Chapter 4 Linear Equations In Two Variables

The linear equation is $3x - y = x - 1$.

The number of solutions for the given linear equation.

The given linear equation is:

$3x - y = x - 1$

To simplify the equation, we can rearrange the terms. Let's bring all the terms involving $x$ and $y$ to one side and constants to the other side.

$3x - x - y = -1$

Combine the terms involving $x$:

$(3x - x) - y = -1$

$2x - y = -1$

This is a linear equation in two variables, $x$ and $y$. It can be written in the standard form $Ax + By + C = 0$ as $2x - y + 1 = 0$, where $A=2$, $B=-1$, and $C=1$. Since $A$ and $B$ are not both zero, this represents a straight line in the Cartesian plane.

A single linear equation in two variables always has infinitely many solutions. Each solution corresponds to a point $(x, y)$ that lies on the line represented by the equation.

For example, we can find some solutions:

If $x = 0$, then $2(0) - y = -1 \implies -y = -1 \implies y = 1$. So $(0, 1)$ is a solution.

If $x = 1$, then $2(1) - y = -1 \implies 2 - y = -1 \implies -y = -1 - 2 \implies -y = -3 \implies y = 3$. So $(1, 3)$ is a solution.

If $y = 0$, then $2x - 0 = -1 \implies 2x = -1 \implies x = -\frac{1}{2}$. So $(-\frac{1}{2}, 0)$ is a solution.

Since we can choose any real value for $x$ and find a corresponding real value for $y$ that satisfies the equation (or vice versa), there are infinitely many such pairs $(x, y)$.

Therefore, the linear equation $3x - y = x - 1$ has infinitely many solutions.

The correct option is (C) Infinitely many solutions.

The form of a linear equation in two variables is $ax + by + c = 0$.

The correct condition on the coefficients $a$, $b$, and $c$ for the given form to be a linear equation in two variables.

A linear equation in two variables, say $x$ and $y$, is defined as an equation that can be written in the form $ax + by + c = 0$, where $a$, $b$, and $c$ are real numbers, and the coefficients of the variables, $a$ and $b$, are not both zero.

The condition that $a$ and $b$ are not both zero means that at least one of them must be non-zero. This can be expressed as $a \neq 0$ or $b \neq 0$. An equivalent way to state this condition is $a^2 + b^2 \neq 0$.

Let's consider the given options:

(A) $a \neq 0, b \neq 0$: This condition implies that both $a$ and $b$ are non-zero. If this is true, the equation $ax+by+c=0$ is indeed a linear equation in two variables. This represents lines that are not parallel to the coordinate axes.

(B) $a = 0, b \neq 0$: This condition implies $a$ is zero and $b$ is non-zero. The equation becomes $by + c = 0$. Since $b \neq 0$, this is a linear equation in two variables (specifically, representing a horizontal line). This case is included in the definition ($a$ and $b$ are not both zero).

(C) $a \neq 0, b = 0$: This condition implies $a$ is non-zero and $b$ is zero. The equation becomes $ax + c = 0$. Since $a \neq 0$, this is a linear equation in two variables (specifically, representing a vertical line). This case is also included in the definition ($a$ and $b$ are not both zero).

(D) $a = 0, c = 0$: This condition implies $a$ is zero and $c$ is zero. The equation becomes $by = 0$. This is a linear equation in two variables only if $b \neq 0$. If $b = 0$ as well, the equation becomes $0 = 0$, which is not considered a linear equation in two variables as it holds for all $x$ and $y$ and does not define a line.

The standard definition requires $a \neq 0$ or $b \neq 0$. None of the options (A), (B), or (C) represent this complete condition, as each only describes a subset of linear equations in two variables.

However, the question asks for "where" the form is a linear equation in two variables. This implies the condition that must hold. Given the options provided, it is possible that the question is specifically referring to the case where both variables $x$ and $y$ have non-zero coefficients, or the options are intended to test understanding of different cases.

Assuming the question is asking for the condition that ensures the equation is undeniably a linear equation with explicit dependence on both variables (i.e., not a special case of a horizontal or vertical line), option (A) $a \neq 0, b \neq 0$ is the most likely intended answer among the choices provided, although it does not encompass all linear equations in two variables according to the strict definition.

Based on the assumption that the question expects the condition where both coefficients are non-zero:

The correct option is (A) a ≠ 0, b ≠ 0.

We are asked to identify the general form of a point located on the y-axis in a Cartesian coordinate system.

The coordinates $(x, y)$ that represent any point on the y-axis.

In the Cartesian coordinate system, the y-axis is a vertical line where every point on this line has an x-coordinate equal to 0.

A point in the Cartesian plane is represented by an ordered pair $(x, y)$, where $x$ is the x-coordinate and $y$ is the y-coordinate.

For any point that lies on the y-axis, its horizontal distance from the origin (which is the x-coordinate) is always zero.

So, for any point on the y-axis, the x-coordinate is $x = 0$. The y-coordinate ($y$) can be any real number, representing the vertical distance of the point from the origin.

Therefore, any point on the y-axis must have the form $(0, y)$.

Let's examine the given options:

(A) $(x, 0)$: This represents a point on the x-axis.

(B) $(x, y)$: This is the general form for any point in the plane.

(C) $(0, y)$: This represents a point where the x-coordinate is 0, which is characteristic of points on the y-axis.

(D) $(y, y)$: This represents a point where the x-coordinate is equal to the y-coordinate, i.e., points on the line $y=x$.

Based on the analysis, the form of any point on the y-axis is $(0, y)$.

The correct option is (C) (0, y).

The linear equation is $2x - 5y = 7$.

The number of solutions for the given linear equation.

The given equation is $2x - 5y = 7$.

This can be rewritten in the standard form of a linear equation in two variables, $ax + by + c = 0$, as $2x - 5y - 7 = 0$.

In this equation, the coefficients of $x$ and $y$ are $a=2$ and $b=-5$, respectively. Since $a$ and $b$ are not both zero ($2 \neq 0$ and $-5 \neq 0$), the equation is indeed a linear equation in two variables.

A single linear equation in two variables represents a straight line in the Cartesian coordinate plane.

Every point $(x, y)$ that lies on this line is a solution to the equation.

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

For example, if we let $x=0$, we get $2(0) - 5y = 7 \implies -5y = 7 \implies y = -\frac{7}{5}$. So, $(0, -\frac{7}{5})$ is a solution.

If we let $y=0$, we get $2x - 5(0) = 7 \implies 2x = 7 \implies x = \frac{7}{2}$. So, $(\frac{7}{2}, 0)$ is a solution.

We can find infinitely many such pairs $(x, y)$ that satisfy the equation.

Therefore, the linear equation $2x - 5y = 7$ has infinitely many solutions.

The correct option is (C) Infinitely many solutions.

The linear equation is $2x + 5y = 7$.

The condition on $x$ and $y$ such that the equation $2x + 5y = 7$ has a unique solution.

The given equation is a linear equation in two variables $x$ and $y$. The equation is $2x + 5y = 7$.

We need to check the number of solutions when $x$ and $y$ belong to different sets of numbers as given in the options.

Let's analyze each option:

(A) Natural numbers

We are looking for solutions $(x, y)$ where $x \in \{1, 2, 3, \dots\}$ and $y \in \{1, 2, 3, \dots\}$.

The equation is $2x + 5y = 7$.

Since $x \ge 1$ and $y \ge 1$, $2x \ge 2$ and $5y \ge 5$.

Let's test values for $y \ge 1$:

If $y = 1$, the equation becomes:

So, $(x, y) = (1, 1)$ is a solution where both $x$ and $y$ are natural numbers. Let's check if there are other solutions.

If $y \ge 2$, then $5y \ge 5(2) = 10$.

The equation is $2x = 7 - 5y$. If $y \ge 2$, then $2x \le 7 - 10 = -3$.

$2x \le -3 \implies x \le -\frac{3}{2}$.

Since $x$ must be a natural number ($x \ge 1$), $x$ cannot be less than or equal to $-\frac{3}{2}$. Thus, there are no solutions with $y \ge 2$.

If we consider natural numbers including 0 $\{0, 1, 2, ...\}$, if $y=0$, $2x=7$, $x=3.5$ (not natural). If $x=0$, $5y=7$, $y=1.4$ (not natural). So $(1,1)$ is still the only solution in this case too.

Therefore, when $x$ and $y$ are natural numbers, the equation $2x + 5y = 7$ has exactly one solution, which is $(1, 1)$. This is a unique solution.

(B) Positive real numbers

We are looking for solutions $(x, y)$ where $x \in \mathbb{R}, x > 0$ and $y \in \mathbb{R}, y > 0$.

From the equation $2x + 5y = 7$, we can express $y$ in terms of $x$: $5y = 7 - 2x \implies y = \frac{7 - 2x}{5}$.

For $y$ to be a positive real number, we need $y > 0$.

We are also given that $x$ must be a positive real number, so $x > 0$.

Thus, for any value of $x$ in the interval $(0, \frac{7}{2})$, there is a corresponding positive real value of $y$. Since there are infinitely many real numbers between $0$ and $\frac{7}{2}$, there are infinitely many solutions in positive real numbers.

(C) Real numbers

We are looking for solutions $(x, y)$ where $x \in \mathbb{R}$ and $y \in \mathbb{R}$.

A single linear equation in two variables with real coefficients ($a=2, b=5$, not both zero) always represents a straight line in the Cartesian plane. Every point on this line is a solution.

Since there are infinitely many points on a line, there are infinitely many solutions when $x$ and $y$ are real numbers.

(D) Rational numbers

We are looking for solutions $(x, y)$ where $x \in \mathbb{Q}$ and $y \in \mathbb{Q}$.

From the equation $2x + 5y = 7$, we can express $y$ in terms of $x$: $y = \frac{7 - 2x}{5}$.

If we choose any rational number for $x$, say $x = \frac{p}{q}$ (where $p, q$ are integers, $q \neq 0$), then $2x$ is also rational. $7 - 2x$ is rational, and $\frac{7 - 2x}{5}$ is also rational.

For example, if $x=0$ (rational), $y = \frac{7 - 2(0)}{5} = \frac{7}{5}$ (rational). Solution $(0, 7/5)$.

If $x=1/2$ (rational), $y = \frac{7 - 2(1/2)}{5} = \frac{7 - 1}{5} = \frac{6}{5}$ (rational). Solution $(1/2, 6/5)$.

Since there are infinitely many rational numbers, and for each rational $x$ we get a rational $y$ satisfying the equation, there are infinitely many solutions where $x$ and $y$ are rational numbers.

Comparing the results from each option, the equation $2x + 5y = 7$ has a unique solution only when $x$ and $y$ are restricted to be natural numbers.

The correct option is (A) Natural numbers.

The linear equation is $2x + 3y = k$.

The point $(2, 0)$ is a solution to the equation.

The value of $k$.

Since the point $(2, 0)$ is a solution to the linear equation $2x + 3y = k$, it means that when we substitute $x=2$ and $y=0$ into the equation, the equation must be satisfied.

Substitute $x=2$ and $y=0$ into the equation $2x + 3y = k$:

$2(2) + 3(0) = k$

$4 + 0 = k$

$4 = k$

Thus, the value of $k$ is 4.

The correct option is (A) 4.

The linear equation in two variables is $2x + 0y + 9 = 0$.

The general form of any solution $(x, y)$ to the given equation.

The given equation is $2x + 0y + 9 = 0$.

The term $0y$ is equal to 0 for any value of $y$. Therefore, the equation simplifies to:

$2x + 9 = 0$

Now, we can solve this simplified equation for $x$:

$2x = -9$

Divide both sides by 2:

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

This equation tells us that any point $(x, y)$ that is a solution must have an x-coordinate equal to $-\frac{9}{2}$. The y-coordinate ($y$) can be any real number, because the value of $y$ does not affect the validity of the equation $x = -\frac{9}{2}$.

So, a solution $(x, y)$ must satisfy $x = -\frac{9}{2}$ and $y = \text{any real number}$.

If we let $m$ represent any real number, then the general form of a solution is $(-\frac{9}{2}, m)$.

Let's examine the given options to see which one matches this form:

(A) $(-\frac{9}{2}, m)$: This form has the x-coordinate as $-\frac{9}{2}$ and the y-coordinate as $m$, where $m$ can be any real number. This matches our finding.

(B) $(n, -\frac{9}{2})$: This form has the y-coordinate as $-\frac{9}{2}$ and the x-coordinate as $n$ (any real number). This does not match the requirement that $x = -\frac{9}{2}$.

(C) $(0, -\frac{9}{2})$: This form has the x-coordinate as 0, which is not $-\frac{9}{2}$.

(D) $(-9, 0)$: This form has the x-coordinate as $-9$, which is not $-\frac{9}{2}$.

Therefore, the general form of any solution to the equation $2x + 0y + 9 = 0$ is $(-\frac{9}{2}, m)$, where $m$ is any real number.

The correct option is (A) $\left( -\frac{9}{2} , m \right)$.

The linear equation is $2x + 3y = 6$.

The point where the graph of the given equation cuts the y-axis.

A point where the graph of an equation cuts the y-axis is called the y-intercept.

Any point on the y-axis has an x-coordinate equal to 0.

To find the point where the graph of $2x + 3y = 6$ cuts the y-axis, we set $x=0$ in the equation and solve for $y$.

Substitute $x=0$ into the equation $2x + 3y = 6$:

$2(0) + 3y = 6$

$0 + 3y = 6$

$3y = 6$

Divide both sides by 3:

$y = \frac{6}{3}$

$y = 2$

So, when $x=0$, the value of $y$ is 2.

The point where the graph cuts the y-axis is $(x, y) = (0, 2)$.

The correct option is (D) (0, 2).

The equation is $x = 7$.

The representation of the equation $x = 7$ as a linear equation in two variables.

A linear equation in two variables $x$ and $y$ is generally written in the form $ax + by + c = 0$ or $ax + by = c$, where $a$, $b$, and $c$ are real numbers, and $a$ and $b$ are not both zero.

The given equation is $x = 7$. This equation only explicitly involves the variable $x$. To write it as an equation in two variables $x$ and $y$, we need to include the variable $y$ in such a way that its inclusion does not change the equation's meaning.

We can do this by adding a term involving $y$ with a coefficient of zero.

Start with the given equation:

$x = 7$

Subtract 7 from both sides to get the form $ax + by + c = 0$:

$x - 7 = 0$

Now, include the $y$ term with a coefficient of 0:

$1 \cdot x + 0 \cdot y - 7 = 0$

Alternatively, keeping the constant term on the right side ($ax + by = c$ form):

$1 \cdot x + 0 \cdot y = 7$

This equation is in the form $ax + by = c$ with $a=1$, $b=0$, and $c=7$. Since $a \neq 0$, it is a linear equation in two variables.

Let's check the given options:

(A) $1.x + 1.y = 7$ means $x + y = 7$. This is different from $x=7$.

(B) $1.x + 0.y = 7$ means $x + 0 = 7$, which simplifies to $x = 7$. This matches the given equation.

(C) $0.x + 1.y = 7$ means $0 + y = 7$, which simplifies to $y = 7$. This is different from $x=7$.

(D) $0.x + 0.y = 7$ means $0 + 0 = 7$, which simplifies to $0 = 7$. This is a false statement and does not represent the equation $x=7$.

Thus, the equation $x = 7$ written in two variables is $1.x + 0.y = 7$.

The correct option is (B) 1.x + 0.y = 7.

We are asked to identify the general form of a point located on the x-axis in a Cartesian coordinate system.

The coordinates $(x, y)$ that represent any point on the x-axis.

In the Cartesian coordinate system, the x-axis is a horizontal line where every point on this line has a y-coordinate equal to 0.

A point in the Cartesian plane is represented by an ordered pair $(x, y)$, where $x$ is the x-coordinate and $y$ is the y-coordinate.

For any point that lies on the x-axis, its vertical distance from the origin (which is the y-coordinate) is always zero.

So, for any point on the x-axis, the y-coordinate is $y = 0$. The x-coordinate ($x$) can be any real number, representing the horizontal distance of the point from the origin.

Therefore, any point on the x-axis must have the form $(x, 0)$.

Let's examine the given options:

(A) $(x, y)$: This is the general form for any point in the plane, not restricted to the x-axis.

(B) $(0, y)$: This represents a point where the x-coordinate is 0, which is characteristic of points on the y-axis.

(C) $(x, 0)$: This represents a point where the y-coordinate is 0, which is characteristic of points on the x-axis.

(D) $(x, x)$: This represents a point where the x-coordinate is equal to the y-coordinate, i.e., points on the line $y=x$.

Based on the analysis, the form of any point on the x-axis is $(x, 0)$.

The correct option is (C) $\left( x, 0 \right)$.

The equation of the line is $y = x$.

The general form of the coordinates $(x, y)$ for any point on the line $y = x$.

A point $(x, y)$ lies on the line $y = x$ if and only if its coordinates satisfy the equation $y = x$.

This means that for any point on this line, the value of the y-coordinate is equal to the value of the x-coordinate.

Let's consider a general point on the line. If we let the x-coordinate be represented by a variable, say $a$, then according to the equation $y=x$, the y-coordinate must also be equal to $a$.

So, any point on the line $y = x$ can be written in the form $(a, a)$, where $a$ can be any real number.

Let's check the given options:

(A) $(a, a)$: The x-coordinate is $a$ and the y-coordinate is $a$. This satisfies the condition $y=x$.

(B) $(0, a)$: The x-coordinate is 0 and the y-coordinate is $a$. This satisfies the equation $y=x$ only if $a=0$, i.e., the origin $(0,0)$. This is not the general form for *any* point on the line.

(C) $(a, 0)$: The x-coordinate is $a$ and the y-coordinate is 0. This satisfies the equation $y=x$ only if $a=0$, i.e., the origin $(0,0)$. This is not the general form for *any* point on the line.

(D) $(a, -a)$: The x-coordinate is $a$ and the y-coordinate is $-a$. This satisfies the equation $y=x$ only if $a = -a$, which means $2a = 0$, or $a=0$. Again, this only gives the origin, not the general form.

The form that represents any point $(x, y)$ satisfying $y=x$ is $(a, a)$ where $x=a$ and $y=a$.

The correct option is (A) (a, a).

We need to find the equation that represents the x-axis in a two-variable coordinate system.

The equation of the x-axis.

In a Cartesian coordinate system, the x-axis is the horizontal line that passes through the origin $(0, 0)$.

Every point located on the x-axis has a y-coordinate of 0, regardless of its x-coordinate.

For example, the points $(1, 0)$, $(-2, 0)$, $(5.5, 0)$, and $(0, 0)$ all lie on the x-axis. Notice that the y-coordinate is always 0 for these points.

The equation of a line is a condition that is satisfied by the coordinates $(x, y)$ of every point on the line and by no other points.

Since the y-coordinate of every point on the x-axis is 0, the equation that describes the x-axis is $y = 0$.

Let's examine the given options:

(A) $x = 0$: This is the equation of the y-axis, where the x-coordinate is always 0.

(B) $y = 0$: This is the equation where the y-coordinate is always 0. This matches the characteristic of the x-axis.

(C) $x + y = 0$: This is the equation of a line passing through the origin with a slope of -1 (or $y = -x$). Points on this line have $x$ and $y$ values that sum to 0 (e.g., $(1, -1), (-2, 2)$). This is not the x-axis.

(D) $x = y$: This is the equation of a line passing through the origin with a slope of 1. Points on this line have equal $x$ and $y$ values (e.g., $(1, 1), (-2, -2)$). This is not the x-axis.

Therefore, the equation of the x-axis is $y = 0$.

The correct option is (B) y = 0.

The equation of the graph is $y = 6$.

The description of the graph of the equation $y = 6$ in a two-variable coordinate system.

The given equation is $y = 6$. This is a linear equation in two variables, which can be written as $0 \cdot x + 1 \cdot y = 6$.

For any point $(x, y)$ that lies on the graph of this equation, the value of the y-coordinate must be 6. The value of the x-coordinate ($x$) can be any real number.

Consider some points that satisfy this equation:

• If $x = 0$, then $y = 6$. Point is $(0, 6)$.
• If $x = 1$, then $y = 6$. Point is $(1, 6)$.
• If $x = -2$, then $y = 6$. Point is $(-2, 6)$.
• If $x = 100$, then $y = 6$. Point is $(100, 6)$.

Plotting these points in the Cartesian plane, we can see that they all lie on a horizontal line. This line is parallel to the x-axis because for every point on the line, the y-coordinate is constant (equal to 6).

The distance of this line from the x-axis is the absolute value of the y-coordinate, which is $|6| = 6$ units.

Since the y-coordinate is positive ($y=6$), the line is located 6 units above the x-axis.

The origin is the point $(0, 0)$. The line $y=6$ is 6 units away from the origin in the positive y direction.

Therefore, the graph of $y = 6$ is a line parallel to the x-axis at a distance of 6 units from the origin.

Let's evaluate the options:

(A) parallel to x-axis at a distance 6 units from the origin: This matches our description.

(B) parallel to y-axis at a distance 6 units from the origin: A line parallel to the y-axis has the form $x = k$. The distance from the origin would be $|k|$. This option describes a vertical line, which $y=6$ is not.

(C) making an intercept 6 on the x-axis: An x-intercept is the point where the line crosses the x-axis, meaning the y-coordinate is 0. If $y=0$, the equation $y=6$ becomes $0=6$, which is false. The line $y=6$ never crosses the x-axis, so it has no x-intercept. This statement is incorrect.

(D) making an intercept 6 on both the axes: The line $y=6$ crosses the y-axis at $(0, 6)$, so the y-intercept is 6. However, as discussed in (C), it does not have an x-intercept. This statement is incorrect.

Based on the analysis, the graph of $y=6$ is a line parallel to the x-axis at a distance 6 units from the origin.

The correct option is (A) parallel to x-axis at a distance 6 units from the origin.

We are given values for $x$ and $y$: $x = 5$ and $y = 2$.

We are also given four linear equations.

The linear equation for which $x = 5$ and $y = 2$ is a solution.

For a point $(x, y)$ to be a solution of a linear equation, the equation must be satisfied when the values of $x$ and $y$ are substituted into it.

We will substitute $x=5$ and $y=2$ into each of the given equations and check which one holds true.

(A) $x + 2y = 7$

Substitute $x=5$ and $y=2$:

$5 + 2(2) = 5 + 4 = 9$

Is $9 = 7$? No. So $(5, 2)$ is not a solution to this equation.

(B) $5x + 2y = 7$

Substitute $x=5$ and $y=2$:

$5(5) + 2(2) = 25 + 4 = 29$

Is $29 = 7$? No. So $(5, 2)$ is not a solution to this equation.

(C) $x + y = 7$

Substitute $x=5$ and $y=2$:

$5 + 2 = 7$

Is $7 = 7$? Yes. So $(5, 2)$ is a solution to this equation.

(D) $5x + y = 7$

Substitute $x=5$ and $y=2$:

$5(5) + 2 = 25 + 2 = 27$

Is $27 = 7$? No. So $(5, 2)$ is not a solution to this equation.

The only equation for which $(5, 2)$ is a solution is $x + y = 7$.

The correct option is (C) x + y = 7.

Three points $(-2, 2)$, $(0, 0)$, and $(2, -2)$ are solutions of a linear equation.

Four linear equations are provided as options.

The linear equation among the given options that has $(-2, 2)$, $(0, 0)$, and $(2, -2)$ as solutions.

For a point $(x, y)$ to be a solution of a linear equation, the coordinates of the point must satisfy the equation when substituted into it.

We will test each of the given linear equations by substituting the coordinates of the three points: $(-2, 2)$, $(0, 0)$, and $(2, -2)$. The correct equation must be satisfied by all three points.

Let's test Option (A): $y - x = 0$

Substitute $(-2, 2)$: $2 - (-2) = 2 + 2 = 4$. Is $4 = 0$? No.

Since the first point does not satisfy the equation, this is not the correct equation.

Let's test Option (B): $x + y = 0$

Substitute $(-2, 2)$: $(-2) + 2 = 0$. Is $0 = 0$? Yes.

Substitute $(0, 0)$: $0 + 0 = 0$. Is $0 = 0$? Yes.

Substitute $(2, -2)$: $2 + (-2) = 2 - 2 = 0$. Is $0 = 0$? Yes.

All three points satisfy this equation.

Let's test Option (C): $-2x + y = 0$

Substitute $(-2, 2)$: $-2(-2) + 2 = 4 + 2 = 6$. Is $6 = 0$? No.

Since the first point does not satisfy the equation, this is not the correct equation.

Let's test Option (D): $-x + 2y = 0$

Substitute $(-2, 2)$: $-(-2) + 2(2) = 2 + 4 = 6$. Is $6 = 0$? No.

Since the first point does not satisfy the equation, this is not the correct equation.

Only the equation $x + y = 0$ is satisfied by all three given points.

The correct option is (B) x + y = 0.

A linear equation in two variables $ax + by + c = 0$.

We are considering the "positive solutions" of this equation.

The quadrant in which the positive solutions of the equation $ax + by + c = 0$ always lie.

A solution to a linear equation in two variables $ax + by + c = 0$ is an ordered pair of numbers $(x, y)$ that satisfies the equation when substituted for $x$ and $y$.

The term "positive solutions" in this context typically refers to solutions where both the x-coordinate and the y-coordinate are positive. That is, $x > 0$ and $y > 0$.

Let's recall the signs of the coordinates in each of the four quadrants of the Cartesian plane:

• 1st Quadrant: $x > 0$ and $y > 0$ (Positive x, Positive y)
• 2nd Quadrant: $x < 0$ and $y > 0$ (Negative x, Positive y)
• 3rd Quadrant: $x < 0$ and $y < 0$ (Negative x, Negative y)
• 4th Quadrant: $x > 0$ and $y < 0$ (Positive x, Negative y)

We are looking for the quadrant where both $x$ and $y$ are positive ($x > 0$ and $y > 0$).

According to the coordinate signs in the quadrants, the region where both the x-coordinate and the y-coordinate are positive is the 1st quadrant.

Therefore, the positive solutions of the equation $ax + by + c = 0$ (meaning solutions where $x > 0$ and $y > 0$) always lie in the 1st quadrant.

The correct option is (A) 1st quadrant.

The linear equation is $2x + 3y = 6$.

The point where the graph of the given equation meets the x-axis.

A point where the graph of an equation meets or cuts the x-axis is called the x-intercept.

Any point on the x-axis has a y-coordinate equal to 0.

To find the point where the graph of $2x + 3y = 6$ meets the x-axis, we set $y=0$ in the equation and solve for $x$.

Substitute $y=0$ into the equation $2x + 3y = 6$:

$2x + 3(0) = 6$

$2x + 0 = 6$

$2x = 6$

Divide both sides by 2:

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

$x = 3$

So, when $y=0$, the value of $x$ is 3.

The point where the graph meets the x-axis is $(x, y) = (3, 0)$.

The correct option is (C) (3, 0).

The linear equation is $y = x$.

We are given four points.

The point among the given options that the graph of the equation $y = x$ passes through.

A point $(x, y)$ lies on the graph of the equation $y = x$ if and only if its coordinates satisfy the equation, i.e., if the y-coordinate is equal to the x-coordinate ($y=x$).

We will check each option by substituting the coordinates of the given point into the equation $y = x$ and see if the equality holds.

Let's test Option (A): $(\frac{3}{2}, \frac{-3}{2})$

Here, $x = \frac{3}{2}$ and $y = \frac{-3}{2}$.

Substitute into $y = x$:

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

This is false ($\frac{-3}{2} \neq \frac{3}{2}$). So, the point $(\frac{3}{2}, \frac{-3}{2})$ does not lie on the graph of $y=x$.

Let's test Option (B): $(0, \frac{3}{2})$

Here, $x = 0$ and $y = \frac{3}{2}$.

Substitute into $y = x$:

$\frac{3}{2} = 0$

This is false ($\frac{3}{2} \neq 0$). So, the point $(0, \frac{3}{2})$ does not lie on the graph of $y=x$.

Let's test Option (C): $(1, 1)$

Here, $x = 1$ and $y = 1$.

Substitute into $y = x$:

$1 = 1$

This is true. So, the point $(1, 1)$ lies on the graph of $y=x$.

Let's test Option (D): $(\frac{-1}{2}, \frac{1}{2})$

Here, $x = \frac{-1}{2}$ and $y = \frac{1}{2}$.

Substitute into $y = x$:

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

This is false ($\frac{1}{2} \neq \frac{-1}{2}$). So, the point $(\frac{-1}{2}, \frac{1}{2})$ does not lie on the graph of $y=x$.

The only point among the options that satisfies the equation $y = x$ is $(1, 1)$.

The correct option is (C) (1, 1).

A linear equation in one or two variables.

The operation of multiplying or dividing both sides of the equation by a non-zero number.

How the solution(s) of the linear equation are affected by this operation.

Consider a linear equation. Let's represent a general linear equation in two variables as:

$ax + by = c$

where $a$, $b$, and $c$ are constants, and $a$ and $b$ are not both zero.

A solution to this equation is a pair of values $(x_0, y_0)$ such that when $x = x_0$ and $y = y_0$, the equation holds true:

Now, let's multiply both sides of the original equation by a non-zero number, say $k$ ($k \neq 0$).

$k(ax + by) = k(c)$

Distributing $k$ on the left side, we get the new equation:

Let's check if the original solution $(x_0, y_0)$ is also a solution to this new equation (ii). Substitute $x=x_0$ and $y=y_0$ into equation (ii):

$kax_0 + kby_0$

We can factor out $k$ from this expression:

$k(ax_0 + by_0)$

From equation (i), we know that $ax_0 + by_0 = c$. Substitute this into the expression:

$k(c) = kc$

So, $kax_0 + kby_0 = kc$. This means that the point $(x_0, y_0)$, which was a solution to the original equation, is also a solution to the new equation obtained by multiplying by $k$.

Conversely, if $(x_1, y_1)$ is a solution to the new equation (ii), then $kax_1 + kby_1 = kc$. Dividing both sides by the non-zero number $k$, we get $\frac{kax_1 + kby_1}{k} = \frac{kc}{k}$, which simplifies to $ax_1 + by_1 = c$. This shows that $(x_1, y_1)$ is also a solution to the original equation (i).

This demonstrates that equations (i) and (ii) have the exact same set of solutions. The operation of multiplying both sides by a non-zero number produces an equivalent equation.

Similarly, dividing both sides of the original equation $ax + by = c$ by a non-zero number $k$ ($k \neq 0$) gives:

$\frac{ax + by}{k} = \frac{c}{k}$

$\frac{a}{k}x + \frac{b}{k}y = \frac{c}{k}$

Using the same logic as with multiplication, we can show that the solutions to this new equation are identical to the solutions of the original equation.

These operations (multiplying or dividing both sides by the same non-zero number) are fundamental properties used when solving equations, as they preserve the equality and thus the solution set.

Therefore, if we multiply or divide both sides of a linear equation with a non-zero number, the solution of the linear equation remains the same.

The correct option is (B) Remains the same.

A point with coordinates $x = 1$ and $y = 2$.

The number of linear equations in two variables ($x$ and $y$) that can be satisfied by the given point $(1, 2)$.

A linear equation in two variables $x$ and $y$ has the general form $ax + by + c = 0$, where $a$, $b$, and $c$ are real numbers, and $a$ and $b$ are not both zero.

For a point $(x_0, y_0)$ to be a solution to the equation $ax + by + c = 0$, it must satisfy the equation when substituted:

$ax_0 + by_0 + c = 0$

We are given the point $(x_0, y_0) = (1, 2)$. Substituting these values into the general equation, we get the condition that the coefficients $a$, $b$, and $c$ must satisfy for the point $(1, 2)$ to be a solution:

This simplifies to:

We need to find how many distinct linear equations $ax + by + c = 0$ exist such that the coefficients $a$, $b$, and $c$ satisfy equation (i), with the constraint that $a$ and $b$ are not both zero (otherwise, it wouldn't be a linear equation in *two* variables $x$ and $y$ in the typical sense that its graph is a line in the xy-plane).

Equation (i) is a single linear equation with three variables ($a$, $b$, and $c$). We can choose values for two of the variables (say, $a$ and $b$) and then find the value of the third variable ($c$) from the equation $c = -a - 2b$.

Since we can choose infinitely many pairs of real numbers for $a$ and $b$ (as long as they are not both zero simultaneously), there will be infinitely many sets of $(a, b, c)$ that satisfy equation (i) and the non-zero condition for $a$ or $b$.

Each valid set of $(a, b, c)$ defines a unique linear equation $ax + by + c = 0$ that is satisfied by the point $(1, 2)$.

Consider some examples:

• If we choose $a=1$ and $b=0$, then $c = -1 - 2(0) = -1$. The equation is $1x + 0y - 1 = 0$, or $x - 1 = 0$. This is a linear equation in two variables (coefficient of $x$ is non-zero). Substituting $(1, 2)$ gives $1 - 1 = 0$, which is true.
• If we choose $a=0$ and $b=1$, then $c = -0 - 2(1) = -2$. The equation is $0x + 1y - 2 = 0$, or $y - 2 = 0$. This is a linear equation in two variables (coefficient of $y$ is non-zero). Substituting $(1, 2)$ gives $2 - 2 = 0$, which is true.
• If we choose $a=1$ and $b=1$, then $c = -1 - 2(1) = -3$. The equation is $1x + 1y - 3 = 0$, or $x + y - 3 = 0$. Substituting $(1, 2)$ gives $1 + 2 - 3 = 0$, which is true.
• If we choose $a=5$ and $b=-3$, then $c = -5 - 2(-3) = -5 + 6 = 1$. The equation is $5x - 3y + 1 = 0$. Substituting $(1, 2)$ gives $5(1) - 3(2) + 1 = 5 - 6 + 1 = 0$, which is true.

Since there are infinitely many possible non-zero pairs of $(a, b)$, there are infinitely many values for $c$, and thus infinitely many distinct linear equations of the form $ax + by + c = 0$ that are satisfied by the point $(1, 2)$.

Geometrically, this means that infinitely many lines can pass through a single point in a plane.

The correct option is (C) Infinitely many.

A point with coordinates of the form $(a, a)$.

The line on which a point of the form $(a, a)$ always lies.

A point $(x, y)$ lies on a line if its coordinates satisfy the equation of the line.

We are given a point $(x, y)$ where the x-coordinate is $a$ and the y-coordinate is $a$. So, $x = a$ and $y = a$. This implies that for any such point, the y-coordinate is equal to the x-coordinate ($y = x$).

Let's check which of the given options represents a line where the x and y coordinates are always equal.

(A) x-axis: The equation of the x-axis is $y = 0$. For a point $(a, a)$ to be on the x-axis, its y-coordinate must be 0, so $a = 0$. This means only the point $(0, 0)$ is on the x-axis and is of the form $(a, a)$. The form $(a, a)$ does not *always* lie on the x-axis for any arbitrary value of $a$.

(B) y-axis: The equation of the y-axis is $x = 0$. For a point $(a, a)$ to be on the y-axis, its x-coordinate must be 0, so $a = 0$. Again, this is only true for the point $(0, 0)$. The form $(a, a)$ does not *always* lie on the y-axis for any arbitrary value of $a$.

(C) On the line $y = x$: The equation of this line states that the y-coordinate is equal to the x-coordinate. For a point $(a, a)$, the x-coordinate is $a$ and the y-coordinate is $a$. Substituting these into the equation $y=x$, we get $a = a$. This equation is true for any real value of $a$. Therefore, any point of the form $(a, a)$ always lies on the line $y = x$.

(D) On the line $x + y = 0$: This equation can be rewritten as $y = -x$. For a point $(a, a)$ to be on this line, its y-coordinate must be the negative of its x-coordinate. Substituting $x=a$ and $y=a$ into the equation $x+y=0$, we get $a + a = 0$, which simplifies to $2a = 0$, or $a = 0$. This means only the point $(0, 0)$ lies on this line and is of the form $(a, a)$. The form $(a, a)$ does not *always* lie on the line $x+y=0$ for any arbitrary value of $a$ (unless $a=0$).

Based on the analysis, the point of the form $(a, a)$ always lies on the line $y = x$ because its coordinates satisfy the equation $y=x$ for any value of $a$.

The correct option is (C) On the line y = x.

A point with coordinates of the form $(a, -a)$.

The line on which a point of the form $(a, -a)$ always lies.

A point $(x, y)$ lies on a line if its coordinates satisfy the equation of the line.

We are given a point $(x, y)$ where the x-coordinate is $a$ and the y-coordinate is $-a$. So, $x = a$ and $y = -a$.

We need to find a linear equation relating $x$ and $y$ that is always true for any point $(a, -a)$, regardless of the value of $a$.

From the given coordinates, we have $x = a$ and $y = -a$. We can establish a relationship between $x$ and $y$ by eliminating the parameter $a$.

Substitute $a = x$ into the second relation $y = -a$:

$y = -(x)$

$y = -x$

This equation can be rewritten by moving the $x$ term to the left side:

$x + y = 0$

This equation relates $x$ and $y$ and is satisfied by any point $(a, -a)$, since substituting $x=a$ and $y=-a$ gives $a + (-a) = a - a = 0$, which is true for any value of $a$.

Let's check the given options:

(A) $x = a$: This is not an equation relating $x$ and $y$. It states that the x-coordinate is equal to the parameter $a$. This doesn't define a single line for all points $(a, -a)$ as $a$ varies.

(B) $y = -a$: This is also not an equation relating $x$ and $y$. It states that the y-coordinate is equal to the negative of the parameter $a$. This doesn't define a single line for all points $(a, -a)$ as $a$ varies.

(C) $y = x$: Substitute the point $(a, -a)$ into this equation: $-a = a$. This is only true if $2a = 0$, which means $a=0$. So, only the point $(0, 0)$ lies on this line and is of the form $(a, -a)$. This is not true for *any* value of $a$.

(D) $x + y = 0$: Substitute the point $(a, -a)$ into this equation: $a + (-a) = 0$. This simplifies to $a - a = 0$, which is $0 = 0$. This is true for any real value of $a$. Therefore, any point of the form $(a, -a)$ always lies on the line $x + y = 0$.

Based on the analysis, the point of the form $(a, -a)$ always lies on the line $x + y = 0$.

The correct option is (D) x + y = 0.

Several statements regarding linear equations and their graphs.

Whether each statement is True or False, and provide justification.

Let's evaluate each statement:

(i) ax + by + c = 0, where a, b and c are real numbers, is a linear equation in two variables.

False.

Justification:

A linear equation in two variables $x$ and $y$ is defined as an equation that can be written in the form $ax + by + c = 0$, where $a$, $b$, and $c$ are real numbers, and at least one of $a$ or $b$ is non-zero (i.e., $a \neq 0$ or $b \neq 0$).

The given statement specifies that $a$, $b$, and $c$ are real numbers but omits the crucial condition that $a$ and $b$ are not both zero. If $a=0$ and $b=0$, the equation becomes $0 \cdot x + 0 \cdot y + c = 0$, or simply $c=0$. This equation does not involve the variables $x$ and $y$ in a linear relationship. For example, if $c=5$, $0=5$ which is false. If $c=0$, $0=0$ which is always true. Neither case represents a linear equation in two variables $x$ and $y$ that defines a line in the plane.

(ii) A linear equation 2x + 3y = 5 has a unique solution.

False.

Justification:

The equation $2x + 3y = 5$ is a single linear equation in two variables, $x$ and $y$. The coefficients of $x$ and $y$ (2 and 3) are not both zero.

The graph of a single linear equation in two variables is a straight line in the Cartesian coordinate plane.

Every point $(x, y)$ that lies on this straight line represents a solution to the equation.

A straight line consists of infinitely many points.

Therefore, a linear equation in two variables has infinitely many solutions, not a unique solution.

(iii) All the points (2, 0), (–3, 0), (4, 2) and (0, 5) lie on the x-axis.

False.

Justification:

A point lies on the x-axis if and only if its y-coordinate is 0.

Let's check the y-coordinate for each given point:

• For the point $(2, 0)$, the y-coordinate is 0. This point lies on the x-axis.
• For the point $(-3, 0)$, the y-coordinate is 0. This point lies on the x-axis.
• For the point $(4, 2)$, the y-coordinate is 2. Since $2 \neq 0$, this point does not lie on the x-axis.
• For the point $(0, 5)$, the y-coordinate is 5. Since $5 \neq 0$, this point does not lie on the x-axis.

Since the points $(4, 2)$ and $(0, 5)$ do not lie on the x-axis, the statement that *all* the points lie on the x-axis is false.

(iv) The line parallel to the y-axis at a distance 4 units to the left of y-axis is given by the equation x = – 4.

True.

Justification:

A line parallel to the y-axis is a vertical line. The equation of a vertical line is of the form $x = k$, where $k$ is a constant.

The y-axis itself has the equation $x = 0$.

The distance of a vertical line $x=k$ from the y-axis is given by $|k|$.

"4 units to the left of y-axis" means that the x-coordinate of every point on the line is 4 units in the negative direction from the origin (or from the y-axis).

The x-coordinate of points on the y-axis is 0. Moving 4 units to the left from $x=0$ results in an x-coordinate of $0 - 4 = -4$.

Thus, every point on this line has an x-coordinate of $-4$.

The equation of this line is $x = -4$.

(v) The graph of the equation y = mx + c passes through the origin.

False.

Justification:

The origin is the point with coordinates $(0, 0)$.

For the graph of the equation $y = mx + c$ to pass through the origin, the coordinates of the origin $(0, 0)$ must satisfy the equation.

Substitute $x=0$ and $y=0$ into the equation $y = mx + c$:

$0 = m(0) + c$

$0 = 0 + c$

$0 = c$

This shows that the equation $y = mx + c$ passes through the origin if and only if the constant term $c$ is equal to 0.

If $c \neq 0$, the line does not pass through the origin. The value of $c$ represents the y-intercept of the line (the y-coordinate where the line crosses the y-axis).

Since the statement claims that the graph *always* passes through the origin, regardless of the value of $c$, it is false.

A table containing pairs of $(x, y)$ coordinates.

The statement: The coordinates of points given in the table represent some of the solutions of the equation $2x + 2 = y$.

Whether the given statement is True or False, and provide justification.

The equation is given as $2x + 2 = y$. We can rewrite this as $y = 2x + 2$.

The points given in the table are:

For a point $(x, y)$ to be a solution of the equation $y = 2x + 2$, substituting the values of $x$ and $y$ into the equation must result in a true statement (Left Hand Side = Right Hand Side).

Let's check each point from the table:

Point (0, 2): $x=0, y=2$

Substitute into $y = 2x + 2$:

$2 = 2(0) + 2$

$2 = 0 + 2$

$2 = 2$

This is true. So, $(0, 2)$ is a solution.

Point (1, 4): $x=1, y=4$

Substitute into $y = 2x + 2$:

$4 = 2(1) + 2$

$4 = 2 + 2$

$4 = 4$

This is true. So, $(1, 4)$ is a solution.

Point (2, 6): $x=2, y=6$

Substitute into $y = 2x + 2$:

$6 = 2(2) + 2$

$6 = 4 + 2$

$6 = 6$

This is true. So, $(2, 6)$ is a solution.

Point (3, 8): $x=3, y=8$

Substitute into $y = 2x + 2$:

$8 = 2(3) + 2$

$8 = 6 + 2$

$8 = 8$

This is true. So, $(3, 8)$ is a solution.

Point (4, 10): $x=4, y=10$

Substitute into $y = 2x + 2$:

$10 = 2(4) + 2$

$10 = 8 + 2$

$10 = 10$

This is true. So, $(4, 10)$ is a solution.

Since all the points listed in the table satisfy the equation $y = 2x + 2$, they are indeed solutions of the equation. The statement claims they represent "some of the solutions," which is accurate as a linear equation in two variables has infinitely many solutions.

The statement is True.

Justification: As shown above, substituting the coordinates of each point $(x, y)$ from the table into the equation $y = 2x + 2$ yields a true equality, confirming that each point is a solution to the equation.

The linear equation $3x + 4y = 12$.

The point is $(0, 3)$.

Whether the point $(0, 3)$ lies on the graph of the equation $3x + 4y = 12$ and justify the answer.

A point $(x, y)$ lies on the graph of a linear equation if its coordinates satisfy the equation when substituted.

We need to check if substituting $x=0$ and $y=3$ into the equation $3x + 4y = 12$ results in a true statement.

Substitute $x=0$ and $y=3$ into the left side of the equation:

$3x + 4y = 3(0) + 4(3)$

$3x + 4y = 0 + 12$

$3x + 4y = 12$

The result of the substitution is 12, which is equal to the right side of the equation.

Since the coordinates of the point $(0, 3)$ satisfy the equation $3x + 4y = 12$, the point lies on the graph of the equation.

The statement is True.

Justification:

When the coordinates $x=0$ and $y=3$ are substituted into the equation $3x + 4y = 12$, we get $3(0) + 4(3) = 0 + 12 = 12$, which is equal to the right hand side of the equation. Therefore, the point $(0, 3)$ satisfies the equation and lies on its graph.

The linear equation $x + 2y = 7$.

The point is $(0, 7)$.

Whether the graph of the equation $x + 2y = 7$ passes through the point $(0, 7)$ and justify the answer.

For the graph of the equation $x + 2y = 7$ to pass through the point $(0, 7)$, the coordinates of the point $(0, 7)$ must satisfy the equation when substituted.

We need to check if substituting $x=0$ and $y=7$ into the equation $x + 2y = 7$ results in a true statement.

Substitute $x=0$ and $y=7$ into the left side of the equation:

LHS = $x + 2y = 0 + 2(7)$

LHS = $0 + 14$

LHS = $14$

The right side of the equation is RHS = 7.

Comparing the LHS and RHS:

Since $14 \neq 7$, the coordinates of the point $(0, 7)$ do not satisfy the equation $x + 2y = 7$.

Therefore, the point $(0, 7)$ does not lie on the graph of the linear equation $x + 2y = 7$, and the graph does not pass through this point.

The statement is False.

Justification:

When the coordinates $x=0$ and $y=7$ are substituted into the equation $x + 2y = 7$, we get $0 + 2(7) = 14$. The right hand side of the equation is 7. Since $14 \neq 7$, the point $(0, 7)$ does not satisfy the equation, and thus does not lie on its graph.

A graph of a straight line in the Cartesian plane.

The statement: The graph represents the linear equation $x + y = 0$.

Whether the given statement is True or False, and provide justification.

The given equation is $x + y = 0$. This can be rewritten as $y = -x$.

For a point $(x, y)$ to be on the graph of this equation, the value of its y-coordinate must be equal to the negative of its x-coordinate.

Let's examine the given graph and identify some points that lie on the line shown:

• The line clearly passes through the origin $(0, 0)$. Let's check if this point satisfies the equation $x + y = 0$: $0 + 0 = 0$. This is true.
• The line passes through the point where $x=1$ and $y=-1$. Let's check: $1 + (-1) = 1 - 1 = 0$. This is true.
• The line passes through the point where $x=-1$ and $y=1$. Let's check: $(-1) + 1 = 0$. This is true.
• The line appears to pass through the point where $x=2$ and $y=-2$. Let's check: $2 + (-2) = 2 - 2 = 0$. This is true.
• The line appears to pass through the point where $x=-2$ and $y=2$. Let's check: $(-2) + 2 = 0$. This is true.

In every point observed on the graph, the y-coordinate is the negative of the x-coordinate, which is the condition required by the equation $y = -x$ or $x + y = 0$.

The graph is a straight line passing through the origin $(0,0)$ and points like $(1,-1)$ and $(-1,1)$, which is consistent with the graph of $y = -x$ (a line with slope -1 passing through the origin).

The statement is True.

Justification:

The equation $x + y = 0$ (or $y = -x$) describes a set of points where the y-coordinate is the negative of the x-coordinate. The given graph is a straight line that passes through the origin $(0, 0)$ and has a slope of $-1$. Points on this line, such as $(0, 0)$, $(1, -1)$, $(-1, 1)$, $(2, -2)$, etc., satisfy the equation $x + y = 0$. Therefore, the graph represents the linear equation $x + y = 0$.

A graph of a straight line in the Cartesian plane (Fig. 4.2).

The statement: The graph represents the linear equation $x = 3$.

Whether the given statement is True or False, and provide justification.

The given equation is $x = 3$. This is a linear equation in two variables, which can be written as $1 \cdot x + 0 \cdot y = 3$.

For a point $(x, y)$ to be on the graph of this equation, the value of its x-coordinate must be 3, regardless of the value of the y-coordinate.

The graph of the equation $x = k$ is a vertical line parallel to the y-axis, passing through the point $(k, 0)$ on the x-axis.

In this case, the equation is $x = 3$. So, the graph should be a vertical line passing through the point $(3, 0)$ on the x-axis.

Let's examine the given graph (Fig. 4.2):

• The line in the graph is a vertical line.
• The line intersects the x-axis at the point where the x-coordinate is 3. The coordinates of this point are $(3, 0)$.
• Let's check another point on the line. The line passes through the point where the x-coordinate is 3 and the y-coordinate is 2, i.e., the point $(3, 2)$. Substituting $x=3$ and $y=2$ into the equation $x=3$ gives $3 = 3$, which is true.
• Another point on the line is where the x-coordinate is 3 and the y-coordinate is -1, i.e., the point $(3, -1)$. Substituting $x=3$ and $y=-1$ into the equation $x=3$ gives $3 = 3$, which is true.

All points on the line shown in the graph have an x-coordinate equal to 3. This matches the definition of the graph of the equation $x = 3$.

The statement is True.

Justification:

The equation $x = 3$ represents the set of all points $(x, y)$ in the Cartesian plane where the x-coordinate is 3, irrespective of the y-coordinate. This set of points forms a vertical line passing through the point $(3, 0)$ on the x-axis and parallel to the y-axis. The provided graph is exactly this vertical line, confirming that it represents the equation $x = 3$.

A table containing pairs of $(x, y)$ coordinates.

The statement: The coordinates of points given in the table represent some of the solutions of the equation $x – y + 2 = 0$.

Whether the given statement is True or False, and provide justification.

The given equation is $x - y + 2 = 0$. This can be rearranged to express $y$ in terms of $x$: $y = x + 2$.

The points given in the table are:

For a point $(x, y)$ to be a solution of the equation $x - y + 2 = 0$, substituting the values of $x$ and $y$ into the equation must result in a true statement ($x - y + 2 = 0$).

Let's check each point from the table:

Point (0, 2): $x=0, y=2$. Substitute into $x - y + 2 = 0$: $0 - 2 + 2 = 0$. $0 = 0$. True. So, $(0, 2)$ is a solution.

Point (1, 3): $x=1, y=3$. Substitute into $x - y + 2 = 0$: $1 - 3 + 2 = 0$. $-2 + 2 = 0$. $0 = 0$. True. So, $(1, 3)$ is a solution.

Point (2, 4): $x=2, y=4$. Substitute into $x - y + 2 = 0$: $2 - 4 + 2 = 0$. $-2 + 2 = 0$. $0 = 0$. True. So, $(2, 4)$ is a solution.

Point (3, -5): $x=3, y=-5$. Substitute into $x - y + 2 = 0$: $3 - (-5) + 2 = 0$. $3 + 5 + 2 = 0$. $8 + 2 = 0$. $10 = 0$. False. So, $(3, -5)$ is not a solution.

Point (4, 6): $x=4, y=6$. Substitute into $x - y + 2 = 0$: $4 - 6 + 2 = 0$. $-2 + 2 = 0$. $0 = 0$. True. So, $(4, 6)$ is a solution.

The statement claims that the coordinates of points *in the table* represent *some* of the solutions. While some points in the table are indeed solutions, the point $(3, -5)$ is not a solution. Therefore, the collection of points *as given in the table* does not exclusively consist of solutions to the equation $x - y + 2 = 0$. If even one point presented as a solution is not a solution, the statement is considered false.

The statement is False.

Justification:

We tested each point from the table by substituting its coordinates into the equation $x - y + 2 = 0$. While points $(0, 2)$, $(1, 3)$, $(2, 4)$, and $(4, 6)$ satisfy the equation, the point $(3, -5)$ does not, as $3 - (-5) + 2 = 3 + 5 + 2 = 10 \neq 0$. Since the table includes a point that is not a solution, the statement that the coordinates of points *in the table* represent some solutions is false.

The statement: Every point on the graph of a linear equation in two variables does not represent a solution of the linear equation.

Whether the given statement is True or False, and provide justification.

The graph of a linear equation in two variables, say $ax + by + c = 0$ (where $a$ and $b$ are not both zero), is defined as the set of all points $(x, y)$ in the Cartesian plane whose coordinates satisfy the equation.

A solution of a linear equation in two variables is an ordered pair $(x_0, y_0)$ such that when $x$ is replaced by $x_0$ and $y$ is replaced by $y_0$, the equation $ax_0 + by_0 + c = 0$ is a true statement.

By definition, the graph of the equation consists precisely of those points whose coordinates are solutions to the equation.

Therefore, every point that lies on the graph of a linear equation in two variables is a solution to that equation. Conversely, every solution to the linear equation in two variables corresponds to a point that lies on its graph.

The statement "Every point on the graph ... does not represent a solution" is the opposite of this fundamental definition.

The statement is False.

Justification:

By the definition of the graph of a linear equation in two variables, the graph is the collection of all points $(x, y)$ whose coordinates satisfy the equation. Any point $(x, y)$ whose coordinates satisfy the equation is a solution to the equation. Thus, every point on the graph of a linear equation in two variables represents a solution of the linear equation.

The statement: The graph of every linear equation in two variables need not be a line.

Whether the given statement is True or False, and provide justification.

A linear equation in two variables, say $x$ and $y$, is defined as an equation that can be written in the form $ax + by + c = 0$, where $a$, $b$, and $c$ are real numbers, and the coefficients $a$ and $b$ are not both zero (i.e., $a^2 + b^2 \neq 0$).

The graph of a linear equation in two variables is the set of all points $(x, y)$ in the Cartesian plane whose coordinates satisfy the equation.

According to the fundamental principles of coordinate geometry, the graph of any equation of the form $ax + by + c = 0$, where $a$ and $b$ are not both zero, is always a straight line.

• If $b \neq 0$, the equation can be rewritten as $by = -ax - c$, or $y = -\frac{a}{b}x - \frac{c}{b}$. This is in the form $y = mx + C$, which represents a straight line with slope $m = -\frac{a}{b}$ and y-intercept $C = -\frac{c}{b}$.
• If $b = 0$, then $a$ must be non-zero (since $a$ and $b$ are not both zero). The equation becomes $ax + c = 0$, which can be rewritten as $ax = -c$, or $x = -\frac{c}{a}$. This is the equation of a vertical line, which is also a straight line.

In all cases where the equation is a linear equation in two variables (by definition), its graph is a straight line.

The statement "The graph of every linear equation in two variables need not be a line" implies that there exists at least one linear equation in two variables whose graph is not a line. This contradicts the geometric interpretation and definition of such equations.

The statement is False.

Justification:

By definition, a linear equation in two variables is an equation of the form $ax + by + c = 0$, where $a, b \in \mathbb{R}$ and $a$ and $b$ are not both zero. It is a fundamental result in coordinate geometry that the graph of every such equation is always a straight line. Conversely, every straight line in the Cartesian plane can be represented by a linear equation in two variables.

The linear equation is $3x + 4y = 12$.

The coordinates of the points where the graph of the equation cuts the x-axis and the y-axis.

The point where the graph of a linear equation cuts the x-axis is called the x-intercept.

Any point on the x-axis has a y-coordinate equal to 0.

To find the x-intercept, we set $y = 0$ in the given equation and solve for $x$.

Substitute $y = 0$ into the equation $3x + 4y = 12$:

$3x + 4(0) = 12$

$3x + 0 = 12$

$3x = 12$

Divide both sides by 3:

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

$x = 4$

So, the graph cuts the x-axis at the point where $x=4$ and $y=0$. The coordinates of the x-intercept are $(4, 0)$.

The point where the graph of a linear equation cuts the y-axis is called the y-intercept.

Any point on the y-axis has an x-coordinate equal to 0.

To find the y-intercept, we set $x = 0$ in the given equation and solve for $y$.

Substitute $x = 0$ into the equation $3x + 4y = 12$:

$3(0) + 4y = 12$

$0 + 4y = 12$

$4y = 12$

Divide both sides by 4:

$y = \frac{12}{4}$

$y = 3$

So, the graph cuts the y-axis at the point where $x=0$ and $y=3$. The coordinates of the y-intercept are $(0, 3)$.

The graph of the equation $3x + 4y = 12$ cuts the x-axis at the point (4, 0) and the y-axis at the point (0, 3).

The linear equation of the first line is $x + y = 5$.

The second line is parallel to the y-axis, at a distance of 2 units from the origin, and in the positive direction of the x-axis.

The point of intersection of the graph of the equation $x + y = 5$ and the second line.

First, let's determine the equation of the second line.

A line parallel to the y-axis is a vertical line. Its equation is of the form $x = k$, where $k$ is a constant.

The distance of this line from the origin (which is the y-axis) is $|k|$. We are told this distance is 2 units.

So, $k$ can be 2 or -2.

We are also told that the line is "in the positive direction of x-axis" from the origin. This means the x-coordinate is positive.

Therefore, $k$ must be positive.

The equation of the second line is $x = 2$.

Now, we need to find the point where the graph of the equation $x + y = 5$ meets the line $x = 2$. This point is the intersection of the two lines.

To find the intersection point, we solve the system of the two linear equations:

$x + y = 5$

$x = 2$

Substitute the value of $x$ from the second equation into the first equation:

Substitute $x=2$ into the equation $x + y = 5$:

$2 + y = 5$

Now, solve for $y$:

$y = 5 - 2$

$y = 3$

The intersection point has coordinates $(x, y) = (2, 3)$.

The graph of the linear equation $x + y = 5$ meets the line $x = 2$ at the point (2, 3).

The linear equation is $2x + 5y = 20$.

The condition on the point $(x, y)$ on the graph is that the x-coordinate is $\frac{5}{2}$ times its ordinate (y-coordinate).

The coordinates of the point $(x, y)$ on the graph of $2x + 5y = 20$ that satisfies the given condition.

The given linear equation is:

The condition that the x-coordinate is $\frac{5}{2}$ times its ordinate can be written as:

We need to find the point $(x, y)$ that satisfies both equation (i) and condition (ii).

We can solve this system of equations by substituting the expression for $x$ from equation (ii) into equation (i).

Substitute $x = \frac{5}{2}y$ into equation (i):

$2 \left( \frac{5}{2}y \right) + 5y = 20$

Simplify the equation:

$\cancel{2} \cdot \frac{5}{\cancel{2}}y + 5y = 20$

$5y + 5y = 20$

Combine the terms involving $y$:

$10y = 20$

Solve for $y$ by dividing both sides by 10:

$y = \frac{20}{10}$

$y = 2$

Now that we have the value of $y$, substitute $y = 2$ into equation (ii) to find the value of $x$:

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

$x = \frac{5}{2}(2)$

$x = 5$

So, the coordinates of the point are $(x, y) = (5, 2)$.

We can verify this point by substituting it back into the original equation (i):

$2(5) + 5(2) = 10 + 10 = 20$

The equation $20 = 20$ is true, so the point $(5, 2)$ lies on the graph of the equation $2x + 5y = 20$.

Also, the x-coordinate is 5 and the y-coordinate (ordinate) is 2. Check the condition $x = \frac{5}{2}y$: $5 = \frac{5}{2}(2) \implies 5 = 5$, which is true.

Thus, the point $(5, 2)$ satisfies both the equation and the given condition.

The required point on the graph is (5, 2).

A straight line parallel to the x-axis and 4 units above it.

Draw the graph of the equation representing this line.

A straight line that is parallel to the x-axis is a horizontal line.

The equation of a horizontal line is generally of the form $y = k$, where $k$ is a constant.

The distance of the line $y = k$ from the x-axis is $|k|$.

We are given that the line is 4 units above the x-axis. This means the y-coordinate of every point on the line is positive and equal to 4.

So, the constant $k$ is 4.

The equation of the line is $y = 4$.

To draw the graph of this equation, we can plot any points $(x, y)$ where $y=4$. Since the x-coordinate can be any real number, some points on the line are:

• If $x=0$, $y=4$. Point is $(0, 4)$.
• If $x=2$, $y=4$. Point is $(2, 4)$.
• If $x=-3$, $y=4$. Point is $(-3, 4)$.

We can plot these points on a Cartesian coordinate system. The point $(0, 4)$ is on the y-axis.

The graph is the straight line passing through these points. It will be a horizontal line, parallel to the x-axis, located 4 units above the x-axis.

Steps to draw the graph:

1. Draw the x-axis and y-axis, intersecting at the origin (0,0).

2. Choose a scale for the axes.

3. Locate the point (0, 4) on the y-axis.

4. Draw a horizontal line through the point (0, 4). This line is parallel to the x-axis.

This line is the graph of $y=4$.

Two linear equations: $y = x$ and $y = -x$.

Draw the graphs of the given equations on the same Cartesian plane and make observations.

To draw the graph of a linear equation, we need to find at least two points that satisfy the equation. It's good practice to find a few points to ensure accuracy.

For the equation $y = x$:

This equation means that the y-coordinate is always equal to the x-coordinate for any point on the line.

Let's choose some values for $x$ and find the corresponding values for $y$:

These points are $(0, 0)$, $(1, 1)$, $(2, 2)$, $(-1, -1)$, and $(-2, -2)$. Plot these points on the Cartesian plane and draw a straight line passing through them. This line represents the graph of $y = x$.

For the equation $y = -x$:

This equation means that the y-coordinate is always equal to the negative of the x-coordinate for any point on the line.

Let's choose some values for $x$ and find the corresponding values for $y$:

These points are $(0, 0)$, $(1, -1)$, $(2, -2)$, $(-1, 1)$, and $(-2, 2)$. Plot these points on the same Cartesian plane and draw a straight line passing through them. This line represents the graph of $y = -x$.

Drawing the Graphs:

1. Draw the x-axis and y-axis intersecting at the origin $(0, 0)$.

2. Plot the points for the equation $y = x$: $(0, 0), (1, 1), (2, 2), (-1, -1), (-2, -2)$. Draw a line through these points. This line goes upwards from left to right, passing through the origin.

3. Plot the points for the equation $y = -x$: $(0, 0), (1, -1), (2, -2), (-1, 1), (-2, 2)$. Draw a line through these points. This line goes downwards from left to right, passing through the origin.

Observation:

From the graphs, we observe the following:

1. Both lines pass through the origin (0, 0).

2. The line $y = x$ lies in the 1st and 3rd quadrants.

3. The line $y = -x$ lies in the 2nd and 4th quadrants.

4. The two lines are perpendicular to each other. This is because the slope of $y=x$ is $m_1 = 1$ and the slope of $y=-x$ is $m_2 = -1$. The product of their slopes is $m_1 \cdot m_2 = 1 \cdot (-1) = -1$, which is the condition for two non-vertical lines to be perpendicular.

The linear equation is $2x + 5y = 19$.

The condition on the point $(x, y)$ on the graph is that its ordinate ($y$-coordinate) is $1\frac{1}{2}$ times its abscissa ($x$-coordinate).

The coordinates of the point $(x, y)$ on the graph of $2x + 5y = 19$ that satisfies the given condition.

The given linear equation is:

The condition that the ordinate ($y$) is $1\frac{1}{2}$ times its abscissa ($x$) can be written as:

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

Convert the mixed number to an improper fraction:

So, the condition is:

We need to find the point $(x, y)$ that satisfies both equation (i) and condition (ii).

We can solve this system of equations by substituting the expression for $y$ from equation (ii) into equation (i).

Substitute $y = \frac{3}{2}x$ into equation (i):

$2x + 5\left(\frac{3}{2}x\right) = 19$

Simplify the equation:

$2x + \frac{15}{2}x = 19$

To combine the terms on the left side, find a common denominator, which is 2:

$\frac{2 \cdot 2}{2}x + \frac{15}{2}x = 19$

$\frac{4}{2}x + \frac{15}{2}x = 19$

Combine the numerators:

$\frac{4x + 15x}{2} = 19$

$\frac{19x}{2} = 19$

Multiply both sides by 2:

$19x = 19 \times 2$

$19x = 38$

Solve for $x$ by dividing both sides by 19:

$x = \frac{38}{19}$

$x = 2$

Now that we have the value of $x$, substitute $x = 2$ into equation (ii) to find the value of $y$:

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

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

$y = 3$

So, the coordinates of the point are $(x, y) = (2, 3)$.

We can verify this point by substituting it back into the original equation (i):

$2(2) + 5(3) = 4 + 15 = 19$. The equation $19 = 19$ is true.

Also, check the condition: ordinate ($y=3$) is $1\frac{1}{2}$ times abscissa ($x=2$). $3 = \frac{3}{2} \times 2 \implies 3 = 3$. The condition is satisfied.

The required point on the graph is (2, 3).

A straight line parallel to the x-axis and at a distance 3 units below it.

Draw the graph of the equation representing this line.

A straight line that is parallel to the x-axis is a horizontal line.

The equation of a horizontal line is generally of the form $y = k$, where $k$ is a constant.

The distance of the line $y = k$ from the x-axis is $|k|$.

We are given that the line is at a distance of 3 units below the x-axis. This means the y-coordinate of every point on the line is 3 units in the negative direction from the x-axis (where $y=0$).

So, the y-coordinate for every point on this line must be $-3$.

Therefore, the constant $k$ is $-3$.

The equation of the line is $y = -3$.

To draw the graph of this equation, we need to plot points $(x, y)$ where $y=-3$. The value of $x$ can be any real number.

Some points on the line are:

• If $x=0$, $y=-3$. Point is $(0, -3)$.
• If $x=2$, $y=-3$. Point is $(2, -3)$.
• If $x=-4$, $y=-3$. Point is $(-4, -3)$.

Steps to draw the graph:

1. Draw the x-axis and y-axis, intersecting at the origin $(0, 0)$.

2. Choose a suitable scale for both axes.

3. Locate the point $(0, -3)$ on the y-axis. This point is 3 units below the origin.

4. Draw a horizontal line that passes through the point $(0, -3)$. This line is parallel to the x-axis.

This horizontal line represents the graph of the equation $y = -3$. Every point on this line will have a y-coordinate of $-3$.

The solutions of the linear equation are represented by points $(x, y)$ such that the sum of their coordinates is 10 units.

Draw the graph of this linear equation.

Let a point on the graph of the linear equation be $(x, y)$.

According to the given condition, the sum of the coordinates of any solution is 10 units.

So, the sum of the x-coordinate and the y-coordinate must be equal to 10.

This gives us the equation:

This is the linear equation whose graph we need to draw.

To draw the graph of a linear equation, we need to find at least two points that satisfy the equation. We can find points by choosing a value for $x$ or $y$ and solving for the other variable.

Let's find two points:

1. Find the y-intercept (set $x=0$):

Substitute $x = 0$ into equation (i):

$0 + y = 10$

$y = 10$

This gives the point $(0, 10)$. This is the point where the graph cuts the y-axis.

2. Find the x-intercept (set $y=0$):

Substitute $y = 0$ into equation (i):

$x + 0 = 10$

$x = 10$

This gives the point $(10, 0)$. This is the point where the graph cuts the x-axis.

We have found two points $(0, 10)$ and $(10, 0)$ that lie on the graph of the equation $x + y = 10$.

To draw the graph:

1. Draw the x-axis and y-axis on a Cartesian plane.

2. Choose a suitable scale for both axes, such that points up to 10 units can be easily plotted.

3. Plot the point $(0, 10)$ on the y-axis.

4. Plot the point $(10, 0)$ on the x-axis.

5. Draw a straight line passing through these two points.

This straight line is the graph of the linear equation $x + y = 10$.

A linear equation whose graph contains points $(x, y)$ such that the ordinate ($y$-coordinate) is 3 times its abscissa ($x$-coordinate).

The linear equation that describes this relationship.

Let $(x, y)$ be any point on the graph of the required linear equation.

The abscissa of the point is $x$.

The ordinate of the point is $y$.

The given condition is that the ordinate is 3 times its abscissa. This can be written as an equation:

So, the equation is:

$y = 3x$

This is a linear equation in two variables, as it can be written in the standard form $ax + by + c = 0$ by rearranging the terms:

$3x - y = 0$

In this form, $a=3$, $b=-1$, and $c=0$. Since $a \neq 0$ and $b \neq 0$, this is a valid linear equation in two variables.

Every point $(x, y)$ that satisfies the condition ($y$ is 3 times $x$) will lie on the graph of this equation.

The linear equation is $y = 3x$ (or $3x - y = 0$).

The linear equation is $3y = ax + 7$.

The point $(3, 4)$ lies on the graph of this equation.

The value of the constant $a$.

Since the point $(3, 4)$ lies on the graph of the equation $3y = ax + 7$, the coordinates of the point must satisfy the equation when substituted for $x$ and $y$.

The coordinates of the given point are $x = 3$ and $y = 4$.

Substitute $x=3$ and $y=4$ into the equation $3y = ax + 7$:

$3(4) = a(3) + 7$

Simplify the left side of the equation:

$12 = 3a + 7$

Now, we need to solve this equation for $a$. Subtract 7 from both sides of the equation:

$12 - 7 = 3a + 7 - 7$

$5 = 3a$

Divide both sides by 3 to isolate $a$:

$\frac{5}{3} = \frac{3a}{3}$

$a = \frac{5}{3}$

Thus, the value of $a$ is $\frac{5}{3}$.

The value of $a$ is $\frac{5}{3}$.

The equation $2x + 1 = x - 3$.

The number of solutions of the given equation on:

(i) the number line

(ii) the Cartesian plane

First, let's solve the given equation for $x$.

$2x + 1 = x - 3$

Subtract $x$ from both sides of the equation:

$2x - x + 1 = x - x - 3$

$x + 1 = -3$

Subtract 1 from both sides of the equation:

$x + 1 - 1 = -3 - 1$

$x = -4$

The solution to the equation in terms of $x$ is $x = -4$.

(i) On the Number Line:

When we consider the equation $x = -4$ on a number line, the solution is a single point corresponding to the value -4.

Since there is only one specific value of $x$ that satisfies the equation, there is a unique solution on the number line.

(ii) On the Cartesian Plane:

When we consider the equation $x = -4$ on a Cartesian plane, we interpret it as a linear equation in two variables, $x$ and $y$. The equation can be written in the form $ax + by + c = 0$ as $1 \cdot x + 0 \cdot y + 4 = 0$.

For any point $(x, y)$ to be a solution to this equation, the x-coordinate must be equal to -4, while the y-coordinate can be any real number.

The solutions are of the form $(-4, y)$, where $y \in \mathbb{R}$.

For example, $(-4, 0)$, $(-4, 1)$, $(-4, 5)$, $(-4, -2)$, etc., are all solutions.

The set of all such points $(-4, y)$ forms a vertical line in the Cartesian plane that passes through the point $(-4, 0)$ on the x-axis and is parallel to the y-axis.

Since there are infinitely many possible real values for $y$, there are infinitely many points on this line.

Therefore, there are infinitely many solutions on the Cartesian plane.

Number of solutions on the number line: One (Unique solution)

Number of solutions on the Cartesian plane: Infinitely many solutions

The linear equation is $x + 2y = 8$.

The solution of the equation that represents a point on:

(i) the x-axis

(ii) the y-axis

A point on the graph of a linear equation is a solution to the equation. We need to find the points that are simultaneously on the line $x + 2y = 8$ and on the specified axis.

(i) Point on the x-axis:

Any point on the x-axis has a y-coordinate equal to 0. To find the point where the graph of $x + 2y = 8$ meets the x-axis, we set $y=0$ in the equation and solve for $x$.

Substitute $y = 0$ into the equation $x + 2y = 8$:

$x + 2(0) = 8$

$x + 0 = 8$

$x = 8$

So, when $y=0$, the value of $x$ is 8.

The point on the x-axis that is a solution to the equation is $(x, y) = (8, 0)$.

(ii) Point on the y-axis:

Any point on the y-axis has an x-coordinate equal to 0. To find the point where the graph of $x + 2y = 8$ meets the y-axis, we set $x=0$ in the equation and solve for $y$.

Substitute $x = 0$ into the equation $x + 2y = 8$:

$0 + 2y = 8$

$2y = 8$

Divide both sides by 2:

$y = \frac{8}{2}$

$y = 4$

So, when $x=0$, the value of $y$ is 4.

The point on the y-axis that is a solution to the equation is $(x, y) = (0, 4)$.

(i) The solution representing a point on the x-axis is (8, 0).

(ii) The solution representing a point on the y-axis is (0, 4).

The linear equation is $2x + cy = 8$.

The equation has a solution where the values of $x$ and $y$ are equal.

The value of the constant $c$.

We are given that the linear equation $2x + cy = 8$ has a solution $(x, y)$ where $x$ and $y$ have equal values. Let this equal value be $k$. So, the solution is of the form $(k, k)$, where $x=k$ and $y=k$ for some real number $k$.

Substitute $x=k$ and $y=k$ into the given equation:

$2(k) + c(k) = 8$

$2k + ck = 8$

Factor out $k$ from the left side:

For this equation to have a solution for $k$, the coefficient of $k$, which is $(2 + c)$, must not be zero. If $2 + c = 0$ (i.e., $c = -2$), equation (i) becomes $k(0) = 8$, which simplifies to $0 = 8$. This is a contradiction, meaning there is no value of $k$ that satisfies the equation when $c = -2$. Thus, if $c = -2$, there is no solution to the original equation where $x=y$.

If $2 + c \neq 0$ (i.e., $c \neq -2$), we can solve equation (i) for $k$:

This means that for any value of $c$ except $c = -2$, there exists a unique value of $k$, and the point $(\frac{8}{2+c}, \frac{8}{2+c})$ is a solution to the equation $2x + cy = 8$ with equal $x$ and $y$ values.

The question asks for "the value of c", suggesting a specific value. The condition $c \neq -2$ gives infinitely many possible values for $c$ for which such a solution exists.

Let's consider another interpretation often associated with finding a specific parameter value related to $x=y$: the line $2x+cy=8$ is symmetric with respect to the line $y=x$. A line is symmetric with respect to $y=x$ if, whenever a point $(x, y)$ is on the line, the point $(y, x)$ is also on the line. For a line $Ax+By=C$, it is symmetric about $y=x$ if the equation $Bx+Ay=C$ represents the same line.

For the equation $2x + cy = 8$, the symmetric equation (by swapping coefficients of $x$ and $y$) is $cx + 2y = 8$.

For the equations $2x + cy = 8$ and $cx + 2y = 8$ to represent the same line, their coefficients must be proportional:

From the proportion $\frac{c}{2} = \frac{8}{8}$, we get:

$\frac{c}{2} = 1$

$c = 2$

Let's check if this value of $c$ also satisfies the first part of the proportion $\frac{2}{c} = 1$:

$\frac{2}{2} = 1$. This is true.

So, when $c = 2$, the equation is $2x + 2y = 8$, which simplifies to $x + y = 4$. This line is symmetric about $y=x$. A line symmetric about $y=x$ will intersect the line $y=x$ unless it is parallel to it. The line $x+y=4$ is not parallel to $y=x$ (slope -1 vs 1). The intersection point(s) of a line and its axis of symmetry lie on the axis of symmetry, hence $x=y$ at the intersection.

The intersection point is found by setting $y=x$ in $x+y=4$: $x+x=4 \implies 2x=4 \implies x=2$. Thus $y=2$. The point is $(2, 2)$. This point is a solution to $x+y=4$ (and $2x+2y=8$) and has equal $x$ and $y$ values.

Therefore, the value $c=2$ results in a line that has a solution with equal $x$ and $y$ values.

Given the phrasing asking for "the value of c" and the calculation showing that such a solution exists for any $c \neq -2$, the interpretation related to symmetry provides a unique value for $c$ that ties the structure of the equation to the condition $x=y$. Assuming this geometric property is implicitly intended, the value of $c$ is 2.

The value of $c$ is 2.

It is given that $y$ varies directly as $x$.

When $x = 4$, $y = 12$.

1. Write a linear equation representing the direct variation.

2. Find the value of $y$ when $x = 5$.

When $y$ varies directly as $x$, the relationship between $y$ and $x$ can be expressed as:

This proportionality can be written as a linear equation by introducing a constant of variation, let's call it $k$:

where $k$ is a non-zero constant.

We are given that when $x = 4$, $y = 12$. We can use these values to find the constant $k$.

Substitute $x=4$ and $y=12$ into equation (i):

$12 = k(4)$

To solve for $k$, divide both sides by 4:

$\frac{12}{4} = k$

$k = 3$

Now that we have the value of the constant of variation, we can write the complete linear equation representing the relationship between $x$ and $y$. Substitute $k=3$ back into equation (i):

This is the required linear equation.

Next, we need to find the value of $y$ when $x = 5$.

Use the linear equation $y = 3x$ and substitute $x=5$:

$y = 3(5)$

$y = 15$

So, the value of $y$ is 15 when $x$ is 5.

The linear equation is $y = 3x$.

The value of $y$ when $x = 5$ is 15.

The linear equation is $2x + 3y = 12$.

1. Draw the graph of the given linear equation.

2. Find the points where the graph cuts the x-axis and the y-axis.

To draw the graph of a linear equation in two variables, we need to find at least two points that satisfy the equation. A straight line passes through these points.

The points where the graph cuts the axes are particularly easy to find.

Finding the x-intercept:

The graph cuts the x-axis at the point where the y-coordinate is 0.

Substitute $y = 0$ into the equation $2x + 3y = 12$:

$2x + 3(0) = 12$

$2x + 0 = 12$

$2x = 12$

Divide both sides by 2:

$x = \frac{12}{2}$

$x = 6$

The point where the graph cuts the x-axis is $(6, 0)$.

Finding the y-intercept:

The graph cuts the y-axis at the point where the x-coordinate is 0.

Substitute $x = 0$ into the equation $2x + 3y = 12$:

$2(0) + 3y = 12$

$0 + 3y = 12$

$3y = 12$

Divide both sides by 3:

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

$y = 4$

The point where the graph cuts the y-axis is $(0, 4)$.

Drawing the graph:

We have found two points on the line: $(6, 0)$ and $(0, 4)$.

1. Draw the x-axis and y-axis on a Cartesian plane, intersecting at the origin $(0, 0)$.

2. Choose a suitable scale for both axes.

3. Plot the point $(6, 0)$ on the x-axis.

4. Plot the point $(0, 4)$ on the y-axis.

5. Draw a straight line passing through the points $(6, 0)$ and $(0, 4)$. Extend the line in both directions beyond these points.

This straight line represents the graph of the linear equation $2x + 3y = 12$.

The graph of the equation $2x + 3y = 12$ cuts the x-axis at the point (6, 0).

The graph of the equation $2x + 3y = 12$ cuts the y-axis at the point (0, 4).

Two points $(x, y)$ that satisfy a linear equation are given in a table:

1. Draw the graph of the linear equation using the given points.

2. Find the point where the graph cuts the x-axis.

3. Find the point where the graph cuts the y-axis.

We are given two points that lie on the graph of a linear equation: $(0, 1)$ and $(1, 3)$. Since a straight line is uniquely determined by two distinct points, we can draw the graph by plotting these two points and drawing a line through them.

Drawing the graph:

1. Draw the x-axis and y-axis on a Cartesian plane, intersecting at the origin $(0, 0)$.

2. Choose a suitable scale for both axes.

3. Plot the first point $(0, 1)$. This point is on the y-axis, 1 unit above the origin.

4. Plot the second point $(1, 3)$. Move 1 unit right from the origin along the x-axis, and then 3 units up parallel to the y-axis.

5. Draw a straight line passing through the points $(0, 1)$ and $(1, 3)$. Extend the line in both directions.

This straight line is the graph of the linear equation satisfied by the given points.

Now, let's find the points where this graph cuts the x-axis and the y-axis.

(i) Point where the graph cuts the x-axis:

The graph cuts the x-axis at the point where the y-coordinate is 0. This is the x-intercept.

We can find the equation of the line first. The points are $(0, 1)$ and $(1, 3)$. The slope $m$ is given by:

$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - 1}{1 - 0} = \frac{2}{1} = 2$

The equation of the line can be found using the point-slope form $y - y_1 = m(x - x_1)$. Using the point $(0, 1)$:

$y - 1 = 2(x - 0)$

$y - 1 = 2x$

$y = 2x + 1$

To find where the graph cuts the x-axis, set $y = 0$ in the equation:

$0 = 2x + 1$

Subtract 1 from both sides:

$-1 = 2x$

Divide by 2:

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

The point where the graph cuts the x-axis is $(-\frac{1}{2}, 0)$.

(ii) Point where the graph cuts the y-axis:

The graph cuts the y-axis at the point where the x-coordinate is 0. This is the y-intercept.

Set $x = 0$ in the equation $y = 2x + 1$:

$y = 2(0) + 1$

$y = 0 + 1$

$y = 1$

The point where the graph cuts the y-axis is $(0, 1)$. This point was already given in the table.

(i) The graph of the linear equation cuts the x-axis at the point $(-\frac{1}{2}, 0)$.

(ii) The graph of the linear equation cuts the y-axis at the point (0, 1).

Autorickshaw fare structure:

$\textsf{₹}10$ for the first kilometer.

$\textsf{₹}4$ per kilometer for subsequent distance covered.

1. Write the linear equation expressing the above statement.

2. Draw the graph of the linear equation.

Let the total distance covered be $x$ kilometers, where $x \ge 0$.

Let the total fare be $y$ Rupees, where $y \ge 0$.

The fare structure distinguishes between the first kilometer and any subsequent distance.

For the first kilometer, the fare is a fixed amount of $\textsf{₹}10$.

For distances greater than 1 kilometer (i.e., when $x > 1$), the total distance $x$ can be thought of as the first 1 kilometer plus the remaining $(x-1)$ kilometers.

The fare for the first 1 kilometer is $\textsf{₹}10$.

The fare for the subsequent $(x-1)$ kilometers is charged at $\textsf{₹}4$ per kilometer. So, the charge for the subsequent distance is $4 \times (x-1)$.

The total fare $y$ for a distance $x > 1$ is the sum of the fare for the first kilometer and the fare for the subsequent distance:

Let's simplify this equation:

This linear equation $y = 4x + 6$ represents the relationship between the total distance $x$ and the total fare $y$ for distances $x \ge 1$. Assuming this is the linear equation intended to be graphed, we proceed to draw its graph.

The linear equation is $y = 4x + 6$.

Drawing the graph of $y = 4x + 6$:

To draw the graph of a linear equation, we need to find at least two points that satisfy the equation. Let's find two points by choosing values for $x$ and calculating the corresponding $y$ values using the equation $y = 4x + 6$.

Let's choose two simple values for $x$, for example, $x=0$ and $x=1$ (noting that $x=0$ might not represent a covered distance in the physical context, but it helps in drawing the line):

• When $x = 0$: $y = 4(0) + 6 = 0 + 6 = 6$. This gives the point $(0, 6)$.
• When $x = 1$: $y = 4(1) + 6 = 4 + 6 = 10$. This gives the point $(1, 10)$.

We can use these two points $(0, 6)$ and $(1, 10)$ to draw the straight line representing the graph of $y = 4x + 6$.

Steps to draw the graph:

1. Draw the x-axis and y-axis on a Cartesian plane, intersecting at the origin (0,0).

2. Choose a suitable scale for both axes, ensuring that points like $(0, 6)$ and $(1, 10)$ can be plotted accurately.

3. Plot the point $(0, 6)$ on the y-axis. This is the y-intercept of the line.

4. Plot the point $(1, 10)$. Move 1 unit right from the origin along the x-axis, and then 10 units up parallel to the y-axis.

5. Draw a straight line passing through the points $(0, 6)$ and $(1, 10)$. Extend the line in both directions.

This straight line is the graph of the linear equation $y = 4x + 6$. In the context of the fare problem, the physically relevant part of the graph is typically for $x \ge 0$.

Work done by a body on application of a constant force is the product of the constant force and the distance travelled by the body in the direction of force.

1. Express this relationship in the form of a linear equation in two variables.

2. Draw the graph of this linear equation.

Let $W$ represent the work done by the body.

Let $d$ represent the distance travelled by the body in the direction of the force.

Let $F$ represent the constant force applied.

According to the given statement, the work done is the product of the constant force and the distance travelled:

To express this in the form of a linear equation in two variables, let's consider Work Done and Distance as the two variables. Let:

Work Done ($W$) be represented by $y$ (ordinate).

Distance Travelled ($d$) be represented by $x$ (abscissa).

The constant force $F$ is a fixed value, let's represent it by a constant $k$.

Substituting these into the formula, we get the linear equation:

This is a linear equation in two variables $x$ and $y$, where $k$ is the constant force.

The linear equation is $y = kx$, where $k$ is the constant force.

Drawing the graph of the linear equation $y = kx$:

The equation $y = kx$ represents a straight line. To draw the graph, we need specific values for $k$. Since the problem does not provide a value for the constant force $F$ (or $k$), we must assume a suitable non-zero value to draw a specific graph. Let's assume the constant force $F = 5$ units. So, the equation becomes $y = 5x$.

To draw the graph of $y = 5x$, we find at least two points that satisfy this equation.

• When $x = 0$: $y = 5(0) = 0$. The point is $(0, 0)$.
• When $x = 1$: $y = 5(1) = 5$. The point is $(1, 5)$.
• When $x = 2$: $y = 5(2) = 10$. The point is $(2, 10)$.

The graph is the straight line passing through these points, including the origin $(0, 0)$.

Steps to draw the graph (assuming $F=5$, so $y=5x$):

1. Draw the x-axis (representing distance $d$) and the y-axis (representing work done $W$) on a Cartesian plane, intersecting at the origin $(0, 0)$. Label the axes appropriately (e.g., 'Distance (x)' and 'Work Done (y)').

2. Choose a suitable scale for both axes.

3. Plot the point $(0, 0)$.

4. Plot another point, such as $(1, 5)$. Move 1 unit right along the x-axis, then 5 units up parallel to the y-axis.

5. Draw a straight line passing through the points $(0, 0)$ and $(1, 5)$. Extend the line in both directions.

This straight line represents the graph of the linear equation $y = 5x$. Note that in the physical context, distance $x$ is typically non-negative ($x \ge 0$), and if the force is positive, work done $y$ will also be non-negative ($y \ge 0$). Therefore, the part of the line in the first quadrant is the most physically relevant.

The linear equation is $y = 9x - 7$.

The points are A (1, 2), B (– 1, – 16), and C (0, – 7).

That the points A (1, 2), B (– 1, – 16), and C (0, – 7) lie on the graph of the equation $y = 9x - 7$.

For a point $(x, y)$ to lie on the graph of the linear equation $y = 9x - 7$, the coordinates of the point must satisfy the equation when substituted.

Let's check each point:

For point A (1, 2):

Substitute $x = 1$ and $y = 2$ into the equation $y = 9x - 7$.

LHS = $y = 2$

RHS = $9x - 7 = 9(1) - 7 = 9 - 7 = 2$

Since LHS = RHS ($2 = 2$), the point A (1, 2) satisfies the equation $y = 9x - 7$.

For point B (– 1, – 16):

Substitute $x = -1$ and $y = -16$ into the equation $y = 9x - 7$.

LHS = $y = -16$

RHS = $9x - 7 = 9(-1) - 7 = -9 - 7 = -16$

Since LHS = RHS ($-16 = -16$), the point B (– 1, – 16) satisfies the equation $y = 9x - 7$.

For point C (0, – 7):

Substitute $x = 0$ and $y = -7$ into the equation $y = 9x - 7$.

LHS = $y = -7$

RHS = $9x - 7 = 9(0) - 7 = 0 - 7 = -7$

Since LHS = RHS ($-7 = -7$), the point C (0, – 7) satisfies the equation $y = 9x - 7$.

Since all three points A (1, 2), B (– 1, – 16), and C (0, – 7) satisfy the equation $y = 9x - 7$, they lie on the graph of the linear equation $y = 9x - 7$.

Two points $(x, y)$ that satisfy a linear equation are given in a table:

1. Write the linear equation satisfied by the given points.

2. Draw the graph of this linear equation.

3. Find the points where the graph cuts the x-axis and the y-axis.

Let the linear equation be of the form $y = mx + c$, where $m$ is the slope and $c$ is the y-intercept.

We are given two points on the line: $(x_1, y_1) = (6, -2)$ and $(x_2, y_2) = (-6, 6)$.

First, calculate the slope $m$ using the formula $m = \frac{y_2 - y_1}{x_2 - x_1}$.

$m = \frac{6 - (-2)}{-6 - 6}$

$m = \frac{6 + 2}{-12}$

$m = \frac{8}{-12}$

Simplify the fraction for $m$:

$m = -\frac{\cancel{8}^{2}}{\cancel{12}_{3}}$

$m = -\frac{2}{3}$

Now, use the point-slope form of the equation of a line, $y - y_1 = m(x - x_1)$, with $m = -\frac{2}{3}$ and one of the points, say $(6, -2)$.

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

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

$y + 2 = -\frac{2}{3}x + \frac{12}{3}$

$y + 2 = -\frac{2}{3}x + 4$

Subtract 2 from both sides to solve for $y$:

$y = -\frac{2}{3}x + 4 - 2$

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

This is the linear equation satisfied by the given points. We can also write it in the standard form $Ax + By = C$ by multiplying by 3:

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

$3y = -2x + 6$

Add $2x$ to both sides:

$2x + 3y = 6$

The linear equation is $y = -\frac{2}{3}x + 2$ (or $2x + 3y = 6$).

Drawing the graph:

We can use the two given points, $(6, -2)$ and $(-6, 6)$, to draw the graph of the linear equation.

1. Draw the x-axis and y-axis on a Cartesian plane, intersecting at the origin $(0, 0)$.

2. Choose a suitable scale for both axes that can accommodate coordinates up to 6 and -6.

3. Plot the point $(6, -2)$. Move 6 units right from the origin along the x-axis, and then 2 units down parallel to the y-axis.

4. Plot the point $(-6, 6)$. Move 6 units left from the origin along the x-axis, and then 6 units up parallel to the y-axis.

5. Draw a straight line passing through the points $(6, -2)$ and $(-6, 6)$. Extend the line in both directions.

This straight line is the graph of the linear equation $2x + 3y = 6$.

Finding the points where the graph cuts the axes:

(i) Point where the graph cuts the x-axis:

The graph cuts the x-axis at the point where the y-coordinate is 0. Substitute $y=0$ into the equation $2x + 3y = 6$:

$2x + 3(0) = 6$

$2x = 6$

Divide both sides by 2:

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

$x = 3$

The graph cuts the x-axis at the point $(3, 0)$.

(ii) Point where the graph cuts the y-axis:

The graph cuts the y-axis at the point where the x-coordinate is 0. Substitute $x=0$ into the equation $2x + 3y = 6$:

$2(0) + 3y = 6$

$3y = 6$

Divide both sides by 3:

$y = \frac{6}{3}$

$y = 2$

The graph cuts the y-axis at the point $(0, 2)$.

(i) The graph of the linear equation cuts the x-axis at the point (3, 0).

(ii) The graph of the linear equation cuts the y-axis at the point (0, 2).

The linear equation is $3x + 4y = 6$.

1. Draw the graph of the given linear equation.

2. Find the points where the graph cuts the x-axis and the y-axis.

To draw the graph of the linear equation $3x + 4y = 6$, we need to find at least two points that satisfy the equation. It is convenient to find the points where the graph intersects the coordinate axes (the x-intercept and the y-intercept).

Finding the x-intercept:

The graph cuts the x-axis at the point where the y-coordinate is 0.

Substitute $y = 0$ into the equation $3x + 4y = 6$:

$3x + 4(0) = 6$

$3x + 0 = 6$

$3x = 6$

Divide both sides by 3:

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

$x = 2$

The point where the graph cuts the x-axis is $(2, 0)$.

Finding the y-intercept:

The graph cuts the y-axis at the point where the x-coordinate is 0.

Substitute $x = 0$ into the equation $3x + 4y = 6$:

$3(0) + 4y = 6$

$0 + 4y = 6$

$4y = 6$

Divide both sides by 4:

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

Simplify the fraction:

$y = \frac{\cancel{6}^{3}}{\cancel{4}_{2}}$

$y = \frac{3}{2}$

The point where the graph cuts the y-axis is $(0, \frac{3}{2})$. (Note: $\frac{3}{2} = 1.5$)

Drawing the graph:

We have found two points on the line: $(2, 0)$ and $(0, \frac{3}{2})$.

1. Draw the x-axis and y-axis on a Cartesian plane, intersecting at the origin $(0, 0)$.

2. Choose a suitable scale for both axes.

3. Plot the point $(2, 0)$ on the x-axis.

4. Plot the point $(0, \frac{3}{2})$ or $(0, 1.5)$ on the y-axis.

5. Draw a straight line passing through the points $(2, 0)$ and $(0, \frac{3}{2})$. Extend the line in both directions beyond these points.

This straight line represents the graph of the linear equation $3x + 4y = 6$.

The graph of the equation $3x + 4y = 6$ cuts the x-axis at the point (2, 0).

The graph of the equation $3x + 4y = 6$ cuts the y-axis at the point (0, $\frac{3}{2}$).

The linear equation for converting Fahrenheit (F) to Celsius (C) is $C = \frac{5F - 160}{9}$.

(i) Temperature in Celsius when Fahrenheit is 86°F.

(ii) Temperature in Fahrenheit when Celsius is 35°C.

(iii) Temperature in Fahrenheit when Celsius is 0°C, and temperature in Celsius when Fahrenheit is 0°F.

(iv) The numerical value of the temperature which is the same in both scales.

The given equation is $C = \frac{5F - 160}{9}$. We can rearrange this equation to express F in terms of C if needed:

$9C = 5F - 160$

$9C + 160 = 5F$

(i) Find C when F = 86°F:

Substitute $F = 86$ into the given equation $C = \frac{5F - 160}{9}$:

$C = \frac{5(86) - 160}{9}$

$C = \frac{430 - 160}{9}$

$C = \frac{270}{9}$

$C = 30$

So, when the temperature is 86°F, the temperature in Celsius is 30°C.

(ii) Find F when C = 35°C:

Substitute $C = 35$ into the rearranged equation $F = \frac{9C + 160}{5}$ (or use the original equation):

Using $F = \frac{9C + 160}{5}$:

$F = \frac{9(35) + 160}{5}$

$F = \frac{315 + 160}{5}$

$F = \frac{475}{5}$

$F = 95$

So, when the temperature is 35°C, the temperature in Fahrenheit is 95°F.

(iii) Find F when C = 0°C and C when F = 0°F:

Find F when C = 0°C:

Substitute $C = 0$ into the rearranged equation $F = \frac{9C + 160}{5}$:

$F = \frac{9(0) + 160}{5}$

$F = \frac{0 + 160}{5}$

$F = \frac{160}{5}$

$F = 32$

When the temperature is 0°C, the temperature in Fahrenheit is 32°F.

Find C when F = 0°F:

Substitute $F = 0$ into the original equation $C = \frac{5F - 160}{9}$:

$C = \frac{5(0) - 160}{9}$

$C = \frac{0 - 160}{9}$

$C = -\frac{160}{9}$

When the temperature is 0°F, the temperature in Celsius is $-\frac{160}{9}$°C.

(iv) Find the numerical value of the temperature which is same in both scales:

We need to find a temperature value $T$ such that $C = T$ and $F = T$. Substitute $C=T$ and $F=T$ into the given equation $C = \frac{5F - 160}{9}$:

$T = \frac{5T - 160}{9}$

Multiply both sides by 9:

$9T = 5T - 160$

Subtract $5T$ from both sides:

$9T - 5T = -160$

$4T = -160$

Divide both sides by 4:

$T = \frac{-160}{4}$

$T = -40$

So, the temperature $-40$ is the same in both Fahrenheit and Celsius scales.

The numerical value of the temperature which is same in both the scales is -40.

The linear equation relating Fahrenheit ($y^\circ$F) and Kelvin ($x^\circ$K) is $y = \frac{9}{5} (x – 273) + 32$.

(i) Temperature of the liquid in Fahrenheit when the temperature is 313°K.

(ii) Temperature of the liquid in Kelvin when the temperature is 158°F.

The given linear equation is:

(i) Find the temperature in Fahrenheit if the temperature is 313°K:

We are given the temperature in Kelvin, so $x = 313$. Substitute this value into equation (i):

$y = \frac{9}{5} (313 – 273) + 32$

First, calculate the value inside the parenthesis:

$313 - 273 = 40$

Now substitute this back into the equation:

$y = \frac{9}{5} (40) + 32$

Multiply $\frac{9}{5}$ by 40:

$y = 9 \times \frac{40}{5} + 32$

$y = 9 \times 8 + 32$

$y = 72 + 32$

Finally, add the numbers:

$y = 104$

Thus, the temperature of the liquid in Fahrenheit is 104°F when the temperature in Kelvin is 313°K.

(ii) If the temperature is 158°F, then find the temperature in Kelvin:

We are given the temperature in Fahrenheit, so $y = 158$. Substitute this value into equation (i):

$158 = \frac{9}{5} (x – 273) + 32$

We need to solve this equation for $x$. First, isolate the term involving $x$. Subtract 32 from both sides of the equation:

$158 - 32 = \frac{9}{5} (x – 273) + 32 - 32$

$126 = \frac{9}{5} (x – 273)$

Now, multiply both sides by $\frac{5}{9}$ to isolate the term $(x - 273)$: (This is equivalent to multiplying by 5 and then dividing by 9)

$\frac{5}{9} \times 126 = \frac{5}{9} \times \frac{9}{5} (x – 273)$

$\frac{5 \times 126}{9} = x – 273$

Calculate the value on the left side: $\frac{126}{9} = 14$.

$5 \times 14 = x – 273$

$70 = x – 273$

Finally, add 273 to both sides to solve for $x$:

$70 + 273 = x – 273 + 273$

$343 = x$

Thus, the temperature of the liquid in Kelvin is 343°K when the temperature in Fahrenheit is 158°F.

The force exerted to pull a cart is directly proportional to the acceleration produced in the body.

Constant mass of the body is 6 kg.

1. Express the statement as a linear equation in two variables.

2. Draw the graph of the equation with the constant mass equal to 6 kg.

3. Read from the graph, the force required when the acceleration is 5 m/sec$^2$ and 6 m/sec$^2$.

According to Newton's second law of motion, the force ($F$) applied to a body is directly proportional to the acceleration ($a$) produced in it, and the constant of proportionality is the mass ($m$) of the body. The relationship is given by the formula:

To express this as a linear equation in two variables, let the two variables be the acceleration and the force.

Let the acceleration be represented by $x$ (abscissa).

Let the force be represented by $y$ (ordinate).

The mass $m$ is given as a constant value, $m = 6$ kg.

Substitute these into the formula $F = m \times a$:

The linear equation is:

This is a linear equation in two variables $x$ and $y$.

Drawing the graph of $y = 6x$:

To draw the graph of a linear equation, we need at least two points that satisfy the equation. Let's find a few points by choosing values for $x$ (acceleration) and calculating the corresponding $y$ (force) values using the equation $y = 6x$. In this physical context, acceleration ($x$) is typically non-negative, which implies force ($y$) is also non-negative.

The graph is the straight line passing through these points. Note that the unit of force is Newtons (N), given that mass is in kg and acceleration is in m/sec$^2$.

Steps to draw the graph:

1. Draw the x-axis (representing acceleration) and the y-axis (representing force) on a Cartesian plane, intersecting at the origin $(0, 0)$. Label the axes appropriately (e.g., 'Acceleration (m/sec$^2$)' and 'Force (N)').

2. Choose a suitable scale for both axes. Since the values go up to 12 on the y-axis, choose a scale that accommodates this.

3. Plot the points $(0, 0)$, $(1, 6)$, $(2, 12)$, etc. (Plotting more points helps ensure the line is correct).

4. Draw a straight line passing through the plotted points. Since acceleration is typically non-negative in this context, the relevant part of the graph starts from the origin and extends into the first quadrant.

Reading from the graph:

We need to find the force ($y$) when the acceleration ($x$) is given. On the graph of $y=6x$, find the given value of $x$ on the x-axis, then move vertically up to the line, and then move horizontally left to the y-axis to read the corresponding value of $y$. Alternatively, we can use the equation $y=6x$ directly, which is equivalent to reading from the graph when the graph is drawn accurately.

(i) Force required when acceleration produced is 5 m/sec$^2$:

Using the equation $y = 6x$, substitute $x=5$:

$y = 6(5)$

$y = 30$

From the graph, locate $x=5$ on the acceleration axis, move up to the line $y=6x$, then move left to the force axis. You should read the value 30 N.

The force required is 30 N.

(ii) Force required when acceleration produced is 6 m/sec$^2$:

Using the equation $y = 6x$, substitute $x=6$:

$y = 6(6)$

$y = 36$

From the graph, locate $x=6$ on the acceleration axis, move up to the line $y=6x$, then move left to the force axis. You should read the value 36 N.

The force required is 36 N.