:

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Condition for Infinitely Many Solutions:

A pair of linear equations $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ has infinitely many solutions if and only if:

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

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Identifying Coefficients:

Comparing equation (1) with $a_1x + b_1y + c_1 = 0$, we get:

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

Comparing equation (2) with $a_2x + b_2y + c_2 = 0$, we get:

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

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Applying the Condition:

For infinitely many solutions, we must have:

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

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

Let's consider the first two parts:

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

$k^2 = 12 \times 3$

$k^2 = 36$

$k = \pm 6$

Now, let's consider the last two parts (assuming $k \neq 0$):

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

Multiply both sides by $-k$:

$3(-1) = 3 - k$

$-3 = 3 - k$

$k = 3 + 3$

$k = 6$

Now, let's consider the first and third parts (assuming $k \neq 0$):

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

$k(-k) = 12(3 - k)$

$-k^2 = 36 - 12k$

$k^2 - 12k + 36 = 0$

$(k - 6)^2 = 0$

$k = 6$

For the system to have infinitely many solutions, the value of $k$ must satisfy all the conditions. The common value obtained from all parts is $k=6$.

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Verification for k=6:

Substitute $k=6$ into the ratios:

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

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

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

Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$ when $k = 6$, the system has infinitely many solutions.

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Conclusion:

The value of $k$ for which the given pair of linear equations has infinitely many solutions is $k = 6$.

We compare the given pairs of linear equations with the standard form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$. The conditions for the nature of solutions are:

• Unique solution: $\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$
• No solution: $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$
• Infinitely many solutions: $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$

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(i) $x - 3y - 3 = 0$ and $3x - 9y - 2 = 0$

Here, $a_1 = 1, b_1 = -3, c_1 = -3$

And $a_2 = 3, b_2 = -9, c_2 = -2$

Compute the ratios:

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

$\frac{b_1}{b_2} = \frac{-3}{-9} = \frac{1}{3}$

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

Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the pair of linear equations has no solution.

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(ii) $2x + y = 5$ and $3x + 2y = 8$

Rewrite the equations in standard form:

$2x + y - 5 = 0$

$3x + 2y - 8 = 0$

Here, $a_1 = 2, b_1 = 1, c_1 = -5$

And $a_2 = 3, b_2 = 2, c_2 = -8$

Compute the ratios:

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

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

Since $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, the pair of linear equations has a unique solution.

Using the cross-multiplication method:

$\frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1}$

$\frac{x}{(1)(-8) - (2)(-5)} = \frac{y}{(-5)(3) - (-8)(2)} = \frac{1}{(2)(2) - (3)(1)}$

$\frac{x}{-8 - (-10)} = \frac{y}{-15 - (-16)} = \frac{1}{4 - 3}$

$\frac{x}{-8 + 10} = \frac{y}{-15 + 16} = \frac{1}{1}$

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

Equating $\frac{x}{2}$ with $\frac{1}{1}$:

$x = 2$

Equating $\frac{y}{1}$ with $\frac{1}{1}$:

$y = 1$

The unique solution is $x = 2, y = 1$.

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(iii) $3x - 5y = 20$ and $6x - 10y = 40$

Rewrite the equations in standard form:

$3x - 5y - 20 = 0$

$6x - 10y - 40 = 0$

Here, $a_1 = 3, b_1 = -5, c_1 = -20$

And $a_2 = 6, b_2 = -10, c_2 = -40$

Compute the ratios:

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

$\frac{b_1}{b_2} = \frac{-5}{-10} = \frac{1}{2}$

$\frac{c_1}{c_2} = \frac{-20}{-40} = \frac{1}{2}$

Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, the pair of linear equations has infinitely many solutions.

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(iv) $x - 3y - 7 = 0$ and $3x - 3y - 15 = 0$

Here, $a_1 = 1, b_1 = -3, c_1 = -7$

And $a_2 = 3, b_2 = -3, c_2 = -15$

Compute the ratios:

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

$\frac{b_1}{b_2} = \frac{-3}{-3} = 1$

Since $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, the pair of linear equations has a unique solution.

Using the cross-multiplication method:

$\frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1}$

$\frac{x}{(-3)(-15) - (-3)(-7)} = \frac{y}{(-7)(3) - (-15)(1)} = \frac{1}{(1)(-3) - (3)(-3)}$

$\frac{x}{45 - 21} = \frac{y}{-21 - (-15)} = \frac{1}{-3 - (-9)}$

$\frac{x}{24} = \frac{y}{-21 + 15} = \frac{1}{-3 + 9}$

$\frac{x}{24} = \frac{y}{-6} = \frac{1}{6}$

Equating $\frac{x}{24}$ with $\frac{1}{6}$:

$x = \frac{24}{6}$

$x = 4$

Equating $\frac{y}{-6}$ with $\frac{1}{6}$:

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

$y = -1$

The unique solution is $x = 4, y = -1$.

(i) Infinite Number of Solutions

The given pair of linear equations is:

$2x + 3y = 7 \implies 2x + 3y - 7 = 0$

$(a - b)x + (a + b)y = 3a + b - 2 \implies (a - b)x + (a + b)y - (3a + b - 2) = 0$

For a pair of linear equations $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ to have an infinite number of solutions, the condition is:

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

Here, $a_1 = 2, b_1 = 3, c_1 = -7$.

And $a_2 = a - b, b_2 = a + b, c_2 = -(3a + b - 2)$.

Applying the condition:

$\frac{2}{a - b} = \frac{3}{a + b} = \frac{-7}{-(3a + b - 2)} = \frac{7}{3a + b - 2}$

From the first two parts:

$\frac{2}{a - b} = \frac{3}{a + b}$

$2(a + b) = 3(a - b)$

$2a + 2b = 3a - 3b$

$5b = a$

From the last two parts:

$\frac{3}{a + b} = \frac{7}{3a + b - 2}$

$3(3a + b - 2) = 7(a + b)$

$9a + 3b - 6 = 7a + 7b$

$9a - 7a + 3b - 7b = 6$

$2a - 4b = 6$

Substitute the value of $a$ from equation (1) into equation (2):

$(5b) - 2b = 3$

$3b = 3$

$b = 1$

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

$a = 5(1)$

$a = 5$

Thus, the pair of linear equations will have an infinite number of solutions when $a = 5$ and $b = 1$.

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(ii) No Solution

The given pair of linear equations is:

$3x + y = 1 \implies 3x + y - 1 = 0$

$(2k - 1)x + (k - 1)y = 2k + 1 \implies (2k - 1)x + (k - 1)y - (2k + 1) = 0$

For a pair of linear equations $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ to have no solution, the condition is:

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

Here, $a_1 = 3, b_1 = 1, c_1 = -1$.

And $a_2 = 2k - 1, b_2 = k - 1, c_2 = -(2k + 1)$.

Applying the condition:

$\frac{3}{2k - 1} = \frac{1}{k - 1} \neq \frac{-1}{-(2k + 1)}$

First, consider the equality part:

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

$3(k - 1) = 1(2k - 1)$

$3k - 3 = 2k - 1$

$3k - 2k = -1 + 3$

$k = 2$

Now, consider the inequality part with $k = 2$:

$\frac{b_1}{b_2} \neq \frac{c_1}{c_2}$

$\frac{1}{k - 1} \neq \frac{-1}{-(2k + 1)} = \frac{1}{2k + 1}$

Substitute $k = 2$:

$\frac{1}{2 - 1} \neq \frac{1}{2(2) + 1}$

$\frac{1}{1} \neq \frac{1}{4 + 1}$

$1 \neq \frac{1}{5}$

This inequality is true for $k = 2$.

Thus, the pair of linear equations will have no solution when $k = 2$.

The given pair of linear equations is:

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1. Substitution Method:

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

$2y = 4 - 3x$

Substitute the value of $y$ from equation (3) into equation (1):

$8x + 5 \left( \frac{4 - 3x}{2} \right) = 9$

Multiply the entire equation by 2 to clear the fraction:

$2(8x) + 5(4 - 3x) = 2(9)$

$16x + 20 - 15x = 18$

$x + 20 = 18$

$x = 18 - 20$

$x = -2$

Substitute the value of $x = -2$ back into equation (3):

$y = \frac{4 - 3(-2)}{2}$

$y = \frac{4 + 6}{2}$

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

$y = 5$

So, by substitution method, the solution is $x = -2, y = 5$.

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2. Cross-Multiplication Method:

Rewrite the equations in the standard form $ax + by + c = 0$:

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

$a_1 = 8, b_1 = 5, c_1 = -9$

$a_2 = 3, b_2 = 2, c_2 = -4$

Using the cross-multiplication formula:

$\frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1}$

Substitute the values:

$\frac{x}{(5)(-4) - (2)(-9)} = \frac{y}{(-9)(3) - (-4)(8)} = \frac{1}{(8)(2) - (3)(5)}$

$\frac{x}{-20 - (-18)} = \frac{y}{-27 - (-32)} = \frac{1}{16 - 15}$

$\frac{x}{-20 + 18} = \frac{y}{-27 + 32} = \frac{1}{1}$

$\frac{x}{-2} = \frac{y}{5} = \frac{1}{1}$

Now, solve for $x$ and $y$:

$\frac{x}{-2} = \frac{1}{1} \implies x = -2$

$\frac{y}{5} = \frac{1}{1} \implies y = 5$

So, by cross-multiplication method, the solution is $x = -2, y = 5$.

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Result:

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

(i) Hostel Charges

Formulation:

Let the fixed monthly hostel charge be $\textsf{₹ } x$.

Let the cost of food per day be $\textsf{₹ } y$.

For student A (20 days food):

For student B (26 days food):

Solution (using Elimination Method):

Subtract equation (1) from equation (2):

$(x + 26y) - (x + 20y) = 1180 - 1000$

$6y = 180$

$y = \frac{180}{6}$

$y = 30$

Substitute the value of $y = 30$ into equation (1):

$x + 20(30) = 1000$

$x + 600 = 1000$

$x = 1000 - 600$

$x = 400$

Result:

The fixed charges are $\textsf{₹ } 400$ and the cost of food per day is $\textsf{₹ } 30$.

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(ii) Fraction Problem

Formulation:

Let the numerator be $x$ and the denominator be $y$. The fraction is $\frac{x}{y}$.

According to the first condition:

$\frac{x-1}{y} = \frac{1}{3}$

$3(x-1) = y$

According to the second condition:

$\frac{x}{y+8} = \frac{1}{4}$

$4x = y + 8$

Solution (using Elimination Method):

Subtract equation (1) from equation (2):

$(4x - y) - (3x - y) = 8 - 3$

$4x - y - 3x + y = 5$

$x = 5$

Substitute the value of $x = 5$ into equation (1):

$3(5) - y = 3$

$15 - y = 3$

$-y = 3 - 15$

$-y = -12$

$y = 12$

Result:

The fraction is $\frac{x}{y}$, which is $\frac{5}{12}$.

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(iii) Test Score Problem

Formulation:

Let the number of right answers be $r$.

Let the number of wrong answers be $w$.

According to the first condition:

According to the second condition:

$4r - 2w = 50$

Divide by 2:

Solution (using Elimination Method):

Subtract equation (2) from equation (1):

$(3r - w) - (2r - w) = 40 - 25$

$3r - w - 2r + w = 15$

$r = 15$

Substitute the value of $r = 15$ into equation (1):

$3(15) - w = 40$

$45 - w = 40$

$-w = 40 - 45$

$-w = -5$

$w = 5$

The total number of questions is the sum of right and wrong answers.

Total questions = $r + w = 15 + 5 = 20$

Result:

There were 20 questions in the test.

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(iv) Highway Cars Problem

Formulation:

Let the speed of the car starting from A be $u$ km/h.

Let the speed of the car starting from B be $v$ km/h.

Assume $u > v$ for them to meet when travelling in the same direction.

Case 1: Same direction

Relative speed = $u - v$ km/h.

Distance = Speed $\times$ Time

$100 = (u - v) \times 5$

Case 2: Towards each other

Relative speed = $u + v$ km/h.

Distance = Speed $\times$ Time

$100 = (u + v) \times 1$

Solution (using Elimination Method):

Add equation (1) and equation (2):

$(u - v) + (u + v) = 20 + 100$

$2u = 120$

$u = 60$

Substitute the value of $u = 60$ into equation (2):

$60 + v = 100$

$v = 100 - 60$

$v = 40$

Result:

The speeds of the two cars are 60 km/h and 40 km/h.

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(v) Rectangle Area Problem

Formulation:

Let the length of the rectangle be $l$ units.

Let the breadth of the rectangle be $b$ units.

The original area is $A = l \times b$.

Condition 1: Length reduced by 5, breadth increased by 3, area reduced by 9.

New length = $l - 5$

New breadth = $b + 3$

New area = $(l - 5)(b + 3)$

New area = Original area - 9

$(l - 5)(b + 3) = lb - 9$

$lb + 3l - 5b - 15 = lb - 9$

$3l - 5b = -9 + 15$

Condition 2: Length increased by 3, breadth increased by 2, area increased by 67.

New length = $l + 3$

New breadth = $b + 2$

New area = $(l + 3)(b + 2)$

New area = Original area + 67

$(l + 3)(b + 2) = lb + 67$

$lb + 2l + 3b + 6 = lb + 67$

$2l + 3b = 67 - 6$

Solution (using Elimination Method):

Multiply equation (1) by 2 and equation (2) by 3 to eliminate $l$.

Subtract equation (3) from equation (4):

$(6l + 9b) - (6l - 10b) = 183 - 12$

$6l + 9b - 6l + 10b = 171$

$19b = 171$

$b = \frac{171}{19}$

$b = 9$

Substitute the value of $b = 9$ into equation (2):

$2l + 3(9) = 61$

$2l + 27 = 61$

$2l = 61 - 27$

$2l = 34$

$l = 17$

Result:

The dimensions of the rectangle are length = 17 units and breadth = 9 units.

Given Equations:

These equations are not linear in $x$ and $y$. We can reduce them to a linear form.

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Reduction to Linear Form:

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

Substituting these into equations (i) and (ii), we get:

Now we have a pair of linear equations in $u$ and $v$.

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Solving for $u$ and $v$ (using Elimination Method):

Multiply equation (iii) by 4 and equation (iv) by 3 to eliminate $v$:

$4 \times (2u + 3v = 13) \implies 8u + 12v = 52$

$3 \times (5u - 4v = -2) \implies 15u - 12v = -6$

Add equation (v) and equation (vi):

$(8u + 12v) + (15u - 12v) = 52 + (-6)$

$23u = 46$

$u = \frac{46}{23}$

$u = 2$

Substitute the value $u = 2$ into equation (iii):

$2(2) + 3v = 13$

$4 + 3v = 13$

$3v = 13 - 4$

$3v = 9$

$v = \frac{9}{3}$

$v = 3$

---

Finding $x$ and $y$:

Now substitute back the values of $u$ and $v$:

$u = \frac{1}{x} \implies 2 = \frac{1}{x} \implies x = \frac{1}{2}$

$v = \frac{1}{y} \implies 3 = \frac{1}{y} \implies y = \frac{1}{3}$

---

Result:

The solution to the given pair of equations is $x = \frac{1}{2}$ and $y = \frac{1}{3}$.

Given Equations:

These equations are not linear in $x$ and $y$. We need to reduce them to a linear form.

---

Reduction to Linear Form:

Let $u = \frac{1}{x - 1}$ and $v = \frac{1}{y - 2}$.

Substituting these into equations (i) and (ii), we get a pair of linear equations in $u$ and $v$:

---

Solving for $u$ and $v$ (using Elimination Method):

Multiply equation (iii) by 3 to make the coefficients of $v$ equal in magnitude:

$3 \times (5u + v = 2) \implies 15u + 3v = 6$

Now, add equation (iv) and equation (v):

$(6u - 3v) + (15u + 3v) = 1 + 6$

$21u = 7$

$u = \frac{7}{21}$

$u = \frac{1}{3}$

Substitute the value $u = \frac{1}{3}$ into equation (iii):

$5 \left( \frac{1}{3} \right) + v = 2$

$\frac{5}{3} + v = 2$

$v = 2 - \frac{5}{3}$

$v = \frac{6 - 5}{3}$

$v = \frac{1}{3}$

---

Finding $x$ and $y$:

Now substitute back the original definitions of $u$ and $v$:

$u = \frac{1}{x - 1} \implies \frac{1}{3} = \frac{1}{x - 1}$

Cross-multiplying gives:

$x - 1 = 3$

$x = 3 + 1$

$x = 4$

$v = \frac{1}{y - 2} \implies \frac{1}{3} = \frac{1}{y - 2}$

Cross-multiplying gives:

$y - 2 = 3$

$y = 3 + 2$

$y = 5$

---

Result:

The solution to the given pair of equations is $x = 4$ and $y = 5$.

Formulation:

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

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

Then, the speed of the boat downstream = $(x + y)$ km/h.

And the speed of the boat upstream = $(x - y)$ km/h.

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

Scenario 1: 30 km upstream and 44 km downstream in 10 hours.

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

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

Total time = 10 hours.

Scenario 2: 40 km upstream and 55 km downstream in 13 hours.

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

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

Total time = 13 hours.

---

Reduction to Linear Form:

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

Substituting these into equations (i) and (ii), we get:

---

Solving for $u$ and $v$ (using Elimination Method):

Multiply equation (iii) by 4 and equation (iv) by 3 to make the coefficients of $u$ equal.

$4 \times (30u + 44v = 10) \implies 120u + 176v = 40$

$3 \times (40u + 55v = 13) \implies 120u + 165v = 39$

Subtract equation (vi) from equation (v):

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

$11v = 1$

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

Substitute $v = \frac{1}{11}$ into equation (iii):

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

$30u + 4 = 10$

$30u = 10 - 4$

$30u = 6$

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

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

---

Finding $x$ and $y$:

Now substitute back the original definitions of $u$ and $v$:

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

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

Add equations (vii) and (viii):

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

$2x = 16$

$x = 8$

Substitute $x = 8$ into equation (viii):

$8 + y = 11$

$y = 11 - 8$

$y = 3$

---

Result:

The speed of the boat in still water is 8 km/h.

The speed of the stream is 3 km/h.

(i) $\frac{1}{2x} + \frac{1}{3y} = 2$ and $\frac{1}{3x} + \frac{1}{2y} = \frac{13}{6}$

Let $u = \frac{1}{x}$ and $v = \frac{1}{y}$. The equations become:

$\frac{u}{2} + \frac{v}{3} = 2 \implies 3u + 2v = 12$ ...(1)

$\frac{u}{3} + \frac{v}{2} = \frac{13}{6} \implies 2u + 3v = 13$ ...(2)

Multiply equation (1) by 2 and equation (2) by 3:

$6u + 4v = 24$ ...(3)

$6u + 9v = 39$ ...(4)

Subtract equation (3) from equation (4):

$(6u + 9v) - (6u + 4v) = 39 - 24$

$5v = 15 \implies v = 3$

Substitute $v = 3$ into equation (1):

$3u + 2(3) = 12 \implies 3u + 6 = 12 \implies 3u = 6 \implies u = 2$

Since $u = \frac{1}{x}$, $2 = \frac{1}{x} \implies x = \frac{1}{2}$.

Since $v = \frac{1}{y}$, $3 = \frac{1}{y} \implies y = \frac{1}{3}$.

Solution: $x = \frac{1}{2}, y = \frac{1}{3}$.

---

(ii) $\frac{2}{\sqrt{x}} + \frac{3}{\sqrt{y}} = 2$ and $\frac{4}{\sqrt{x}} - \frac{9}{\sqrt{y}} = -1$

Let $u = \frac{1}{\sqrt{x}}$ and $v = \frac{1}{\sqrt{y}}$. The equations become:

$2u + 3v = 2$ ...(1)

$4u - 9v = -1$ ...(2)

Multiply equation (1) by 3:

$6u + 9v = 6$ ...(3)

Add equation (2) and equation (3):

$(4u - 9v) + (6u + 9v) = -1 + 6$

$10u = 5 \implies u = \frac{1}{2}$

Substitute $u = \frac{1}{2}$ into equation (1):

$2(\frac{1}{2}) + 3v = 2 \implies 1 + 3v = 2 \implies 3v = 1 \implies v = \frac{1}{3}$

Since $u = \frac{1}{\sqrt{x}}$, $\frac{1}{2} = \frac{1}{\sqrt{x}} \implies \sqrt{x} = 2 \implies x = 4$.

Since $v = \frac{1}{\sqrt{y}}$, $\frac{1}{3} = \frac{1}{\sqrt{y}} \implies \sqrt{y} = 3 \implies y = 9$.

Solution: $x = 4, y = 9$.

---

(iii) $\frac{4}{x} + 3y = 14$ and $\frac{3}{x} - 4y = 23$

Let $u = \frac{1}{x}$. The equations become:

$4u + 3y = 14$ ...(1)

$3u - 4y = 23$ ...(2)

Multiply equation (1) by 4 and equation (2) by 3:

$16u + 12y = 56$ ...(3)

$9u - 12y = 69$ ...(4)

Add equation (3) and equation (4):

$(16u + 12y) + (9u - 12y) = 56 + 69$

$25u = 125 \implies u = 5$

Substitute $u = 5$ into equation (1):

$4(5) + 3y = 14 \implies 20 + 3y = 14 \implies 3y = -6 \implies y = -2$

Since $u = \frac{1}{x}$, $5 = \frac{1}{x} \implies x = \frac{1}{5}$.

Solution: $x = \frac{1}{5}, y = -2$.

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(iv) $\frac{5}{x - 1} + \frac{1}{y - 2} = 2$ and $\frac{6}{x - 1} - \frac{3}{y - 2} = 1$

Let $u = \frac{1}{x - 1}$ and $v = \frac{1}{y - 2}$. The equations become:

$5u + v = 2$ ...(1)

$6u - 3v = 1$ ...(2)

Multiply equation (1) by 3:

$15u + 3v = 6$ ...(3)

Add equation (2) and equation (3):

$(6u - 3v) + (15u + 3v) = 1 + 6$

$21u = 7 \implies u = \frac{1}{3}$

Substitute $u = \frac{1}{3}$ into equation (1):

$5(\frac{1}{3}) + v = 2 \implies \frac{5}{3} + v = 2 \implies v = 2 - \frac{5}{3} = \frac{1}{3}$

Since $u = \frac{1}{x - 1}$, $\frac{1}{3} = \frac{1}{x - 1} \implies x - 1 = 3 \implies x = 4$.

Since $v = \frac{1}{y - 2}$, $\frac{1}{3} = \frac{1}{y - 2} \implies y - 2 = 3 \implies y = 5$.

Solution: $x = 4, y = 5$.

---

(v) $\frac{7x - 2y}{xy} = 5$ and $\frac{8x + 7y}{xy} = 15$

Rewrite the equations (assuming $x \neq 0, y \neq 0$):

$\frac{7x}{xy} - \frac{2y}{xy} = 5 \implies \frac{7}{y} - \frac{2}{x} = 5$

$\frac{8x}{xy} + \frac{7y}{xy} = 15 \implies \frac{8}{y} + \frac{7}{x} = 15$

Let $u = \frac{1}{x}$ and $v = \frac{1}{y}$. The equations become:

$7v - 2u = 5 \implies -2u + 7v = 5$ ...(1)

$8v + 7u = 15 \implies 7u + 8v = 15$ ...(2)

Multiply equation (1) by 7 and equation (2) by 2:

$-14u + 49v = 35$ ...(3)

$14u + 16v = 30$ ...(4)

Add equation (3) and equation (4):

$(-14u + 49v) + (14u + 16v) = 35 + 30$

$65v = 65 \implies v = 1$

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

$7u + 8(1) = 15 \implies 7u + 8 = 15 \implies 7u = 7 \implies u = 1$

Since $u = \frac{1}{x}$, $1 = \frac{1}{x} \implies x = 1$.

Since $v = \frac{1}{y}$, $1 = \frac{1}{y} \implies y = 1$.

Solution: $x = 1, y = 1$.

---

(vi) $6x + 3y = 6xy$ and $2x + 4y = 5xy$

Assuming $x \neq 0, y \neq 0$, divide both equations by $xy$:

$\frac{6x}{xy} + \frac{3y}{xy} = 6 \implies \frac{6}{y} + \frac{3}{x} = 6$

$\frac{2x}{xy} + \frac{4y}{xy} = 5 \implies \frac{2}{y} + \frac{4}{x} = 5$

Let $u = \frac{1}{x}$ and $v = \frac{1}{y}$. The equations become:

$3u + 6v = 6 \implies u + 2v = 2$ ...(1)

$4u + 2v = 5$ ...(2)

Subtract equation (1) from equation (2):

$(4u + 2v) - (u + 2v) = 5 - 2$

$3u = 3 \implies u = 1$

Substitute $u = 1$ into equation (1):

$1 + 2v = 2 \implies 2v = 1 \implies v = \frac{1}{2}$

Since $u = \frac{1}{x}$, $1 = \frac{1}{x} \implies x = 1$.

Since $v = \frac{1}{y}$, $\frac{1}{2} = \frac{1}{y} \implies y = 2$.

Also check if $x=0, y=0$ is a solution. $6(0)+3(0)=6(0)(0) \implies 0=0$. $2(0)+4(0)=5(0)(0) \implies 0=0$. So $x=0, y=0$ is also a solution. However, the method of dividing by $xy$ assumes $x,y \neq 0$. The non-zero solution is sought.

Non-zero Solution: $x = 1, y = 2$.

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(vii) $\frac{10}{x + y} + \frac{2}{x - y} = 4$ and $\frac{15}{x + y} - \frac{5}{x - y} = -2$

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

$10u + 2v = 4 \implies 5u + v = 2$ ...(1)

$15u - 5v = -2$ ...(2)

From equation (1), $v = 2 - 5u$. Substitute this into equation (2):

$15u - 5(2 - 5u) = -2$

$15u - 10 + 25u = -2$

$40u = 8 \implies u = \frac{8}{40} = \frac{1}{5}$

Substitute $u = \frac{1}{5}$ back into $v = 2 - 5u$:

$v = 2 - 5(\frac{1}{5}) = 2 - 1 = 1$

Since $u = \frac{1}{x + y}$, $\frac{1}{5} = \frac{1}{x + y} \implies x + y = 5$ ...(3)

Since $v = \frac{1}{x - y}$, $1 = \frac{1}{x - y} \implies x - y = 1$ ...(4)

Add equations (3) and (4):

$(x + y) + (x - y) = 5 + 1$

$2x = 6 \implies x = 3$

Substitute $x = 3$ into equation (3):

$3 + y = 5 \implies y = 2$

Solution: $x = 3, y = 2$.

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(viii) $\frac{1}{3x + y} + \frac{1}{3x - y} = \frac{3}{4}$ and $\frac{1}{2(3x + y)} - \frac{1}{2(3x - y)} = -\frac{1}{8}$

Let $u = \frac{1}{3x + y}$ and $v = \frac{1}{3x - y}$. The equations become:

$u + v = \frac{3}{4}$ ...(1)

$\frac{u}{2} - \frac{v}{2} = -\frac{1}{8}$

Multiply the second equation by 2: $u - v = -\frac{1}{4}$ ...(2)

Add equation (1) and equation (2):

$(u + v) + (u - v) = \frac{3}{4} + (-\frac{1}{4})$

$2u = \frac{2}{4} = \frac{1}{2} \implies u = \frac{1}{4}$

Substitute $u = \frac{1}{4}$ into equation (1):

$\frac{1}{4} + v = \frac{3}{4} \implies v = \frac{3}{4} - \frac{1}{4} = \frac{2}{4} = \frac{1}{2}$

Since $u = \frac{1}{3x + y}$, $\frac{1}{4} = \frac{1}{3x + y} \implies 3x + y = 4$ ...(3)

Since $v = \frac{1}{3x - y}$, $\frac{1}{2} = \frac{1}{3x - y} \implies 3x - y = 2$ ...(4)

Add equations (3) and (4):

$(3x + y) + (3x - y) = 4 + 2$

$6x = 6 \implies x = 1$

Substitute $x = 1$ into equation (3):

$3(1) + y = 4 \implies 3 + y = 4 \implies y = 1$

Solution: $x = 1, y = 1$.

(i) Ritu's Rowing Speed

Formulation:

Let the speed of Ritu's rowing in still water be $x$ km/h.

Let the speed of the current be $y$ km/h.

Speed downstream = $(x + y)$ km/h.

Speed upstream = $(x - y)$ km/h.

We use the formula: Time = $\frac{\text{Distance}}{\text{Speed}}$.

Condition 1: Downstream 20 km in 2 hours.

$2 = \frac{20}{x + y} \implies x + y = \frac{20}{2}$

Condition 2: Upstream 4 km in 2 hours.

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

Solution (using Elimination Method):

Add equation (1) and equation (2):

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

$2x = 12$

$x = 6$

Substitute the value of $x = 6$ into equation (1):

$6 + y = 10$

$y = 4$

Result:

Ritu's speed of rowing in still water is 6 km/h and the speed of the current is 4 km/h.

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(ii) Embroidery Work

Formulation:

Let the time taken by 1 woman alone to finish the work be $w$ days.

Let the time taken by 1 man alone to finish the work be $m$ days.

Work done by 1 woman in 1 day = $\frac{1}{w}$.

Work done by 1 man in 1 day = $\frac{1}{m}$.

Condition 1: 2 women and 5 men finish the work in 4 days.

Work done by 2 women and 5 men in 1 day = $\frac{1}{4}$.

$2 \left(\frac{1}{w}\right) + 5 \left(\frac{1}{m}\right) = \frac{1}{4}$

Condition 2: 3 women and 6 men finish the work in 3 days.

Work done by 3 women and 6 men in 1 day = $\frac{1}{3}$.

$3 \left(\frac{1}{w}\right) + 6 \left(\frac{1}{m}\right) = \frac{1}{3}$

Let $u = \frac{1}{w}$ and $v = \frac{1}{m}$. The equations become:

Solution (using Elimination Method):

Multiply equation (1) by 9 and equation (2) by 8:

Subtract equation (4) from equation (3):

$(72u + 180v) - (72u + 144v) = 9 - 8$

$36v = 1 \implies v = \frac{1}{36}$

Substitute $v = \frac{1}{36}$ into equation (1):

$8u + 20 \left(\frac{1}{36}\right) = 1$

$8u + \frac{5}{9} = 1$

$8u = 1 - \frac{5}{9} = \frac{4}{9}$

$u = \frac{4}{9 \times 8} = \frac{1}{18}$

Since $u = \frac{1}{w}$, $w = \frac{1}{u} = 18$.

Since $v = \frac{1}{m}$, $m = \frac{1}{v} = 36$.

Result:

The time taken by 1 woman alone is 18 days.

The time taken by 1 man alone is 36 days.

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(iii) Roohi's Travel

Formulation:

Let the speed of the train be $t$ km/h.

Let the speed of the bus be $b$ km/h.

Total distance = 300 km.

We use the formula: Time = $\frac{\text{Distance}}{\text{Speed}}$.

Condition 1: 60 km by train and (300 - 60 = 240) km by bus takes 4 hours.

Condition 2: 100 km by train and (300 - 100 = 200) km by bus takes 4 hours and 10 minutes.

10 minutes = $\frac{10}{60} = \frac{1}{6}$ hours.

Total time = $4 + \frac{1}{6} = \frac{25}{6}$ hours.

Let $u = \frac{1}{t}$ and $v = \frac{1}{b}$. The equations become:

$60u + 240v = 4$

Divide by 4: $15u + 60v = 1$ ...(1)

$100u + 200v = \frac{25}{6}$

Multiply by 6: $600u + 1200v = 25$

Divide by 25: $24u + 48v = 1$ ...(2)

Solution (using Substitution Method):

From equation (1): $15u = 1 - 60v \implies u = \frac{1 - 60v}{15}$.

Substitute this into equation (2):

$24 \left( \frac{1 - 60v}{15} \right) + 48v = 1$

Simplify $\frac{24}{15} = \frac{8}{5}$:

$\frac{8(1 - 60v)}{5} + 48v = 1$

Multiply by 5:

$8(1 - 60v) + 240v = 5$

$8 - 480v + 240v = 5$

$8 - 240v = 5$

$3 = 240v$

$v = \frac{3}{240} = \frac{1}{80}$

Substitute $v = \frac{1}{80}$ into equation (1):

$15u + 60 \left( \frac{1}{80} \right) = 1$

$15u + \frac{6}{8} = 1$

$15u + \frac{3}{4} = 1$

$15u = 1 - \frac{3}{4} = \frac{1}{4}$

$u = \frac{1}{4 \times 15} = \frac{1}{60}$

Since $u = \frac{1}{t}$, $t = \frac{1}{u} = 60$.

Since $v = \frac{1}{b}$, $b = \frac{1}{v} = 80$.

Result:

The speed of the train is 60 km/h and the speed of the bus is 80 km/h.

Formulation:

Let the age of Ani be $x$ years.

Let the age of Biju be $y$ years.

The ages of Ani and Biju differ by 3 years. This gives two possible cases:

Case 1: Ani is older than Biju.

Case 2: Biju is older than Ani.

Ani's father Dharam's age = $2 \times$ Ani's age = $2x$ years.

Biju is twice as old as his sister Cathy, so Cathy's age = $\frac{\text{Biju's age}}{2} = \frac{y}{2}$ years.

The ages of Cathy and Dharam differ by 30 years. Since Dharam is a father and Cathy is a sister, it is reasonable to assume Dharam is older than Cathy.

Dharam's age - Cathy's age = 30

$2x - \frac{y}{2} = 30$

Multiply by 2 to clear the fraction:

We need to solve the system for both cases.

---

Solution for Case 1:

Using equations (i) and (iii):

Subtract equation (i) from equation (iii):

$(4x - y) - (x - y) = 60 - 3$

$4x - y - x + y = 57$

$3x = 57$

$x = \frac{57}{3}$

$x = 19$

Substitute $x = 19$ into equation (i):

$19 - y = 3$

$y = 19 - 3$

$y = 16$

In this case, Ani's age is 19 years and Biju's age is 16 years.

---

Solution for Case 2:

Using equations (ii) and (iii):

Add equation (ii) and equation (iii):

$(-x + y) + (4x - y) = 3 + 60$

$3x = 63$

$x = \frac{63}{3}$

$x = 21$

Substitute $x = 21$ into equation (ii):

$y - 21 = 3$

$y = 24$

In this case, Ani's age is 21 years and Biju's age is 24 years.

---

Result:

There are two possible solutions:

1. Ani's age is 19 years and Biju's age is 16 years.

2. Ani's age is 21 years and Biju's age is 24 years.

Formulation:

Let the capital of the first friend be $\textsf{₹ } x$.

Let the capital of the second friend be $\textsf{₹ } y$.

First Statement: "Give me a hundred, friend! I shall then become twice as rich as you".

If the second friend gives $\textsf{₹ } 100$ to the first friend:

First friend's capital = $x + 100$

Second friend's capital = $y - 100$

According to the statement: $x + 100 = 2(y - 100)$

$x + 100 = 2y - 200$

$x - 2y = -200 - 100$

Second Statement: "If you give me ten, I shall be six times as rich as you".

If the first friend gives $\textsf{₹ } 10$ to the second friend:

First friend's capital = $x - 10$

Second friend's capital = $y + 10$

According to the statement: $y + 10 = 6(x - 10)$

$y + 10 = 6x - 60$

$-6x + y = -60 - 10$

---

Solution (using Substitution Method):

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

Substitute this expression for $y$ into equation (1):

$x - 2(6x - 70) = -300$

$x - 12x + 140 = -300$

$-11x = -300 - 140$

$-11x = -440$

$x = \frac{-440}{-11}$

$x = 40$

Substitute the value of $x = 40$ back into equation (3):

$y = 6(40) - 70$

$y = 240 - 70$

$y = 170$

---

Result:

The capital of the first friend is $\textsf{₹ } 40$.

The capital of the second friend is $\textsf{₹ } 170$.

Formulation:

Let the uniform speed of the train be $s$ km/h.

Let the scheduled time taken be $t$ hours.

Let the distance covered be $d$ km.

We know the relationship: Distance = Speed $\times$ Time

$d = s \times t$

Condition 1: Speed is $(s + 10)$ km/h, Time is $(t - 2)$ hours.

Distance $d = (s + 10)(t - 2)$

Using (*), $st = (s + 10)(t - 2)$

$st = st - 2s + 10t - 20$

$0 = -2s + 10t - 20$

$2s - 10t = -20$

Divide by 2:

Condition 2: Speed is $(s - 10)$ km/h, Time is $(t + 3)$ hours.

Distance $d = (s - 10)(t + 3)$

Using (*), $st = (s - 10)(t + 3)$

$st = st + 3s - 10t - 30$

$0 = 3s - 10t - 30$

---

Solution (using Elimination Method):

We have the pair of linear equations:

Multiply equation (1) by 2:

Subtract equation (3) from equation (2):

$(3s - 10t) - (2s - 10t) = 30 - (-20)$

$3s - 10t - 2s + 10t = 30 + 20$

$s = 50$

Substitute the value $s = 50$ into equation (1):

$50 - 5t = -10$

$-5t = -10 - 50$

$-5t = -60$

$t = \frac{-60}{-5}$

$t = 12$

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Finding the Distance:

Now, calculate the distance using the formula $d = st$.

$d = 50 \times 12$

$d = 600$

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Result:

The distance covered by the train is 600 km.

Formulation:

Let the number of rows be $r$.

Let the number of students in each row be $s$.

The total number of students in the class = Number of rows $\times$ Number of students per row

Total number of students = $r \times s$.

Condition 1: If 3 students are extra in a row $(s + 3)$, there would be 1 row less $(r - 1)$.

Total number of students = $(r - 1)(s + 3)$.

Since the total number of students remains the same:

$rs = (r - 1)(s + 3)$

$rs = rs + 3r - s - 3$

$0 = 3r - s - 3$

Condition 2: If 3 students are less in a row $(s - 3)$, there would be 2 rows more $(r + 2)$.

Total number of students = $(r + 2)(s - 3)$.

Since the total number of students remains the same:

$rs = (r + 2)(s - 3)$

$rs = rs - 3r + 2s - 6$

$0 = -3r + 2s - 6$

$3r - 2s = -6$

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Solution (using Elimination Method):

We need to solve the following pair of linear equations:

Subtract equation (2) from equation (1):

$(3r - s) - (3r - 2s) = 3 - (-6)$

$3r - s - 3r + 2s = 3 + 6$

$s = 9$

Substitute the value $s = 9$ into equation (1):

$3r - 9 = 3$

$3r = 3 + 9$

$3r = 12$

$r = \frac{12}{3}$

$r = 4$

So, the number of rows is 4 and the number of students in each row is 9.

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Finding the Total Number of Students:

Total number of students = $r \times s$

Total number of students = $4 \times 9$

Total number of students = $36$

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Result:

The number of students in the class is 36.

Given:

In $\triangle ABC$, the relationships between the angles are:

We also know the angle sum property of a triangle:

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To Find:

The measures of the three angles: $\angle A$, $\angle B$, and $\angle C$.

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Solution:

From equations (1) and (2), we have:

$3 \angle B = 2 (\angle A + \angle B)$

$3 \angle B = 2 \angle A + 2 \angle B$

$3 \angle B - 2 \angle B = 2 \angle A$

$\angle B = 2 \angle A$

This implies:

Now, substitute equations (1) and (4) into the angle sum property (equation (3)):

$\angle A + \angle B + \angle C = 180^\circ$

$\left( \frac{\angle B}{2} \right) + \angle B + (3 \angle B) = 180^\circ$

Multiply by 2 to clear the fraction:

$\angle B + 2 \angle B + 6 \angle B = 360^\circ$

$9 \angle B = 360^\circ$

$\angle B = \frac{360^\circ}{9}$

$\angle B = 40^\circ$

Now find $\angle C$ using equation (1):

$\angle C = 3 \angle B = 3 \times 40^\circ$

$\angle C = 120^\circ$

Find $\angle A$ using equation (4):

$\angle A = \frac{\angle B}{2} = \frac{40^\circ}{2}$

$\angle A = 20^\circ$

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Result:

The three angles of the triangle are:

$\angle A = 20^\circ$

$\angle B = 40^\circ$

$\angle C = 120^\circ$

Given Equations:

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Finding Points for Graphing:

To draw the graphs, we need coordinates of points satisfying each equation.

For equation (i): $5x - y = 5 \implies y = 5x - 5$

x	y = 5x - 5	Point
0	-5	A(0, -5)
1	0	B(1, 0)
2	5	(2, 5)

Point A(0, -5) is the intersection of line (i) with the y-axis.

For equation (ii): $3x - y = 3 \implies y = 3x - 3$

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

Point C(0, -3) is the intersection of line (ii) with the y-axis.

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Graphical Representation and Triangle Vertices:

Plot the points for each equation on graph paper and draw the lines.

Line 1 ($y = 5x - 5$) passes through A(0, -5) and B(1, 0).

Line 2 ($y = 3x - 3$) passes through C(0, -3) and B(1, 0).

(Note: Graphical plot needs to be done on graph paper. The description below explains the result.)

The graph shows the two lines intersecting at point B(1, 0).

The triangle is formed by these two lines and the y-axis (the line $x=0$).

The vertices of this triangle are:

• The point of intersection of the two lines: B(1, 0).
• The point where the first line ($5x - y = 5$) intersects the y-axis: A(0, -5).
• The point where the second line ($3x - y = 3$) intersects the y-axis: C(0, -3).

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Result:

The coordinates of the vertices of the triangle formed by the lines $5x - y = 5$, $3x - y = 3$, and the y-axis are (1, 0), (0, -5), and (0, -3).

(i) $px + qy = p - q$ and $qx - py = p + q$

Given equations:

Using the Elimination Method:

Multiply equation (1) by $p$ and equation (2) by $q$:

Add equation (3) and equation (4):

$(p^2x + pqy) + (q^2x - pqy) = (p^2 - pq) + (pq + q^2)$

$p^2x + q^2x = p^2 + q^2$

$(p^2 + q^2)x = p^2 + q^2$

Assuming $p^2 + q^2 \neq 0$, we get $x = 1$.

Substitute $x = 1$ into equation (1):

$p(1) + qy = p - q$

$p + qy = p - q$

$qy = -q$

Assuming $q \neq 0$, we get $y = -1$.

Solution: $x = 1, y = -1$.

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(ii) $ax + by = c$ and $bx + ay = 1 + c$

Given equations:

Using the Elimination Method:

Multiply equation (1) by $a$ and equation (2) by $b$:

Subtract equation (4) from equation (3):

$(a^2x + aby) - (b^2x + aby) = ac - (b + bc)$

$a^2x - b^2x = ac - b - bc$

$(a^2 - b^2)x = c(a - b) - b$

Assuming $a^2 \neq b^2$, $x = \frac{c(a - b) - b}{a^2 - b^2}$.

Multiply equation (1) by $b$ and equation (2) by $a$:

Subtract equation (5) from equation (6):

$(abx + a^2y) - (abx + b^2y) = (a + ac) - bc$

$a^2y - b^2y = a + ac - bc$

$(a^2 - b^2)y = a + c(a - b)$

Assuming $a^2 \neq b^2$, $y = \frac{a + c(a - b)}{a^2 - b^2}$.

Solution: $x = \frac{c(a - b) - b}{a^2 - b^2}, y = \frac{a + c(a - b)}{a^2 - b^2}$.

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(iii) $\frac{x}{a} - \frac{y}{b} = 0$ and $ax + by = a^2 + b^2$

Given equations:

Using the Substitution Method:

From equation (1): $\frac{x}{a} = \frac{y}{b} \implies x = \frac{ay}{b}$.

Substitute this expression for $x$ into equation (2):

$a \left( \frac{ay}{b} \right) + by = a^2 + b^2$

$\frac{a^2y}{b} + by = a^2 + b^2$

Multiply by $b$ to clear the fraction:

$a^2y + b^2y = b(a^2 + b^2)$

$(a^2 + b^2)y = b(a^2 + b^2)$

Assuming $a^2 + b^2 \neq 0$, we get $y = b$.

Substitute $y = b$ back into $x = \frac{ay}{b}$:

$x = \frac{a(b)}{b}$

$x = a$

Solution: $x = a, y = b$.

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(iv) $(a - b)x + (a + b)y = a^2 - 2ab - b^2$ and $(a + b)(x + y) = a^2 + b^2$

Given equations:

Expand the second equation:

Using the Elimination Method:

Subtract equation (1) from equation (2):

$[(a + b)x + (a + b)y] - [(a - b)x + (a + b)y] = (a^2 + b^2) - (a^2 - 2ab - b^2)$

$(a + b)x - (a - b)x = a^2 + b^2 - a^2 + 2ab + b^2$

$(a + b - a + b)x = 2b^2 + 2ab$

$2bx = 2b(b + a)$

Assuming $b \neq 0$, we get $x = a + b$.

Substitute $x = a + b$ into equation (2):

$(a + b)(a + b) + (a + b)y = a^2 + b^2$

$(a + b)^2 + (a + b)y = a^2 + b^2$

$a^2 + 2ab + b^2 + (a + b)y = a^2 + b^2$

$2ab + (a + b)y = 0$

$(a + b)y = -2ab$

Assuming $a + b \neq 0$, we get $y = \frac{-2ab}{a + b}$.

Solution: $x = a + b, y = \frac{-2ab}{a + b}$.

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(v) $152x - 378y = -74$ and $-378x + 152y = -604$

Given equations:

Using the method of adding and subtracting equations:

Add equation (1) and equation (2):

$(152 - 378)x + (-378 + 152)y = -74 - 604$

$-226x - 226y = -678$

Divide by -226:

Subtract equation (2) from equation (1):

$(152 - (-378))x + (-378 - 152)y = -74 - (-604)$

$(152 + 378)x + (-530)y = -74 + 604$

$530x - 530y = 530$

Divide by 530:

Now solve the simpler system of equations (3) and (4).

Add equation (3) and equation (4):

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

$2x = 4 \implies x = 2$

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

$2 + y = 3 \implies y = 1$

Solution: $x = 2, y = 1$.

Given:

A cyclic quadrilateral ABCD with angles:

$\angle A = 4y + 20$

$\angle B = 3y - 5$

$\angle C = -4x$

$\angle D = -7x + 5$

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Property of Cyclic Quadrilateral:

We know that the sum of opposite angles in a cyclic quadrilateral is $180^\circ$.

Therefore,

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Formulating Equations:

Using equation (i):

$(4y + 20) + (-4x) = 180$

$4y + 20 - 4x = 180$

$-4x + 4y = 180 - 20$

$-4x + 4y = 160$

Divide by 4:

Using equation (ii):

$(3y - 5) + (-7x + 5) = 180$

$3y - 5 - 7x + 5 = 180$

$-7x + 3y = 180$

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Solving the Equations:

We have the pair of linear equations:

$-x + y = 40$ ...(iii)

$-7x + 3y = 180$ ...(iv)

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

$y = x + 40$

Substitute this value of $y$ into equation (iv):

$-7x + 3(x + 40) = 180$

$-7x + 3x + 120 = 180$

$-4x = 180 - 120$

$-4x = 60$

$x = \frac{60}{-4}$

$x = -15$

Substitute the value of $x = -15$ back into $y = x + 40$:

$y = -15 + 40$

$y = 25$

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Calculating the Angles:

Now substitute the values of $x$ and $y$ into the expressions for the angles:

$\angle A = 4y + 20 = 4(25) + 20 = 100 + 20 = 120^\circ$

$\angle B = 3y - 5 = 3(25) - 5 = 75 - 5 = 70^\circ$

$\angle C = -4x = -4(-15) = 60^\circ$

$\angle D = -7x + 5 = -7(-15) + 5 = 105 + 5 = 110^\circ$

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Result:

The angles of the cyclic quadrilateral are:

$\angle A = 120^\circ$

$\angle B = 70^\circ$

$\angle C = 60^\circ$

$\angle D = 110^\circ$