NCERT Exemplar Class 9 Maths Solutions Chapter 4 Linear Equations in Two Variables 2021

NCERT Exemplar Class 9 Maths Solutions Chapter 4 Linear Equations in Two Variables

Edited By Ravindra Pindel | Updated on Aug 31, 2022 11:41 AM IST

NCERT exemplar Class 9 Maths solutions chapter 4 provide students with detailed answers for the chapter 4 Linear Equation in two variables. These NCERT Solutions are important as the concepts used here will be applied in mathematics courses of higher Classes. NCERT exemplar solutions for Class 9 Maths chapter 4 gives a good number of practice questions.

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NCERT Exemplar Class 9 Maths Solutions Chapter 4: Exercise 4.1
NCERT Exemplar Class 9 Maths Solutions Chapter 4: Exercise 4.2
NCERT Exemplar Class 9 Maths Solutions Chapter 4: Exercise 4.3
NCERT Exemplar Class 9 Maths Solutions Chapter 4: Exercise 4.4
NCERT Exemplar Solutions Class 9 Maths Chapter 4 Important Topics:
NCERT Class 9 Exemplar Solutions for Other Subjects:
NCERT Class 9 Maths Exemplar Solutions for Other Chapters:
Features of NCERT Exemplar Class 9 Maths Solutions Chapter 4:

These NCERT exemplar Class 9 Maths solutions chapter 4 are also useful for competitive exams like JEE Main, advanced and NEET and other state and class 9 CBSE syllabus. The Class 9 Maths NCERT exemplar chapter 4 solutions are useful to solve Class 9 Physics problems.

NCERT Exemplar Class 9 Maths Solutions Chapter 4: Exercise 4.1

Question:1

The linear equation $2x -5y = 7$ has
(A) A unique solution
(B) Two solutions
(C) Infinitely many sol

Answer:
(C) Infinitely many solutions
Solution:
Linear equation: The equation of two variables which gives a straight line graph is called linear equation.
Here the linear equation is 2x – 5y = 7
Let y = 0 then the value of x is:
$2x - 5(0) = 7$
$2x = 7$
$x=\frac{7}{2}$
Now, let y = 1 then the value of x is:
$2x - 5(1) = 7$
$2x = 12$
$x = \frac{12}{2}=6$
$x =6$
Here for different values of y, we are getting different values of x
Therefore the equation has infinitely many solutions

Question:2

The equation 2x+5y=7 has a unique solution, if x, y are :
(A) Natural numbers
(B) Positive real numbers
(C) Real numbers
(D) Rational numbers

Answer:
(A) Natural numbers
Solution:
Unique solution: for Unique solution, there should exist only one pair of x and y.
In natural numbers, there exists only one pair which is (1, 1)
2x + 5y = 7
(2) + (5) = 7
7 = 7
In positive real numbers there exists more than one pair.
For example: (0.5, 1.2), (1.5, 0.8)…
In real numbers there exists more than one pair.
For example: (–0.5, 1.6), (–1.5, 2)…
Also in rational numbers, there exists more than one pair.
For example:
$\left (\frac{3}{2},\frac{4}{5} \right ),\left ( \frac{1}{2},\frac{6}{5} \right )$ …
Hence in natural numbers there exists only one pair therefore the solution is natural numbers.

Question:3

If (2, 0) is a solution of the linear equation 2x + 3y = k, then the value of k is
(A) 4
(B) 6
(C) 5
(D) 2

Solution:
In this question we have been given a pair (2, 0)
Here the value of x is 2 and the value of y is 0.
The linear equation is: 2x+ 3y = k
Now put the value of x and y in this equation we get
2x + 3y = k
2(2) + 3(0) = k
4 + 0 = k
4 = k
Hence the value of k is equal to 4

Question:4

Any solution of the linear equation 2x + 0y + 9 = 0 in two variables is of the form
(A) $\left (-\frac{9}{2},m \right )$
(B) $\left (n,-\frac{9}{2} \right )$
(C) $\left (0,-\frac{9}{2} \right )$
(D) (–9, 0)

Answer:
(A) $\left (-\frac{9}{2},m \right )$
Solution:
Here the linear equation is 2x + 0y + 9 = 0
Let,
2x = – 9
$x = -\frac{9}{2}$
In equation (1), y is multiplied by zero so we can put any value of y. Therefore there will be infinite many values of y.
Hence the solution of this question is $\left (-\frac{9}{2},m \right )$
Here m stands for infinite many values.

Question:5

The graph of the linear equation 2x + 3y = 6 cuts the y-axis at the point
(A) (2, 0)
(B) (0, 3)
(C) (3, 0)
(D) (0, 2)

Answer:
(D) (0, 2)
Solution:
Here the given equation is 2x + 3y = 6
We know that at y-axis x = 0
Now put x = 0 in the given equation we get
2x + 3y = 6
2(0) + 3y = 6
3y = 6
$y= \frac{6}{3}=2$

Question:6

The equation x = 7, in two variables, can be written as
(A) 1.x + 1.y = 7
(B) 1.x + 0.y = 7
(C) 0.x + 1.y = 7
(D) 0.x + 0.y = 7

Answer:
(B) 1.x + 0.y = 7
Solution:
The given equation is x = 7
In the given equation there is no second variable.
If we want to write the given equation in the form of two variables then the value of second variable must be zero.
So, the solution of the given equation is 1.x + 0.y = 7
Therefore option (B) is correct.

Question:7

Any point on the x-axis is of the form
(A) (x, y)
(B) (0, y)
(C) (x, 0)
(D) (x, x)

Answer:
(C) (x, 0)
Solution:
We know that at x-axis, y coordinate is zero
So, equation of x-axis is y = 0
Therefore any point on the x-axis is of the form (x, 0).
Hence the solution of the given question is (x, 0)

Question:8

Any point on the line y = x is of the form
(A) (a, a)
(B) (0, a)
(C) (a, 0)
(D) (a, – a)

Answer:
(A) (a, a)
Solution:
The given line is y = x
This can be written as:
y – x = 0
Putting y = 1,
1 – x = 0
x = 1
Point is (1, 1)
Putting y = 2,
2 – x = 0
x = 2
Point is (2, 2)
Hence every point is in the form of (a, a)
Therefore option (A) is correct.

Question:9

The equation of x-axis is of the form
(A) x = 0
(B) y = 0
(C) x + y = 0
(D) x = y

Answer:
(B) y = 0
Solution:
We know that at x-axis, y coordinate is zero
i.e., y = 0
Hence the equation of x-axis is y = 0
Therefore option (B) is correct.

Question:10

The graph of y = 6 is a line
(A) parallel to x-axis at a distance 6 units from the origin
(B) parallel to y-axis at a distance 6 units from the origin
(C) making an intercept 6 on the x-axis.
(D) making an intercept 6 on both the axes.

Answer:
(A) parallel to x-axis at a distance 6 units from the origin
Solution:
The graph of y = 6 is a line as shown below:

Hence in the above graph is a line parallel to x-axis at a distance 6 units from the origin.
Therefore option (A) is correct.

Question:11

x = 5, y = 2 is a solution of the linear equation
(A) x + 2 y = 7
(B) 5x + 2y = 7
(C) x + y = 7
(D) 5 x + y = 7

Answer:
(C) x + y = 7
Solution:
(A) Putting x = 5, y = 2 in x + 2y = 7 we get
LHS = 5 + 2 (2)
= 5 + 4
$= 9 \neq 7 (RHS)$
(B) Putting x = 5, y = 2 in 5x + 2y = 7 we get
LHS = 5(5) + 2(2)
= 25 + 4
$= 29 \neq 7 (RHS)$
(C) Putting x = 5, y = 2 in x + y = 7 we get
LHS = 5 + 2
= 7 = 7 (RHS)
(D) Putting x = 5, y = 2 in 5x + y = 7 we get
LHS = 5(5) + 2
= 25 + 2
$= 27 \neq 7 (RHS)$
Hence only x + y = 7 satisfies the given values of x and y
Therefore option (C) is correct.

Question:12

If a linear equation has solutions (–2, 2), (0, 0) and (2, – 2), then it is of the form
(A) y – x = 0
(B) x + y = 0
(C) –2x + y = 0
(D) –x + 2y = 0

Answer:
(B) x + y = 0
Solution:
(A) Put the given points in y – x = 0
Putting (–2, 2)
LHS = 2 – (–2)
= 2 + 2
$= 4\neq 0 (RHS)$
The given equation does not satisfy this point, so there is no need to check for other points.
(B) Now put the given points in x + y = 0
Putting (–2, 2)
LHS = –2 + 2 = 0 (RHS)
Putting (0, 0)
LHS = 0 + 0 = 0 (RHS)
Putting (2, –2)
LHS = 2 – 2 = 0 (RHS)
The given equation is satisfying all the points
(C) Put the given points in –2x + y = 0
Putting (–2, 2)
LHS = (–2) (–2) + (2)=6, which is not equal to RHS

Question:13

The positive solutions of the equation ax + by + c = 0 always lie in the
(A) 1st quadrant
(B) 2nd quadrant
(C) 3rd quadrant
(D) 4th quadrant

Answer:
(A) 1st quadrant
Solution:
In 1st quadrant the value of x and y are positive
In 2nd quadrant the value of x is negative and y is positive
In 3rd quadrant the value of x is negative and y is also negative
In 4th quadrant the value of x is positive and y is negative.

Hence for the positive solution of the given equation, both the values of x and y must be positive. Therefore option (A) is correct.

Question:14

The graph of the linear equation 2x + 3y = 6 is a line which meets the x-axis at the point:
(A) (0, 2)
(B) (2, 0)
(C) (3, 0)
(D) (0, 3)

Answer:
(C) (3, 0)
Solution:
At x-axis, y = 0
Put y = 0 in 2x + 3y = 6
2x + 3(0) = 6
2x = 6
$x =\frac{6}{2} = 3$
x = 3
So the required point is (3, 0)
The graph of equation 2x + 3y = 6 is

x	0	3
y	2	0

Question:15

The graph of the linear equation y = x passes through the point
(A) $\left (\frac{3}{2}, -\frac{3}{2} \right )$
(B) $\left (0, \frac{3}{2} \right )$
(C) (1, 1)
(D) $\left (\frac{1}{2}, -\frac{1}{2} \right )$

Answer:
(C) (1, 1)
Solution:
The given linear equation is y = x
i.e., y – x = 0
First point is $\left (\frac{3}{2}, -\frac{3}{2} \right )$
Putting the value of $x=\frac{3}{2}$ in the above equation we get
$y-\frac{3}{}2=0$
$y=\frac{3}{2}$
We get $y=\frac{3}{2}$ which does not satisfy our y-coordinate
Second point is $\left (0, \frac{3}{2} \right )$
Putting the value of x = 0 in the given equation, we get
y = 0
We get y = 0 which does not satisfy our y-coordinate
Third point is (1, 1)
Putting the value of x = 1 in the given equation we get
y – 1 = 0
y = 1
We get y = 1 which satisfies our y-coordinate
Fourth point is $\left (\frac{1}{2}, -\frac{1}{2} \right )$
Putting the value of $x=\frac{1}{2}$ in the given equation we get
$y-\left (-\frac{1}{2} \right )=0$
$y=-\frac{1}{2}$
We get $y=-\frac{1}{2}$ which does not satisfy our y-coordinate
Hence only one point which is (1, 1) satisfies the given linear equation.
Therefore option (C) is correct.

Question:16

If we multiply or divide both sides of a linear equation with a non-zero number, then the solution of the linear equation :
(A) Changes
(B) Remains the same
(C) Changes in case of multiplication only
(D) Changes in case of division only

Answer:
(B) Remains the same
Solution:
Let the linear equation is x + y = 2 ……… (1)
Now put x = 1 in equation (1) we get
y = 1
Here x = 1, y = 1 is the solution of the equation
Now multiply both sides of equation (1) by 2
2x + 2y = 4
Put x = 1, y = 1 in left hand side we get
1(1) + 2(1) = 2 + 2 = 4 = RHS
Hence on multiply an equation by a non-zero number the solution remains same
Now divide both sides of equation (1) by 2
$\frac{x}{2}+\frac{y}{2}=1$ Put x = 1, y = 1 in L.H.S. we get
$\frac{1}{2}+\frac{1}{2}=\frac{1+1}{2}=\frac{2}{2}=1= R.H.S$ Also on divide an equation by non-zero number the solution remains same.

Question:17

How many linear equations in x and y can be satisfied by x = 1 and y = 2?
(A) Only one
(B) Two
(C) Infinitely many
(D) Three

Answer:
(C) Infinitely many
Solution:
Let the line’s equation be ax + by + c = 0
Now put x = 1 and y = 2 in above equation
a + 2 b + c = 0
Here only one equation is not possible because a, b and c are real numbers. Hence different values of a, b and c can satisfy the equation a + 2b + c = 0
For example, (a = 1, b = –1, c = 1)
(a = 2, b = –2, c = 2) and many more.
Similarly two and three equations is not the correct answer.
Therefore infinitely many linear equations in x and y can be satisfied by x = 1 and y = 2.
Therefore option (C) is correct.

Question:18

The point of the form (a, a) always lies on:
(A) x-axis
(B) y-axis
(C) On the line y = x
(D) On the line x + y = 0

Answer:
(C) On the line y = x
Solution:
First of all we will discuss about x-axis
On x-axis, the value of y is always zero
Therefore point (a, a) cannot lie on x-axis as value of y-coordinate is given as ‘a’.
Now we will discuss about y-axis
On y-axis, the value of x is always zero.
Therefore point of the form (a, a) cannot lie on y-axis as value of x-coordinate is given as ‘a’.
On the line y = x, for each and every value of x there is an equal value of y
For example for x = 1, we get y = 1
Therefore point of the form (a, a) lies on the line y = x.
On the line x + y = 0
Let the value of x is 1
We get y = –1
Hence point of the form (a, a) does not lies on the line x + y = 0.
Therefore, the point of the form (a, a) always lies on the line y = x
Therefore option (C) is correct.

Question:19

The point of the form (a, – a) always lies on the line
(A) x = a
(B) y = – a
(C) y = x
(D) x + y = 0

Answer:
(D) x + y = 0
Solution:
The given point of the form (a, –a) shows that for every value of x there must be equal value of y with opposite sign.
In x = a, only the value of x is given
Here we do not know the value of y, therefore it is not possible to determine (a, –a) point.
In y = –a, only the value of y is given
Here we do not know the value of x, therefore it is not possible to determine (a, –a) point.
On the line y = x, for each and every value of x there is an equal value of y
For example for x = 1, we get y = 1
Therefore point of the form (a, a) lies on the line y = x.
On the line x + y = 0, i.e., x = -y
for each and every value of x there is an equal value of y (which is negative)
For example for x = 1, we get y = -1
Therefore point of the form (a, -a) lies on the line x + y = 0.
Therefore option (D) is correct.

NCERT Exemplar Class 9 Maths Solutions Chapter 4: Exercise 4.2

Question:1

Write whether the following statement is True or False. Justify your answer.
The point (0, 3) lies on the graph of the linear equation 3x + 4y = 12

Answer:
True
Solution:
The given linear equation is 3x + 4y = 12 and the point is (0, 3)
Here the value of x is 0 and y is 3
Now put the value of x and y in above equation. We get,
LHS = 3(0) + 4 (3)
= 0 + 12
= 12 = RHS
Hence the given point (0, 3) satisfies the linear equation 3x + 4y = 12.
Therefore the given statement is True.

Question:2

Write whether the following statement is True or False. Justify your answer.
The graph of the linear equation x + 2y = 7 passes through the point (0, 7).

Answer:
False
Solution:
First of all, let us plot the graph of the linear equation x + 2y = 7
For x = 0,
0 + 2y = 7
$y=\frac{7}{2}$
For y = 0,
x + 2 (0) = 7
x = 7
For x = 1,
1 + 2y = 7
2y = 6
y = 3
So we have,

x	0	7	1
y	$\frac{7}{2}$	0	3

The above graph shows that the linear equation does not passes through the point (0, 7)
Hence the given statement is false.

Question:3

Write whether the following statement is True or False. Justify your answer.
The graph given below represents the linear equation x + y = 0.

Answer:
True
Solution:
Given graph is:

The points of the graph are (–1, 1) and (–3, 3)
Given linear equation is x + y = 0
Putting x = –1 in above equation
–1 + y = 0
y = 1
Putting x = –3 in above equation
-3 + y = 0
y = 3

Question:4

Write whether the following statement is True or False. Justify your answer.
The graph given below represents the linear equation x = 3
Figure:

Answer:
True
Solution:
Since x = 3 is a vertical line. That is there is no y-intercept, also the slope is not defined.
The given graph is:

Therefore the above graph represents the linear equation x = 3 as it passes through x = 3 and there is no y-intercept, also the slope is not defined.
Hence the given statement is True.

Question:5

Write whether the following statement is True or False. Justify your answer.
The coordinates of points in the table:

x	0	1	2	3	4
y	2	3	4	-5	6

represent some of the solutions of the equation x – y + 2 = 0.
Answer:
False
Solution:
The given equation is x – y + 2 = 0
Put x = 0 in the given equation
0 – y + 2 = 0
y = 2, i.e., the point is (0, 2)
Put x = 1 in the given equation
1 – y + 2 = 0
y = 3, i.e., the point is (1, 3)
Put x = 2 in the given equation
2 – y + 2 = 0
4 = y, i.e., the point is (2, 4)
Put x = 3 in the given equation
3 – y + 2 = 0
5 = y, i.e., the point is (3, 5)
Put x = 4 in the given equation
4 – y + 2 = 0
6 = y, i.e., the point is (4, 6)
The coordinates of points in the table:

x	0	1	2	3	4
y	2	3	4	-5	6

It is given that the above represents some of the solutions of the equation x – y + 2 = 0.
Here all the points satisfy the given equation except (3, -5).
Therefore the given statement is false because one of the table entry is an incorrect solution for the given equation.

Question:6

Write whether the following statement is True or False. Justify your answer.
Every point on the graph of a linear equation in two variables does not represent a solution of the linear equation.

Answer:
False
Solution:
The graph of a linear equation in two variables is a line which can be plotted by finding two solutions (a₁, b₁) and (a₂, b₂).
Hence the graph of a linear equation is constructed by joining these points. So any point on the graph must be a solution of this linear equation.
Hence the given statement is False.

Question:7

Write whether the following statement is True or False. Justify your answer.
The graph of every linear equation in two variables need not be a line.

Answer:
False
Solution:
We know that the standard form of any linear equation is ax + by + c = 0
If we put different value of x we get different values of y corresponding to the values of x.
So we have x proportional to y.
Thus we can say $\frac{x}{y}$ = constant.
Now, the general equation of a line is y=mx+c, which also gives a direct proportionality between x and y.
So if we plot a graph with the help of such points, it will always be a line
Therefore the given statement is false.

NCERT Exemplar Class 9 Maths Solutions Chapter 4: Exercise 4.3

Question:1

Draw the graphs of linear equations y = x and y = – x on the same cartesian plane. What do you observe?

Answer:

First of all the us plot the graph of linear equation y = x and y = –x
The corresponding x and y-coordinates can be written as:

Plotting the above in the same Cartesian plane, we get:

Here we observe that the graph of both the lines is straight line graph and meet only at the origin.

Question:2

Determine the point on the graph of the linear equation 2x + 5y = 19, whose ordinate is times its abscissa.

Answer:
(2, 3)
Solution:
The given linear equation is 2x + 5y = 19 … (i)
Here it is given that the ordinate is $\frac{3}2$ times its abscissa
$y=\frac{3}2x$ ... (1)
Put the value of y in above equation (i) we get
$2x +5\left ( \frac{3}{2} x\right )=19$ $2x +\frac{15x}{2}=19$ $\frac{4x +15x}{2}=19$ $19 x = 19 \times 2$
$x = \frac{19 \times 2}{19}=2$
x = 2
Put the value of x in equation (i)
$y=\frac{3}{2}\times 2$
y = 3
Hence the point is (2, 3)

Question:3

Draw the graph of the equation represented by a straight line which is parallel to the x-axis and at a distance 3 units below it.

Answer:

We know that the straight line which is parallel to the x-axis is on the y-intercept, i.e. there is no x-intercept on it.
Now let us plot a straight line on y-intercept at a distance 3 units below the x-axis

Question:4

Draw the graph of the linear equation whose solutions are represented by the points having the sum of the coordinates as 10 units.

Answer:

It is given that sum of the ordinates is 10 units
x + y = 10 …(1)
We can find such points as:

x	1	2	3
y	9	8	7

The points are (1, 9), (2, 8) , (3, 7) and so on.
Now plotting the graph of the linear equation

Question:5

Write the linear equation such that each point on its graph has an ordinate 3 times its abscissa.

Answer:

It is given that the ordinate is 3 times its abscissa.
That is y = 3x … (i)
Such points are:

x	0	1	2
y	0	3	6

The points are (0, 0), (1, 3) and (2, 6)
Now plot the graph of equation (1)

Hence y = 3x is the required solution such that each point on its graph has ordinate 3 times its abscissa.

Question:6

If the point (3, 4) lies on the graph of 3y = ax + 7, then find the value of a.

Answer:
$a=\frac{5}{3}$
Solution:
The given linear equation is 3y = ax + 7 and the point is (3, 4)
If the point (3, 4) lies on the graph of equation 3y = ax + 7 then it will surely satisfy the equation
Put value of x = 3 and y = 4 in equation 3y = ax + 7
3(4) = a(3) + 7
12 = 3a + 7
12 – 7 = 3a
5 = 3a
$\frac{5}{3}=a$

Question:7

How many solution(s) of the equation 2x + 1 = x – 3 are there on the number line?

How many solution(s) of the equation 2x + 1 = x – 3 are there on the Cartesian plane?

Answer:
Only one solution
Solution:
The given equation is
2x + 1 = x – 3
2x – x = – 3 – 1
x = –4
Representation on number line:

On number line, there is only one solution

Answer:
Infinite solutions
Solution:
The given equation is
2x + 1 = x – 3
2x – x = – 3 – 1
x = –4
Cartesian plane representation:

There are infinite many points on line x = –4 in Cartesian plane so there are infinite solutions.

Question:8

Find the solution of the linear equation x + 2y = 8 which represents a point on x-axis.

Find the solution of the linear equation x + 2y = 8 which represents a point on y-axis.

Answer:
(8, 0)
Solution:
The given equation is
x + 2y = 8
at x-axis, we have y = 0
Now put the value of y = 0 in the given equation
x + 2(0) = 8
x = 8
So the required point is (8, 0)

Answer:
(0, 4)
Solution:
The given equation is
x + 2y = 8
At y-axis, we have x = 0
Now put the value of x = 0 in the given equation
0 + 2y = 8
$y=\frac{8}{2}=4$
y = 4
So the required point is (0, 4)

Question:9

For what value of c, the linear equation 2x + cy = 8 has equal values of x and y for its solution.

Answer:

Answer:
$c=\frac{8-2y}{y}(y\neq 0)$
Solution:
The given linear equation is
2x + cy = 8 … (i)
and it is also given that the value of x and y is equal
i.e., x = y
Putting x = y in the equation (i) we get
2y + cy = 8
cy = 8– 2y
$c=\frac{8-2y}{y}(y\neq 0)$
Hence the correct answer is $c=\frac{8-2y}{y}(y\neq 0)$

Question:10

Let y varies directly as x. If y = 12 when x = 4, then write a linear equation. What is the value of y when x = 5?

Answer:
y = 3x, y = 15 when x = 5.
Solution:
It is given that y varies directly as x
$y\propto x$
y = kx (Here k is constant)
For finding k, we can use the given condition y = 12 when x = 4
12 = k(4)
$\frac{12}{4}=k$
k = 3
Hence the equation is y = 3x
When x = 5 then the value of y is:
y = 3(5)
y = 15
Hence y = 15 when x = 5.

NCERT Exemplar Class 9 Maths Solutions Chapter 4: Exercise 4.4

Question:1

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

Answer:

The given linear equation is y = 9x – 7 and the points are A(1, 2), B(–1, –16) and C(0, –7)
If the points lie on the graph then it must be satisfy the equation y = 9x – 7
Put x = 1, y = 9 – 7
y = 2
Put x = –1 y = –9 – 7
y = –16
Put x = 0 y = 0 – 7
y = -7
Here all the points satisfy the equation y = 9x – 7.
Therefore all the points lie on the graph of the linear equation y = 9x – 7
Hence proved

Question:2

The following observed values of x and y are thought to satisfy a linear equation. Write the linear equation :

x	6	-6
y	-2	6

Draw the graph using the values of x, y as given in the above table. At what points the graph of the linear equation
(i) cuts the x-axis
(ii) cuts the y-axis

Answer:

Let the linear equation is y = mx + c … (i), and it satisfies the points (6, –2), (–6, 6)
So, we have:
–2 = 6m + c … (ii)
6 = – 6m + c …(iii)
Adding equation (ii) & (iii) we get
4 = 2c
$\frac{4}{2}=c$
c = 2
On putting the value of c = 2 in equation (ii) we get
–2 = 6 m + 2
– 2 – 2 = 6 m
–4 = 6 m
$-\frac{4}{6}=m$
$m=-\frac{2}{3}$
Now, equation (i) becomes y=2x/3 + 2
which gives
3y + 2x = 6 … (iv)
Which is required equation and its graph is

Putting y = 0 in equation (iv), we get
2x = 6
x = 3
Hence the graph cut x-axis at (3, 0)
Putting x = 0 in equation (iv), we get
3y = 6
y = 2
Hence the graph cut y axis at (0, 2)

Question:3

Draw the graph of the linear equation 3x + 4y = 6. At what points, the graph cuts the x-axis and the y-axis.

Answer:

The given equation is 3x + 4y = 6 …(i)
For x =0,
0 + 4y = 6
$y = \frac{3}{2}=1.5$
For y = 0,
3x = 6
x = 2
For x = -2,
3 (-2) + 4y = 6
- 6 - 6 = -4y
y = 3
Plotting the points, we get:

Hence the graph cut x-axis at (2, 0) and the graph cut y-axis at (0, 1.5)

Question:4

The linear equation that converts Fahrenheit (F) to Celsius (C) is given by the relation $\frac{5F-160}{9}$ If the temperature is 86°F, what is the temperature in Celsius?

The linear equation that converts Fahrenheit (F) to Celsius (C) is given by the relation $\frac{5F-160}{9}$ .If the temperature is $35^{\circ}C$ , what is the temperature in Fahrenheit (F)?

The linear equation that converts Fahrenheit (F) to Celsius (C) is given by the relation $\frac{5F-160}{9}$ .If the temperature is $0^{\circ}C$ what is the temperature in Fahrenheit and if the temperature is $0^{\circ}F$ , what is the temperature in Celsius?

The linear equation that converts Fahrenheit (F) to Celsius (C) is given by the relation $C=\frac{5F-160}{9}$ .What is the numerical value of the temperature which is same in both the scales?

Answer:
$C =30^{\circ}$
Solution:
Given, $\frac{5F-160}{9}$
Put F = 86° in equation (1)
$C =\frac{5(86)-160}{9}$
$C =\frac{430-160}{9}=\frac{270}{9}=30^{\circ}$
Hence, $C =30^{\circ}$

Answer:
$95^{\circ}F$
Solution:
Given, $\frac{5F-160}{9}$
Put C = $35^{\circ}C$ in above equation
$35^{\circ}=\frac{5F-160}{9}$
315 = 5F – 160
315 + 160 = 5F
475 = 5F
$F=\frac{475}{5}=95^{\circ}$

Answer:
$32^{\circ}F,-17.78^{\circ}C$
Solution:
Given $\frac{5F-160}{9}$ …(1)
Put $C=0^{\circ}$ in equation (1)
$0=\frac{5F-160}{9}$
5F = 160
$F=\frac{160}{5}=32^{\circ}$
Now put F = 0 in (1), we get
$C=\frac{160}{9}=-17.78^{\circ}C$

Answer:
– 40⁰
Solution:
Given that $C=\frac{5F-160}{9}$
It is asked that the numerical value of the temperature which is same in both the scales.
It means that the value of $^{\circ}C$ and $^{\circ}F$ are same
$\therefore F =\frac{5F-160}{9}$
9F – 5F = –160
4F = –160
F = –40⁰

Question:5

If the temperature of a liquid can be measured in Kelvin units as $x^{\circ}K$ or in Fahrenheit units as $y^{\circ}F$ , the relation between the two systems of measurement of temperature is given by the linear equation $y =\frac{9}{5}(x-273)+32$ .(i) Find the temperature of the liquid in Fahrenheit if the temperature of the liquid is $313^{\circ}K$ . (ii) If the temperature is $158^{\circ}F$ , then find the temperature in Kelvin.

Answer:
$104^{\circ}F$

Given that,

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

Put the value of x = $313^{\circ}K$ in equation (1), we get

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

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

$y =\frac{360}{5}+32$

$y =104^{\circ}F$

Given that,

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

$y=158^{\circ}$ in equation (1)

we get

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

$158^{\circ} =\frac{9(x-273)+160}{5}$

790 – 160 = 9x – 2457630 + 2457 = 9x
3087 = 9x
343 = x
Hence answer is 343K

Question:6

The force exerted to pull a cart is directly proportional to the acceleration produced in the body. Express the statement as a linear equation of two variables and draw the graph of the same by taking the constant mass equal to 6 kg. Read from the graph, the force required when the acceleration produced is 5 m/sec².

The force exerted to pull a cart is directly proportional to the acceleration produced in the body. Express the statement as a linear equation of two variables and draw the graph of the same by taking the constant mass equal to 6 kg. Read from the graph, the force required when the acceleration produced is 6 m/sec².

Answer:

It is given that force is directly proportional to acceleration
i.e., $F \propto a$
F = ma {Here m is proportionality constant and also mass of body}
It is also given that m = 6
$\therefore F=6a$

F	6	12
a	1	2

The given equation is F = 6a
When acceleration is 5 m/sec²
Put a = 5
$F = 6 \times 5 = 30$ kgm/sec²
Hence the required answer is 30 kgm/sec²

It is given that force is directly proportional to acceleration
i.e., $F \propto a$
F = ma {Here m is proportionality constant and also mass of body}
It is also given that m = 6
$\therefore F=6a$

F	6	12
a	1	2

The given equation is
F = 6a
When acceleration is 6 m/sec²
Put a = 6
$F = 6 \times 6 = 36$ kgm/sec²
Hence the required answer is 36 kgm/sec²

NCERT Exemplar Solutions Class 9 Maths Chapter 4 Important Topics:

The major pointers of this chapter covered in the NCERT exemplar Class 9 Maths solutions chapter 4 are mentioned below:

◊ Making two-variable linear equations for any given situation or statement.

◊ We will learn to solve these linear equations to find out the values of variables.

◊ NCERT exemplar Class 9 Maths solutions chapter 4 include methods to solve these linear equations to determine the values of variables.

◊ These linear equations can be drawn on a Cartesian plane as straight lines.

◊ We can find out the values of variables by geometric representations of these equations.

◊ We learn about linear equations representing lines parallel to X-axis or Y-axis and the line passing through the origin.

NCERT Class 9 Exemplar Solutions for Other Subjects:

NCERT Class 9 Maths Exemplar Solutions for Other Chapters:

Chapter 1 Number System

Chapter 2 Polynomials

Chapter 3 Coordinate geometry

Chapter 5 Introduction to Euclid’s Geometry

Chapter 6 Lines and Angles

Chapter 7 Triangles

Chapter 8 Quadrilaterals

Chapter 9 Area of Parallelograms and Triangles

Chapter 10 Circles

Chapter 11 Constructions

Chapter 12 Heron’s Formula

Chapter 13 Surface Areas and Volumes

Chapter 14 Statistics and Probability

Features of NCERT Exemplar Class 9 Maths Solutions Chapter 4:

These Class 9 Maths NCERT exemplar chapter 4 solutions will help the students grasp the basics of linear equations in two variables. It gives two relations between Binomial of two variables of a single degree. With the help of those two equations, we have to determine the values of two variables individually. The chapter on Linear Equations in Two Variables can be studied and practiced using these NCERT exemplar Class 9 Maths chapter 4 solutions Linear Equations in Two Variables and will be enough to solve other books such as A textbook of Mathematics by Monica Kapoor, NCERT Class 9 Maths, RD Sharma Class 10 Maths, RS Aggarwal Class 9 Maths et cetera. NCERT exemplar Class 9 Maths solutions chapter 4 pdf download is a free-to-use feature. The solutions can be downloaded using any tools that are available to convert the page to pdf.

Check the solutions of questions given in the book

Chapter No.	Chapter Name
Chapter 1	Number Systems
Chapter 2	Polynomials
Chapter 3	Coordinate Geometry
Chapter 4	Linear Equations In Two Variables
Chapter 5	Introduction to Euclid's Geometry
Chapter 6	Lines And Angles
Chapter 7	Triangles
Chapter 8	Quadrilaterals
Chapter 9	Areas of Parallelograms and Triangles
Chapter 10	Circles
Chapter 11	Constructions
Chapter 12	Heron’s Formula
Chapter 13	Surface Area and Volumes
Chapter 14	Statistics
Chapter 15	Probability

Also, Read NCERT Solution Subject Wise

Check NCERT Notes Subject Wise

Also Check NCERT Books and NCERT Syllabus here

Frequently Asked Questions (FAQs)

1. Can we solve any two equations of two variables?

No, we cannot solve any two equations of two variables all the time, however, yes, if these two equations are linear, we can always solve them provided they have a solution.

2. Is it necessary for a linear equation of two variables to have a solution?

No, two equations written in the linear form of two variables can have zero solution, infinite solution and precisely one unique solution.

3. Can we solve the linear equations of three variables?

Yes, we can solve linear equations of three variables. However, the same method we use in linear equations of two variables will not be used as later we will study solving these equations using a matrix method.

4. What is the mark distribution of Linear equation in two variables for the final examination?

Generally 8-10% of the marks of the final paper account for Linear equation in two variables. The class 9 maths NCERT exemplar solutions of chapter 4 are equipped with the detailed solutions to ace the final examinations.

Also Read

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NCERT Class 9 Maths Chapter 8 Notes Quadrilaterals - Download PDF

NCERT Class 9 Maths Chapter 2 Notes Polynomials - Download PDF

NCERT Class 9 Maths Chapter 1 Notes Number System - Download Free PDF

NCERT Class 9 Maths Chapter 5 Notes Introduction To Euclid’s Geometry - Download PDF

NCERT Class 9 Maths Chapter 6 Notes Lines and Angles - Download PDF

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

NCERT Exemplar Class 9 Maths Solutions Chapter 4 Linear Equations in Two Variables

NCERT Exemplar Class 9 Maths Solutions Chapter 4: Exercise 4.1

NCERT Exemplar Class 9 Maths Solutions Chapter 4: Exercise 4.2

NCERT Exemplar Class 9 Maths Solutions Chapter 4: Exercise 4.3

NCERT Exemplar Class 9 Maths Solutions Chapter 4: Exercise 4.4

NCERT Exemplar Solutions Class 9 Maths Chapter 4 Important Topics:

NCERT Class 9 Exemplar Solutions for Other Subjects:

NCERT Class 9 Maths Exemplar Solutions for Other Chapters:

Features of NCERT Exemplar Class 9 Maths Solutions Chapter 4:

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