Linear Equations in Two Variables

Exercise No. 4.1

Multiple Choice Questions:

Write the correct answer in each of the following:
1. The linear equation 2x – 5y = 7 has
(A) A unique solution
(B) Two solutions
(C) Infinitely many solutions
(D) No solution

Solution:
The given linear equation 2x – 5y = 7 has two variable. Since, a linear equation with two
variable has infinitely many solution.
Hence, the correct option is (C).

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

Solution:
The given equation 2x + 5y = 7 has a unique solution, if x, y are natural number.
Hence, the correct option is (a).

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:
Put x = 2 and y = 0 in the linear equation 2x + 3y = k.
2 2  3 0  k
k 4
Hence, the correct option is (A).

4. Any solution of the linear equation 2x + 0y + 9 = 0 in two variables is of

the form
(A)   , m 
9
 2 
(B)  n,  
9
 2

(C)  0,  
9
 2
(D) (– 9, 0)

Solution:
Consider the linear equation:
2x + 0y + 9 = 0
Now,
2x = -9
9
x
2
Since, the coefficient of y is 0 in the given equation. So, the solution can be given as
 9 
 ,m .
 2 
Hence, the correct option is (A).

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)

Solution:
The graph of the linear equation 2x + 3y = 6 cuts the y-axis at the point where x-coordinate is
zero.
So, put x = 0 in 2x + 3y = 6 , get:
2x  3y  6
2  0  3 y  6
3y  6
y2
So, the point is (0, 2).
Hence, the correct option is (D).

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
Solution:
The equation x = 7 in two variable can be written as 1x  0 y  7 .
Hence, the correct option is (B).

7. Any point on the x-axis is of the form

(A) (x, y)
(B) (0, y)
(C) (x, 0)
(D) (x, x)

Solution:
We know that any point on the x-axis has its ordinate 0. Since, the point will be (x, 0).
Hence, the correct option is (C).

8. Any point on the line y = x is of the form

(A) (a, a)
(B) (0, a)
(C) (a, 0)
(D) (a, – a)

Solution:
Any point on the line y = x is of the form (a, a) because the line y = x will have x and y
coordinate same.
Hence, the correct option is (a).

9. The equation of x-axis is of the form

(A) x = 0
(B) y = 0
(C) x + y = 0
(D) x = y

Solution:
The equation of x-axis is of the form y = 0.
Hence, the correct option is (B).

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.

Solution:
The graph of y = 6 is a line parallel to x-axis because it does not contain x.
Hence, the correct option is (A).

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

Solution:
x = 5, y = 2 is a solution of the linear equation x + y = 7 because 5 + 2 = 7.
Hence, the correct option is (C).

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

Solution:
The point (-2, 2) and (2, -2) have x and y-coordinate of opposite signs.
Now, the sum of the x and y-coordinate is: x + y = - 2 + 2 = 0 i.e., y = - x will have x and y
coordinate opposite signs and also satisfies the point (0, 0).
Hence, the correct option is (B).

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

Solution:
We know that I quadrant has consists of all the point (x, y) positive. So, the positive solution
of the equation ax + by + c = 0 lie in I quadrant.
Hence, the correct option is (A).

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)
Solution:
The graph of the linear equation 2x + 3y = 6 is a line which meets the x-axis at the point
where y = 0. Now, put y = 0 in the given equation, get:
2x  3y  6
2x  y  0  6
2x  6
x3
So, the point is (3, 0) which is lies on the line 2x + 3y = 6.
Hence, the correct option is (C).

15. The graph of the linear equation y = x passes through the point
3 3
(A)  , 
2 2 
(B)  0, 
3
 2
(C) 1,1
1 1
(D)  , 
 2 2

Solution:
The graph of the linear equation y = x passes through the point where x and y coordinate will
have same that is (1, 1).
Hence, the correct option is (C).

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

Solution:
If we multiply or divide both sides of a linear equation with a non-zero number, then the
solution of the linear equation remains the same.

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
Solution:
There are infinitely linear equations in x and y can be satisfied by x = 1 and y = 2. For
example: x + y = 3 or 2y – x = 3.
Hence, the correct option is (C).

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

Solution:
The point of the form (a, a) always lies on y = x, because x and y coordinate are same.
Hence, the correct option is (C).

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

Solution:
The point of the form (a, – a) always lies on the line y = -x or x + y = 0 because x and y
coordinate have opposite signs.
Hence, the correct option is (D).
 Exercise No. 4.2
Short Answer Questions with Reasoning:
Write whether the following statements are True or False? Justify your
answers:
1. The point (0, 3) lies on the graph of the linear equation 3x + 4y = 12.

Solution:
Consider the equation:
3x + 4y = 12

Putting x = 0 and y = 3 in the equation, get:

3  0  4  3  12
12  12 True
So, the point (0, 3) satisfies the equation 3x + 4y = 12.
Hence, the given statement is true.

2. The graph of the linear equation x + 2y = 7 passes through the point (0,
7).

Solution:
Consider the linear equation:
x + 2y = 7

Putting x = 0 and y = 7 in the given equation x + 2y = 7, get:

x  2y  7
0  2 7  7
14  7 Which is False
So, the point (0, 7) does not satisfy the equation.
Hence, the given statement is false.

3. The graph given below represents the linear equation x + y = 0.

Solution:
The given equation is x + y = 0 that is y = - x. Now, any point on the given graph has x and y
–coordinate of opposite signs as, the point (-1, 1) and (-3, 3). So, these points satisfy the
given equation.
Hence, the given statement is true.

4. The graph given below represents the linear equation x = 3 (see Fig. 4.2).

Solution:
See the given graph, the equation of the given line is x = a that is x= 3 parallel to the y-axis
and to the right of y-axis, if a>0.
Hence, the given statement is true.

5. 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.

Solution:
Consider the equation:
x – y + 2 = 0.

The point (0, 2), (1, 3), (2, 4) and (4, 6) satisfy the given equation then they have solution of
it. Now, the point (3, -5) does not satisfy the given equation as 3-(-5)+2= 0, that is 3+5+2=0
or 10 = 0 that is false.
Hence, the given statement is false.

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

Solution:
Every point on the graph of a linear equation in two variables does represent a solution of the
linear equation. So, the given statement is false.

7. The graph of every linear equation in two variables need not be a line.
Solution:
The graph of every linear equation in two variables is always a line. So, the given statement
is false.
 Exercise No. 4.3
Short Answer Questions:
1. Draw the graphs of linear equations y = x and y = – x on the same cartesian
plane.
What do you observe?

Solution:
The point on the graph y = x will have same signs of x and y coordinate but the point on the
graph y = -x will have opposite signs of x and y coordinate.
The graph of the linear equation y = x and y = -x on the same Cartesian plane is show below.
We observed that both graph are passes through the same point (0, 0) and they are mirror
image of each other about the y-axis.

2. Determine the point on the graph of the linear equation 2x + 5y = 19, whose
ordinate is 1 1 times its abscissa.
2

Solution:
Consider the linear equation:
2x + 5y = 19
Let the coordinate of the point (2, 3).
Putting x = 2 and y = 3
2x  5 y  2  2  5 3
 4  15
 19

Therefore, the point (2, 3) is the solution of the given equation.

Now, abscissa of the point is 2and ordinate is 3.
Then,
1 3
2 1  2  3
2 2
1
Hence, the ordinate of the point (2, 3) is 1 times its abscissa.
2

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.

Solution:
The graph of the equation y = - 3 is parallel to the x-axis and at a distance 3
units and passing through the point (0, -3) as show in the figure given below.
 Linear Equations in Two Variables

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

Solution:
The graph of the linear equation whose solutions are represented by the points having the sum
of the coordinates as 10 units is x + y = 10.
When x= 0 then y = 10 and when y = 0 then x = 10.
Now, plot these two point (0, 10) and (10, 0) on the graph of the paper and joint them to get
the straight line.
Hence, the graph of x + y = 10 is a straight line as show in the figure given below.
 NCERT Exemplar Solutions for Class 9 Math’s
Chapter 4
Linear Equations in Two Variables

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

Solution:
The equation of linear equation such that point on its graph has an ordinate 3 times is y= 3x.

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

Solution:
Consider the linear equation:
3y = ax + 7
The point (3, 4) lies on the graph of 3y = ax + 7. So, this point will be satisfy the given equation,
get:
3 y  ax  7
12  3  a  7
12  7  3a
3a  5
5
a
3
 5
Hence, the value of a is .
3

7. How many solution(s) of the equation 2x + 1 = x – 3 are there on the:

(i) Number line
(ii) Cartesian plane

Solution:
(i) Consider the linear equation:
2x + 1 = x – 3
2x-x = -3-1
x = -4
Hence, the x = - 4 is the solution if the given equation in the number line.
(ii) The given linear equation 2x + 1 = x – 3 have infinitely many solution which are lies
on the cartesion plane.

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

on
(i) x-axis (ii) y-axis

Solution:
(i) The linear equation which lies on the x-axis has its ordinate 0.
Now, Putting y = 0 in the equation x + 2y = 8, get:
x  2y  8
x  2 0  8
x 8
(ii) The linear equation which lies on the y-axis has its abscissa 0.
Now, Putting x = 0 in the equation x + 2y = 8, get:
x  2y  8
0  2y  8
8
y
2
y4
9. For what value of c, the linear equation 2x + cy = 8 has equal values of x
and y for its solution.

Solution:
Consider the linear equation:
2x + cy = 8
According to the question, when x = y. Putting y = x, get:
2 x  cx  8
cx  8  2 x
8  2x
c ,x  0
x
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?

Solution:
Given:
y varies directly as x
y x
y  kx

Putting y = 12, when x = 4, get:

12  k 4
12
k
4
k 3
Therefore, the equation is y = 3x.
Now, the value of y when x = 5 is y  3  5  15 .
 Exercise No. 4.4

Long Answer Questions:

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.

Solution:
For A (1, 2), we have 2 = 9(1) – 7 = 9 – 7 = 2
For B(-1, -16), we have -16 = 9(-1) – 7 = -9 – 7 = -16
For C(0, -7), we have -7 = 9(0) – 7 = 0 – 7 = -7

The line y = 9x – 7 is satisfied by the point A (1, 2), B (– 1, – 16) and C (0, – 7). Hence, the
point A (1, 2), B (– 1, – 16) and C (0, – 7) are solution of the linear equation y = 9x – 7.

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

Solution:
Consider the linear equation:
3x + 4y = 6
Both the point (6, -2) and (-6, 6) satisfy the linear equation 3x + 4y = 6. Now, plot the point
(6, -2) and (-6, 6) on the graph paper and then joint these two points and obtain a line. See the
below graph it is cut the x-axis at point (3, 0) and y-axis at (0, 2).
 Linear Equations in Two Variables

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

graph cuts the x-axis and the y-axis.

Solution:
Consider the linear equation:
3x + 4y = 6

The solution of given(above) linear equation cab be expressed as a table as follows:

x 2 -2 0
y 0 3 1.5

Now, plot that above point on the graph paper and then join the points, see below:

Therefore, the graph cuts the x-axis at (2, 0) and y-axis at (0, 1.5).

4. The linear equation that converts Fahrenheit (F) to Celsius (C) is given by
the relation
5F  160
C
9
(i) If the temperature is 86°F, what is the temperature in Celsius?
(ii) If the temperature is 35°C, what is the temperature in Fahrenheit?
(iii) If the temperature is 0°C what is the temperature in Fahrenheit and if
the temperature is 0°F, what is the temperature in Celsius?
(iv) What is the numerical value of the temperature which is same in both
the scales?

Solution:
Consider the linear equation:
5 F  160
C 
9
(i) Putting F  86o , get:
5 F  160
C
9
5  86o  160

9
430  160

9
270

9
 30o
Hence, the temperature in Celsius is 3 0 o .

(ii) Putting C  35o , get:

5 F  160
C
9
5 F  160
35o 
9
315  5 F  160
o

5 F  315  160
 415
475
F
5
F  95o
Hence, the temperature in Fahrenheit is 95F .

(iii) Putting C  0 o , get:

 5 F  160
C
9
5 F  160
0
9
0  5 F  160
5 F  160
160
F
5
F  32o

Now, putting F = 0 o , get:

5 F  160
C
9
5  0  160
C
9
o
 160 
  
 9 
If the temperature is 0F , then the temperature in Fahrenheit is 3 2 o and if the
o
 160 
temperature is 0 F then the temperature in Celsius is    C.
 9 

(iv) Putting C  F , get:

5 F  160
C
9
CF
5 F  160
F
9
9 F  5 F  160
4 F  160
160
F
4
F  40o
Therefore, the numerical value of the temperature which is same in both the scales is
40 .
The linear equation that converts kelvin (x) to Fahrenheit (y) is given by the relation:
9
y  x  273  32
5

5. If the temperature of a liquid can be measured in kelvin units as x° K or

in Fahrenheit units as y° F. The relation between the two systems of
measurement of temperature is given by the linear equation
 Linear Equations in Two Variables
(i) Find the temperature of the liquid in Fahrenheit if the temperature of the
body is 313°K.
(ii) If the temperature is 158° F, then find the temperature in Kelvin.

Solution:
9
y  x  273  32
5
(i) When the temperature of the liquid is x  313o K .
9
y   313  273  32
5
9
  40  32
5
 72o  32o
 104o F
When the temperature of the liquid is y 158 F .
o
(ii)
9
158   x  273  32
5
9
  x  273
5
 158  32
5
x  273  126 
9
 70
x  273  70
x  273  70
x  343o K

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 (i) 5 m /s 2 , (ii) 6 m /s 2 .

Solution:
According to the question:
yx
y  mx
Now, make a table as follows by writing the values of y below the corresponding value of x.

x 0 1 2
y 0 6 12
Plot the point (0, 0), (1, 6) and (2, 12) on a graph paper. Now, joint all the points and obtain a
line, see below.