Chapter 4 Linear Equations In Two Variables (Class 9 - Maths NCERT Exemplar Solutions)

Given:

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

To Find:

The number of solutions for the given equation.

Solution:

First, we simplify the given equation by grouping the like terms on one side:

$3x - x - y = -1$

The simplified equation is a linear equation in two variables ($x$ and $y$). Geometrically, a linear equation in two variables represents a straight line on the Cartesian plane.

Every point $(x, y)$ that lies on this straight line is a solution to the equation. Since a straight line is made up of an infinite number of points, the linear equation has infinitely many solutions.

Correct Option: (C) Infinitely many solutions

Given:

A general linear equation $ax + by + c = 0$.

To Find:

The conditions on the coefficients $a$ and $b$ such that the equation remains a linear equation in two variables.

Solution:

For an equation to be classified as a linear equation in two variables, both variables (in this case $x$ and $y$) must be present in the equation.

If $a = 0$, the term $ax$ becomes zero, and the equation reduces to $by + c = 0$, which is a linear equation in only one variable ($y$).

Similarly, if $b = 0$, the term $by$ becomes zero, and the equation reduces to $ax + c = 0$, which is a linear equation in only one variable ($x$).

Therefore, for the equation to be a linear equation in two variables, both $a$ and $b$ must be non-zero simultaneously.

Correct Option: (A) a ≠ 0, b ≠ 0

Given:

A point lying on the $y$-axis.

To Find:

The general form of the coordinates of any point on the $y$-axis.

Solution:

In a Cartesian coordinate system, any point is represented as $(x, y)$, where $x$ is the abscissa (distance from the $y$-axis) and $y$ is the ordinate (distance from the $x$-axis).

If a point lies on the $y$-axis, its perpendicular distance from the $y$-axis itself is zero. This implies that its $x$-coordinate (abscissa) must be $0$.

The $y$-coordinate can be any real number, which we represent as $y$. Thus, the coordinates of any point on the $y$-axis take the form $(0, y)$.

Correct Option: (C) (0, y)

Given:

A linear equation in two variables:

Solution:

The given equation is a linear equation in two variables ($x$ and $y$).

We can express $y$ in terms of $x$ as follows:

For every real value of $x$, there is a corresponding real value of $y$. Since there are infinitely many real numbers that can be substituted for $x$, there are infinitely many pairs $(x, y)$ that satisfy the equation.

In the Cartesian plane, a linear equation in two variables represents a straight line, and every point on this line is a solution to the equation. Since a line contains infinitely many points, the equation has infinitely many solutions.

Correct Option: (C) Infinitely many solutions

Given:

The linear equation:

Solution:

We need to find the set of numbers for which the equation has only one solution.

Let's test Natural numbers (where $x, y \in \{1, 2, 3, ...\}$):

If $x = 1$:

If $x = 2$:

If $x \geq 4$, then $2x \geq 8$, which would make $5y$ negative, but $y$ must be a natural number ($y \geq 1$).

Thus, $(1, 1)$ is the unique solution in the set of natural numbers.

For Real, Positive Real, or Rational numbers, there would be infinitely many solutions.

Correct Option: (A) Natural numbers

Given:

Linear equation: $2x + 3y = k$

Solution point: $(x, y) = (2, 0)$

To Find:

The value of $k$.

Solution:

Since $(2, 0)$ is a solution of the given equation, it must satisfy the equation when substituted.

Substituting $x = 2$ and $y = 0$ in the equation:

Correct Option: (A) 4

Given:

The linear equation in two variables:

Solution:

We can simplify the given equation to find the value of $x$:

In the equation $2x + 0y + 9 = 0$, the coefficient of $y$ is $0$. This means that no matter what value we choose for $y$, the term $0y$ will always be $0$.

Let $y = m$, where $m$ is any real number.

Then the solution $(x, y)$ will always be in the form $\left( -\frac{9}{2}, m \right)$.

Correct Option: (A) $\left( -\frac{9}{2} , m \right)$

Given:

The linear equation: $2x + 3y = 6$

To Find:

The point where the graph cuts the $y$-axis.

Solution:

A graph cuts the $y$-axis at a point where the $x$-coordinate is zero.

Substituting $x = 0$ in the given equation:

The point is $(x, y) = (0, 2)$.

Correct Option: (D) (0, 2)

Given:

An equation in one variable:

Solution:

A linear equation in two variables $x$ and $y$ is generally expressed in the form $ax + by + c = 0$ or $ax + by = d$.

In the given equation $x = 7$, the variable $y$ is missing. To represent this in two variables, we can assume the coefficient of $y$ to be $0$.

The term $x$ can be written as $1 \cdot x$ and the absence of $y$ can be represented as $0 \cdot y$.

Substituting these into the equation, we get:

This matches the form required for a linear equation in two variables where the value of $x$ remains 7 regardless of the value of $y$.

Correct Option: (B) 1.x + 0.y = 7

Solution:

In the Cartesian coordinate system, the x-axis is the horizontal line where the vertical displacement is zero.

For any point located on the $x$-axis, the perpendicular distance from the $x$-axis (which is the $y$-coordinate) must be $0$.

Therefore, while the $x$-coordinate can be any real number $x$, the $y$-coordinate is strictly $0$.

The general coordinates for such a point are expressed as:

Correct Option: (C) (x, 0)

Given:

Equation of the line:

Solution:

The equation $y = x$ implies that for any point $(x, y)$ lying on this line, the ordinate ($y$-coordinate) is always equal to the abscissa ($x$-coordinate).

If we let the $x$-coordinate be a real number $a$, then according to the equation:

Thus, the point must be of the form $(a, a)$.

Correct Option: (A) (a, a)

Solution:

The $x$-axis is a horizontal line passing through the origin $(0,0)$.

Every point on the $x$-axis has a $y$-coordinate (ordinate) equal to $0$. Whether the point is $(1, 0)$, $(-5, 0)$, or $(\pi, 0)$, the value of $y$ never changes.

This constant property defines the equation of the line.

Hence, the equation representing the $x$-axis is:

Correct Option: (B) y = 0

Given:

Equation of the graph:

Solution:

The equation $y = 6$ can be written in two variables as $0 \cdot x + 1 \cdot y = 6$.

For any value of $x$, the value of $y$ is always $6$. Let's look at some points on this graph:

$(0, 6), (1, 6), (-2, 6), (5, 6)$ etc.

When these points are plotted on a Cartesian plane and joined, they form a horizontal line.

Since the line is horizontal, it is parallel to the x-axis. Since the $y$-value is constant at $6$, the line is at a distance of $6$ units from the $x$-axis (and thus from the origin along the $y$-direction).

Correct Option: (A) parallel to x-axis at a distance 6 units from the origin

Given:

Values: $x = 5$ and $y = 2$.

Solution:

To find which equation has $(5, 2)$ as a solution, we substitute these values into each given option:

For (A): $x + 2y = 7$

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

For (C): $x + y = 7$

For (D): $5x + y = 7$

Since the values satisfy the equation in option (C), it is the correct answer.

Correct Option: (C) x + y = 7

Given:

Solutions of the linear equation are $(-2, 2)$, $(0, 0)$, and $(2, -2)$.

Solution:

Let's check which equation form is satisfied by all three points:

For the point $(0, 0)$, all options (A), (B), (C), and (D) are satisfied as they all pass through the origin.

Let's test the point $(2, -2)$ where $x = 2$ and $y = -2$:

Option (A): $y - x = -2 - 2 = -4 \neq 0$

Option (B): $x + y = 2 + (-2) = 0$

Option (C): $-2x + y = -2(2) + (-2) = -4 - 2 = -6 \neq 0$

Option (D): $-x + 2y = -2 + 2(-2) = -2 - 4 = -6 \neq 0$

We can verify option (B) with the point $(-2, 2)$ as well:

Since all points satisfy $x + y = 0$, the equation is of this form.

Correct Option: (B) x + y = 0

Solution:

The term "positive solutions" refers to the ordered pairs $(x, y)$ where both the coordinates are positive.

In the Cartesian plane:

1. In the 1st Quadrant, $x > 0$ and $y > 0$.

2. In the 2nd Quadrant, $x < 0$ and $y > 0$.

3. In the 3rd Quadrant, $x < 0$ and $y < 0$.

4. In the 4th Quadrant, $x > 0$ and $y < 0$.

Since a positive solution implies that both $x$ and $y$ must be positive real numbers, such points must always lie in the first quadrant.

Correct Option: (A) 1st quadrant

Given:

Equation: $2x + 3y = 6$

Solution:

A line meets the x-axis at a point where the $y$-coordinate is zero.

Substituting $y = 0$ in the given equation:

The coordinates of the point are $(3, 0)$.

Correct Option: (C) (3, 0)

Given:

The linear equation: $y = x$

Solution:

The equation $y = x$ states that for any point on its graph, the $x$-coordinate and $y$-coordinate must be exactly equal.

Let's evaluate the given points:

(A) $\left( \frac{3}{2}, -\frac{3}{2} \right)$: Here $x = \frac{3}{2}$ and $y = -\frac{3}{2}$. Since $\frac{3}{2} \neq -\frac{3}{2}$, this point does not lie on the line.

(B) $\left( 0, \frac{3}{2} \right)$: Here $x = 0$ and $y = \frac{3}{2}$. Since $0 \neq \frac{3}{2}$, this point does not lie on the line.

(C) $(1, 1)$: Here $x = 1$ and $y = 1$. Since $1 = 1$, the coordinates satisfy the equation $y = x$.

(D) $\left( -\frac{1}{2}, \frac{1}{2} \right)$: Here $x = -\frac{1}{2}$ and $y = \frac{1}{2}$. Since $-\frac{1}{2} \neq \frac{1}{2}$, this point does not lie on the line.

Thus, the graph passes through $(1, 1)$.

Correct Option: (C) (1, 1)

Given:

A linear equation and a non-zero number $k$.

Solution:

Consider a general linear equation in two variables:

According to the properties of equality, if we multiply both sides of an equation by a non-zero constant $k$, we get:

Similarly, dividing by $k$ (where $k \neq 0$):

Equations (i), (ii), and (iii) are equivalent equations. This means they represent the same line on a graph and are satisfied by the same set of ordered pairs $(x, y)$.

Therefore, the solution set of the linear equation does not change.

Correct Option: (B) Remains the same

Given:

A point $(x, y) = (1, 2)$.

Solution:

A linear equation in two variables represents a straight line in the Cartesian plane. The question asks how many such lines can pass through the point $(1, 2)$.

From the fundamental principles of geometry, we know that through a single given point, infinitely many lines can be drawn.

Examples of such equations satisfied by $(1, 2)$ are:

1. $x + y = 3$ (as $1 + 2 = 3$)

2. $y = 2x$ (as $2 = 2 \cdot 1$)

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

4. $x - y = -1$ (as $1 - 2 = -1$)

By changing the coefficients of $x$ and $y$, we can generate an unlimited number of linear equations that are satisfied by the coordinates $(1, 2)$.

Correct Option: (C) Infinitely many

Given:

A point with coordinates $(a, a)$.

Solution:

In the given coordinate pair $(a, a)$:

Comparing the values of $x$ and $y$:

This means that for any real number $a$, the point $(a, a)$ satisfies the equation $y = x$. Such points include $(1, 1)$, $(2, 2)$, $(-3, -3)$, etc.

These points always lie on the line where the ordinate is equal to the abscissa.

Correct Option: (C) On the line y = x

Given:

A point with coordinates $(a, -a)$.

Solution:

In the coordinate pair $(a, -a)$:

Let's find the relationship between $x$ and $y$ by adding them:

This indicates that the sum of the coordinates is always zero. Any point where the $y$-coordinate is the negative of the $x$-coordinate will satisfy the linear equation $x + y = 0$.

Examples include $(5, -5)$, $(-2, 2)$, and $(0, 0)$.

Correct Option: (D) x + y = 0

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

Result: False

Justification:

A linear equation in two variables $x$ and $y$ is of the form $ax + by + c = 0$, but there is a necessary condition that both $a$ and $b$ cannot be zero simultaneously.

If $a = 0$ and $b = 0$, the equation reduces to $c = 0$, which is no longer an equation in two variables. Thus, the statement is false as it lacks this condition.

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

Result: False

Justification:

A linear equation in two variables represents a straight line on a Cartesian plane. Every point on this line is a solution to the equation.

Since a line consists of an infinite number of points, a linear equation in two variables has infinitely many solutions.

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

Result: False

Justification:

A point $(x, y)$ lies on the $x$-axis only if its ordinate (y-coordinate) is zero.

1. For $(2, 0)$ and $(-3, 0)$, the ordinate is $0$. (They lie on the $x$-axis).

2. For $(4, 2)$, the ordinate is $2 \neq 0$. (It lies in the I quadrant).

3. For $(0, 5)$, the abscissa is $0$ and ordinate is $5$. (It lies on the $y$-axis).

Therefore, all points do not lie on the $x$-axis.

Statement (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.

Result: True

Justification:

An equation of a line parallel to the $y$-axis is always of the form $x = a$, where $a$ is the distance from the $y$-axis.

If the line is to the left of the $y$-axis, the distance is measured in the negative direction of the $x$-axis.

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

Result: False

Justification:

A line passes through the origin $(0, 0)$ if the coordinates $(0, 0)$ satisfy the equation. Substituting $x = 0$ and $y = 0$ in $y = mx + c$:

$0 = m(0) + c$

The equation $y = mx + c$ passes through the origin only if $c = 0$. If $c \neq 0$, it does not pass through the origin.

Given:

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

The points to check are $(0, 2), (1, 4), (2, 6), (3, 8),$ and $(4, 10)$.

Result: True

Justification:

We check if each $(x, y)$ pair from the table satisfies the given equation $y = 2x + 2$.

1. For x = 0:

2. For x = 1:

3. For x = 2:

4. For x = 3:

5. For x = 4:

Since all the pairs of values satisfy the given equation, the statement that these points represent some of the solutions is True.

Result: True

Justification:

A point lies on the graph of a linear equation if its coordinates satisfy the given equation. Here, we substitute $x = 0$ and $y = 3$ in the equation $3x + 4y = 12$.

L.H.S. $= 3(0) + 4(3)$

L.H.S. $= 0 + 12 = 12$

Since the point $(0, 3)$ satisfies the equation, it lies on the graph of the equation.

Result: False

Justification:

To check if the graph passes through $(0, 7)$, we substitute $x = 0$ and $y = 7$ in the equation $x + 2y = 7$.

L.H.S. $= 0 + 2(7)$

However, the R.H.S. of the equation is $7$.

Since the point does not satisfy the equation, the graph of the equation does not pass through the point $(0, 7)$.

Result: True

Justification:

From the given graph (Fig. 4.1), we can identify points such as $(-1, 1)$, $(-3, 3)$, and the origin $(0, 0)$ lying on the line. Let us check if these points satisfy the equation $x + y = 0$.

1. For $(-1, 1)$: $-1 + 1 = 0$ (Satisfied)

2. For $(-3, 3)$: $-3 + 3 = 0$ (Satisfied)

3. For $(0, 0)$: $0 + 0 = 0$ (Satisfied)

Since the points on the graph satisfy the given equation, the graph correctly represents $x + y = 0$.

Result: True

Justification:

The graph shows a straight line parallel to the $y$-axis. Any line parallel to the $y$-axis is represented by an equation of the form $x = a$, where $a$ is the distance from the $y$-axis.

In the given Fig 4.2, the line passes through the point $(3, 0)$ on the $x$-axis and all points on this line have the same abscissa, which is $3$. Therefore, the equation of this line is $x = 3$.

Result: False

Justification:

The given equation is $x - y + 2 = 0$, which can be rewritten as $y = x + 2$. Let us check the coordinates from the provided table:

1. For $x = 0$, $y = 0 + 2 = 2$ (Matches)

2. For $x = 1$, $y = 1 + 2 = 3$ (Matches)

3. For $x = 2$, $y = 2 + 2 = 4$ (Matches)

4. For $x = 3$, $y = 3 + 2 = 5$. However, the table gives $y = -5$.

Since all the points in the table do not satisfy the equation, the statement is false.

Result: False

Justification:

By the definition of the graph of a linear equation, every point lying on the line represents a pair of values $(x, y)$ that satisfy the equation. Therefore, every point on the graph must be a solution of that linear equation.

Result: False

Justification:

A linear equation in two variables of the form $ax + by + c = 0$ is geometrically represented by a straight line. The name "linear" itself implies that its graphical representation is always a line.

Given:

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

To Find:

The points where the graph of this equation intersects (cuts) the $x$-axis and the $y$-axis.

Solution:

1. To find the point where the graph cuts the x-axis:

We know that for any point on the $x$-axis, the $y$-coordinate is always $0$. Therefore, we substitute $y = 0$ in the given equation:

$3x + 4(0) = 12$

$3x = 12$

So, the graph cuts the $x$-axis at the point (4, 0).

2. To find the point where the graph cuts the y-axis:

We know that for any point on the $y$-axis, the $x$-coordinate is always $0$. Therefore, we substitute $x = 0$ in the given equation:

$3(0) + 4y = 12$

$4y = 12$

So, the graph cuts the $y$-axis at the point (0, 3).

Given:

Linear equation: $x + y = 5$.

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

To Find:

The point of intersection of these two lines.

Solution:

The equation of a line parallel to the $y$-axis at a distance '$a$' from the origin is $x = a$. Since the distance is $2$ units in the positive direction:

To find the point where this line meets $x + y = 5$, we substitute $x = 2$ into the equation:

$2 + y = 5$

Final Answer: The graph meets the given line at the point (2, 3).

Given:

Linear equation: $2x + 5y = 20$.

Condition: $x$-coordinate is $\frac{5}{2}$ times the $y$-coordinate (ordinate).

To Find:

The coordinates of the point satisfying this condition.

Solution:

Let the ordinate ($y$-coordinate) of the point be $y$.

According to the given condition:

Now, substitute the value of $x$ from equation (i) into the given equation $2x + 5y = 20$:

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

$10y = 20$

$y = 2$

Now, substitute $y = 2$ back into equation (i) to find $x$:

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

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

Given:

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

To Find:

The equation of the line and its graph.

Solution:

Any line parallel to the $x$-axis is represented by an equation of the form $y = k$, where $k$ is the distance from the $x$-axis.

Since the line is $4$ units above the $x$-axis, the value of $k$ is $4$.

To draw the graph, we identify that for any value of $x$, the value of $y$ will always remain $4$. Some points on this line are $(0, 4)$, $(1, 4)$, $(-2, 4)$, etc.

The graph is a horizontal line passing through the point $(0, 4)$ on the $y$-axis.

Given:

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

To Find:

Plot the graphs on the same Cartesian plane and state the observation.

Solution:

First, we find a few solutions for both equations to plot the lines.

For $y = x$:

For $y = -x$:

Observations:

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

2. The lines are perpendicular to each other, intersecting at an angle of $90^\circ$.

3. The line $y = x$ bisects the first and third quadrants, while $y = -x$ bisects the second and fourth quadrants.

Given:

Linear equation: $2x + 5y = 19$

Condition: Ordinate ($y$) is $1\frac{1}{2}$ times its abscissa ($x$).

To Find:

The coordinates of the point $(x, y)$.

Solution:

According to the given condition, we can write the relationship as:

Now, we substitute the value of $y$ from equation (i) into the linear equation $2x + 5y = 19$:

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

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

Taking the L.C.M. on the left side:

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

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

$19x = 19 \times 2$

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

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

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

Given:

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

To Find:

The equation of the line and its graph.

Solution:

We know that the equation of any line parallel to the $x$-axis is of the form $y = k$.

Since the line is 3 units below the $x$-axis, it lies in the negative direction of the $y$-axis.

To plot this graph, we note that the ordinate ($y$) is always $-3$ regardless of the value of $x$. Points such as $(0, -3)$, $(2, -3)$, and $(-2, -3)$ lie on this line.

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

Given:

Sum of the coordinates of a point $(x, y)$ is $10$ units.

To Find:

The linear equation and its graph.

Solution:

Let the coordinates of a point be $x$ and $y$. According to the given condition:

To draw the graph, we find at least two solutions for the equation:

1. If $x = 0$, then $0 + y = 10 \Rightarrow y = 10$. Point is $(0, 10)$.

2. If $y = 0$, then $x + 0 = 10 \Rightarrow x = 10$. Point is $(10, 0)$.

3. If $x = 5$, then $5 + y = 10 \Rightarrow y = 5$. Point is $(5, 5)$.

Final Answer: The equation is $x + y = 10$.

Given:

Condition: Ordinate ($y$) is $3$ times the abscissa ($x$).

To Find:

The linear equation representing this condition.

Solution:

By the definition of coordinates, we have:

Abscissa $= x$

Ordinate $= y$

As per the given condition:

This can also be written in the general form $ax + by + c = 0$ as:

Final Answer: The required linear equation is $y = 3x$ or $3x - y = 0$.

Given:

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

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

To Find:

The value of the constant $a$.

Solution:

Since the point $(3, 4)$ lies on the graph of the equation, its coordinates must satisfy the equation. We substitute $x = 3$ and $y = 4$ into the given equation:

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

Subtracting $7$ from both sides:

$12 - 7 = 3a$

$5 = 3a$

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

Given:

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

Solution:

First, we solve the given equation for $x$:

$2x - x = -3 - 1$

(i) On the Number Line:

On a number line, the equation $x = -4$ represents a unique point. Therefore, there is only one solution.

(ii) On the Cartesian Plane:

In a Cartesian plane, the equation $x = -4$ can be treated as a linear equation in two variables, written as:

This represents a straight line parallel to the $y$-axis. Every point on this line is a solution to the equation. Since a line contains an infinite number of points, there are infinitely many solutions.

Given:

Linear equation: $x + 2y = 8$.

Solution:

(i) On the x-axis:

We know that for any point on the $x$-axis, the $y$-coordinate is $0$.

Substituting $y = 0$ in the equation:

$x + 2(0) = 8$

$x = 8$

So, the point is (8, 0).

(ii) On the y-axis:

We know that for any point on the $y$-axis, the $x$-coordinate is $0$.

Substituting $x = 0$ in the equation:

$0 + 2y = 8$

$2y = 8$

$y = 4$

So, the point is (0, 4).

Given:

Linear equation: $2x + cy = 8$.

Condition: $x$ and $y$ have equal values in the solution ($x = y$).

Solution:

Let $x = y = k$, where $k$ is a real number ($k \neq 0$ for a unique $c$). Substituting $y = x$ in the equation:

$2x + c(x) = 8$

From this, we can express $c$ in terms of $x$:

$2 + c = \frac{8}{x}$

Given:

$y$ varies directly as $x$ ($y \propto x$).

$y = 12$ when $x = 4$.

To Find:

1. The linear equation.

2. The value of $y$ when $x = 5$.

Solution:

Since $y$ varies directly as $x$, we have:

Substitute $y = 12$ and $x = 4$ to find $k$:

$12 = k(4)$

Thus, the required linear equation is:

Now, to find the value of $y$ when $x = 5$, we substitute $x = 5$ in equation (i):

$y = 3(5)$

Final Answer: The linear equation is $y = 3x$ and the value of $y$ when $x = 5$ is 15.

Given:

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

To Find:

1. The points where the graph cuts the $x$-axis and the $y$-axis.

2. Draw the graph of the equation.

Solution:

To find the points where the graph cuts the axes, we follow these steps:

1. For the x-axis: We know that for any point on the $x$-axis, the $y$-coordinate is always $0$. Substituting $y = 0$ in the equation:

$2x + 3(0) = 12$

$2x = 12$

Thus, the point is (6, 0).

2. For the y-axis: We know that for any point on the $y$-axis, the $x$-coordinate is always $0$. Substituting $x = 0$ in the equation:

$2(0) + 3y = 12$

$3y = 12$

Thus, the point is (0, 4).

Table for Graph:

Final Answer: The graph cuts the $x$-axis at (6, 0) and the $y$-axis at (0, 4).

Given:

The values of $x$ and $y$ satisfying a linear equation are:

To Find:

1. Draw the graph of the linear equation.

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

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

Solution:

Step 1: Drawing the graph

We plot the points $A(1, 1)$ and $B(2, 3)$ on the Cartesian plane. By joining these two points and extending the line in both directions, we obtain the graph of the linear equation.

Step 2: Finding the Equation

To find the exact points of intersection with the axes, we first determine the linear equation. Let the equation be $y = mx + c$. Using the given points:

Now, substituting $m = 2$ and point $(1, 1)$ into $y = mx + c$:

$1 = 2(1) + c$

$1 = 2 + c$

Thus, the linear equation is:

Step 3: Finding Intercepts

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

On the $x$-axis, the $y$-coordinate is $0$. Substituting $y = 0$ in equation (i):

$2x = 1$

Therefore, the graph cuts the x-axis at the point (0.5, 0).

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

On the $y$-axis, the $x$-coordinate is $0$. Substituting $x = 0$ in equation (i):

Therefore, the graph cuts the y-axis at the point (0, –1).

Final Answer:

(i) The point where the graph cuts the $x$-axis is (0.5, 0).

(ii) The point where the graph cuts the $y$-axis is (0, –1).

Given:

Fare for the first km $= \textsf{₹} 10$.

Fare for subsequent distance $= \textsf{₹} 4$ per km.

To Find:

Linear equation and its graph.

Solution:

Let the total distance covered be $x$ km and the total fare be $\textsf{₹} y$.

According to the condition:

Fare for the first km $= 10$

Remaining distance $= (x - 1)$ km

Fare for remaining distance $= 4(x - 1)$

Total Fare ($y$) $= 10 + 4(x - 1)$

$y = 10 + 4x - 4$

Table for Graph:

Note: Distance cannot be negative, so the graph starts from $x=0$ (theoretically) or $x=1$ (practically).

Given:

Work done $=$ Constant Force $\times$ Distance.

Solution:

Let the work done be $y$ units and the distance travelled be $x$ units.

Let the constant force be $k$ units.

The linear equation is:

To draw the graph, let us assume the constant force $k = 3$ units.

Then the equation becomes $y = 3x$.

Points to plot:

1. If $x = 0, y = 0$. Point is $(0, 0)$.

2. If $x = 1, y = 3$. Point is $(1, 3)$.

3. If $x = 2, y = 6$. Point is $(2, 6)$.

The graph is a straight line passing through the origin.

Given:

Linear equation: $y = 9x - 7$

Points: $A(1, 2)$, $B(-1, -16)$, and $C(0, -7)$.

To Prove:

Points $A, B,$ and $C$ lie on the graph of the given linear equation.

Proof:

A point lies on the graph of an equation if its coordinates satisfy the equation.

1. For Point A(1, 2):

Substituting $x = 1$ and $y = 2$ in the equation $y = 9x - 7$:

$L.H.S. = 2$

$R.H.S. = 9(1) - 7 = 9 - 7 = 2$

2. For Point B(–1, –16):

Substituting $x = -1$ and $y = -16$ in the equation $y = 9x - 7$:

$L.H.S. = -16$

$R.H.S. = 9(-1) - 7 = -9 - 7 = -16$

3. For Point C(0, –7):

Substituting $x = 0$ and $y = -7$ in the equation $y = 9x - 7$:

$L.H.S. = -7$

$R.H.S. = 9(0) - 7 = 0 - 7 = -7$

Since all three points satisfy the equation, they all lie on the graph of $y = 9x - 7$.

Given:

Two solutions of a linear equation: $(6, -2)$ and $(-6, 6)$.

To Find:

1. The linear equation.

2. The graph of the equation.

3. Points where it cuts the $x$-axis and $y$-axis.

Solution:

Let the linear equation be $ax + by = c$. We can find the equation using the two-point form:

$y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$

Substituting $(x_1, y_1) = (6, -2)$ and $(x_2, y_2) = (-6, 6)$:

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

$y + 2 = \frac{8}{-12}(x - 6)$

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

$3(y + 2) = -2(x - 6)$

$3y + 6 = -2x + 12$

Points on the Axes:

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

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

$2x + 3(0) = 6$

So, the point is (3, 0).

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

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

$2(0) + 3y = 6$

So, the point is (0, 2).

Graph:

Given:

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

To Find:

1. The graph of the equation.

2. Points where the graph cuts the $x$-axis and the $y$-axis.

Solution:

To draw the graph, we find the points of intersection with the axes.

1. Intersection with the x-axis:

On the $x$-axis, $y = 0$. Substituting $y = 0$ in $3x + 4y = 6$:

$3x + 4(0) = 6$

$3x = 6$

So, the point is (2, 0).

2. Intersection with the y-axis:

On the $y$-axis, $x = 0$. Substituting $x = 0$ in $3x + 4y = 6$:

$3(0) + 4y = 6$

$4y = 6$

So, the point is (0, 1.5).

Table of solutions:

Graph:

Final Answer: The graph cuts the $x$-axis at (2, 0) and the $y$-axis at (0, 1.5).

Given:

The relation between Celsius ($C$) and Fahrenheit ($F$) is:

Solution (i):

Given temperature is $86^\circ\text{F}$. Substituting $F = 86$ in equation (i):

$C = \frac{5(86) - 160}{9}$

$C = \frac{430 - 160}{9}$

$C = \frac{270}{9}$

The temperature in Celsius is $30^\circ\text{C}$.

Solution (ii):

Given temperature is $35^\circ\text{C}$. Substituting $C = 35$ in equation (i):

$35 = \frac{5F - 160}{9}$

$35 \times 9 = 5F - 160$

$315 = 5F - 160$

$5F = 315 + 160$

The temperature in Fahrenheit is $95^\circ\text{F}$.

Solution (iii):

Case A: If temperature is $0^\circ\text{C}$, find $F$.

$0 = \frac{5F - 160}{9}$

$0 = 5F - 160$

Case B: If temperature is $0^\circ\text{F}$, find $C$.

$C = \frac{5(0) - 160}{9}$

Solution (iv):

Let the temperature be $x$ such that $C = F = x$. Substituting this in equation (i):

$x = \frac{5x - 160}{9}$

$9x = 5x - 160$

$9x - 5x = -160$

$4x = -160$

Hence, $-40^\circ$ is the numerical value which is same in both Celsius and Fahrenheit scales.

Given:

The relation is $y = \frac{9}{5}(x - 273) + 32$, where $x$ is in Kelvin and $y$ is in Fahrenheit.

Solution (i):

Given $x = 313^\circ\text{K}$. Substituting this value in the equation:

$y = \frac{9}{5}(313 - 273) + 32$

$y = \frac{9}{5}(40) + 32$

$y = 9(8) + 32$

The temperature is $104^\circ\text{F}$.

Solution (ii):

Given $y = 158^\circ\text{F}$. Substituting this value in the equation:

$158 = \frac{9}{5}(x - 273) + 32$

$158 - 32 = \frac{9}{5}(x - 273)$

$126 = \frac{9}{5}(x - 273)$

$\frac{126 \times 5}{9} = x - 273$

$14 \times 5 = x - 273$

$70 = x - 273$

The temperature is $343^\circ\text{K}$.

Given:

Force ($F$) is directly proportional to acceleration ($a$).

Constant of proportionality is mass ($m = 6$ kg).

Solution:

Let Force be $y$ units and Acceleration be $x$ units. According to Newton's Second Law ($F = ma$):

To draw the graph, we find some solutions for $y = 6x$:

From the linear equation and the graph:

(i) When acceleration ($x$) is $5 \text{ m/sec}^2$:

(ii) When acceleration ($x$) is $6 \text{ m/sec}^2$:

Final Answer: The required force values are $30 \text{ N}$ and $36 \text{ N}$ respectively.