Lesson: Linear Equations In Two Variables

Exercise 4.1 (19 Multiple Choice Questions and

Answers)

Question: 1

The linear equation 2x – 5y = 7 has:

(a) A unique solution

(b) Two solutions

(c) Infinitely many solutions

(d) No solution

Solution:

c

Since, 2x – 5y = 7 is a linear equation, we get a line

when we join all the points which satisfy the equation.

So, there are infinite numbers of points on this line

which satisfy the equation 2x – 5y = 7.

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

Solution:

a

In case of natural numbers, there is only one solution

i.e. (1, 1) which satisfies the equation 2x + 5y = 7. But

in case of positive real numbers, real numbers or

rational numbers, there will be infinite number of

solutions for the equation.

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:

a

As (2, 0) is a solution of the linear equation

2x +3y = k,

2 2 3 0 k

4 k

Question: 4

Any solution of the linear equation 2x + 0y + 9 = 0 in

two variables is of the form:

(a)

9

,

2

m

(b)

9

,

2

n

(c)

9

0,

2

(d) (−9, 0)

Solution:

a

Given linear equation is 2x + 0y + 9 = 0.

2 9 x

9

2

x

So, the value of and that of can be any real

9

2

x

number.

Question: 5

The graph of the linear equation 2x + 3y = 6 cuts the y

− axis at the point:

(a) (2, 0)

(b) (9, 3)

(c) (3, 0)

(d) (0, 2)

Solution:

d

The given linear equation is 2x + 3y = 6. If the line cut

the y − axis, the value of x at the point of intersection

will be zero. Hence, the point is (0, 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

Solution:

b

The given linear equation is x = 7.

Here, the coefficient of y in the equation is zero.

Hence, the required equation is 1. x + 0. y = 7.

Question: 7

Any point on the x − axis is of the form:

(a) (x, y)

(b) (0, y)

(c) (x, 0)

(d) (x, x)

Solution:

c

The y – coordinate of any point on the x – axis is zero.

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)

Solution:

a

The given equation y = x shows that the x –coordinate

of the point is equal to the y – coordinate of the point.

Question: 9

Any equation of x − axis of the form:

(a) x = 0

(b) y = 0

(c) x + y = 0

(d) x = y

Solution:

b

The value of the y – coordinate for all the points at the

x – axis is zero.

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.

Solution:

a

The graph of y = 6 is a line parallel to the x − axis at a

distance 6 units from the origin.

Question: 11

x = 5, y = 2 is a solution of the linear equation

(a) x + 2y = 7

(b) 5x + 2y = 7

(c) x + y = 7

(d) 5x + y = 7

Solution:

c

For option (a), the given equation is x + 2y = 7.

LHS = x + 2y

= 5 + 2 × 2

= 9 ≠ 2

Hence, LHS ≠ RHS

For option (b), the given equation is 5x + 2y = 7.

LHS = 5x + 2y

= 5 × 5 + 2 × 2

= 29 ≠ 7

Hence, LHS ≠ RHS

For option (c), the given equation is x + y = 7.

LHS = x + y

= 5 + 2

= 7

Hence, LHS ≠ RHS

For option (d), the given equation is 5x + y = 7.

LHS = 5x + y

= 5 × 5 + 2

= 27 ≠ 7

Hence, LHS ≠ RHS

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

Solution:

b

Let us consider a linear equation ax + by + c = 0 … (i)

Since, (−2, 2), (0, 0) and (2, −2) are the solutions of

the linear equation, therefore these points satisfy

eq (i).

Corresponding to the point (−2, 2), the equation is

−2x + 2b + c = 7 … (ii)

Corresponding to the point (0, 0), the equation is

0 + 0 + c = 0 … (iii)

= 0 c

Corresponding to the point (2, −2), the equation is

2a − 2b + c = 7 … (iv)

From eqs. (ii) and (iii),

c = 0 and −2a + 2b + c = 0

2 2 a b

2

2

b

a b

On putting a = b and c = 0 in eq. (i)

bx + by + 0 = 0

+ 0 bx by

+( ) 0 b x y

+ 0, 0 x y b

Hence, x + y = 0 is the required form of the linear

equation.

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

Solution:

a

The first quadrant contains only positive values of x

and y. So, the positive solutions of the equation

ax + by + c = 0 always lie in the 1

st

quadrant.

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)

Solution:

c

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

At the x − axis, the value of y is zero.

2 3 0 6 x

2 6 x

6

3

2

x

Hence, the coordinate on the x − axis is (3, 0).

Question 15

The graph of the linear equation y = x passes

through the point:

(a)

3 3

,

2 2

(b)

3

0,

2

(c) (1, 1)

(d)

1 1

,

2 2

Solution:

c

The given equation is y = x.

In this equation the value of x is equal to the value of

y. So, at any point both the coordinates are equal.

Hence, the required point is (1, 1).

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) Doubles

(b) Remains the same

(c) Changes in case of multiplication only

(d) Changes in case of division only

Solution:

b

By property, 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 i.e.,

the solutions of the linear equation remains

unchanged.

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

Solution:

c

Let the linear equation be ax + bx + c = 0.

On putting x = 1 and x = 2, in this equation we get

x + 2b + c = 0, where a, b and c, are real number.

Here, different values of a, b and c, satisfy

a + 2b + c = 0.

Hence, infinitely many linear equations in x and y can

be satisfied by x = 1 and x = 2.

Question: 18

The point of the form (a, a) always lies on:

(a) a − axis

(b) y − axis

(c) On the line y − x

(d) On the line y + x = 0

Solution:

c

If the value of x is equal to value of y. The point of the

form (a, a) always lies on the line y = x.

Question: 19

The point of the form (a, −a) always lies on the line:

(a) x = a

(b) x = −a

(c) y = x

(d) x + y = 0

Solution:

d

The given equation is x + y = 0.

This can be written as x = –y.

So, x and y are equal and of opposite sign. Hence, the

point of the form (a, –a) always lies on the line

x + y = 0

Exercise 4.2

Question: 1

Write whether the following statements are True or

False? Justify your answer:

(a) The point (0, 3) lies on the graph of the linear

equation 3x + 4y = 12.

(b) The graph of the linear equation x + 2y = 7 passes

through the point (0, 7).

(c) The graph given below represents the linear

equation x + y = 0.

(d) The graph given below represents the linear

equation x = 3

(e) The coordinates of points in the table

represent some of the solutions of the equation

x – y + 2 = 0.

X

0

1

2

3

4

Y

2

3

4

5

6

(f) Every point on the graph of a linear equation in two

variables does not represent a solution of the linear

equation.

(g) The graph of every linear equation in two variables

need not be a line.

Solution:

(a) 3x + 4y = 12 …… (i)

Putting x = 0 and y = 3, we get,

3 × 0 + 4 × 3

= 0 +12

= 12

∴ LHS = RHS

True, since (0, 3) satisfies the equation 3x + 4y = 12.

(b) LHS = x +12y

Putting x = 0, y = 7

= 0 + 2 × 7

= 12

∴ ≠ RHS

False, since (0, 7) does not satisfies the equation.

(c) LHS = x +y

Putting x = −3 and y = 3, we get

= −3 + 3

= 0 = RHS

Again, at the point (−1, 1) putting x = −1 and y = 1,

we get

−1 + 1 = 0 = RHS

True, since (–1, 1) and (–3, 3) satisfy the given

equation and the two points determine a unique line.

(d) The given linear equation is x = 3

This can be written as x + 0.y = 3

x

3

3

3

y

0

1

2

It is a line parallel to the y – axis at a distance of 3

units in the +ve direction.

True, since this graph is a line parallel to y – axis at a

distance of 3 units (to the right) from it.

(e) The coordinates of the points are (0, 2), (1, 3),

(2, 4), (3, –5), and (4, 6).

Given equation is x – y + 2 = 0.

At the point (0, 2),

0 2– 2 22 = 0

It satisfies the given equation

At the point (1, 3),

1 3 2 = 3 3 0 0 = 0.

It satisfies the given equation

At the point (3, –5),

3 ( 5)+ 2 = 3+ 5+ 2 = 0 10 0

It does not satisfy the given equation.

At the point (4, 6),

4 6 2 = 6 6 0 0 0.

It satisfies the given equation.

Hence, the point (3, 5) does not satisfy the given

equation.

(f) False, since every point on the graph of the

equation represents a solution.

(g) False, since the graph of a linear equation in

two variables is always a line.

(vi) False, since every point on the graph of the

equation represents a solution.

(vii) False, since the graph of a linear equation in two

variables is always a line.

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?

Solution:

y = x

x

0

1

2

y

0

1

2

y = −x

y

0

1

−1

x

0

−1

1

Graphs of each equation is a line passing through

(0, 0).

Question: 2

Determine the point on the graph of the linear

equation 2x + 5y = 19, whose ordinate is

times its abscissa

1

1

2

Solution:

a

Given, 2x + 5y = 19 … (1)

As the ordinate is times its abscissa, .

1

1

2

3

2

y

x

Hence eq (1) becomes,

3

2 5 19

2

x

x

15

2 19

2

x

x

4 15

19

2

x x

19

19

2

x

∴ x = 2

3 3 2

3

2 2

x

y

Thus, the point on the graph 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.

Solution:

Any line parallel to the x − axis and at a distance of 3

units below it is given by y = –3.

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.

Solution:

The sum of the coordinates is 10 units.

Let x and y be two coordinates, then

x + y = 10.

x

0

10

y

10

0

Question: 5

Write the linear equation such that each point on its

graph has an ordinate 3 times its abscissa.

Solution:

Let the abscissa of the point be x.

According to the question, the ordinate (y) = 3 ×

abscissa

= 3 y x

When x = 1, then y = 3 and when y = 2, then

y = 3 × 2 = 6.

x

1

2

y

3

6

Here, we find two points A(1, 3) and (2, 6).

So, we draw the graph by plotting the points and

joining the line AB.

Hence, y = 3x is the required equation such that point

on its graph has an 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.

Solution:

Given the graph,

3y = ax + 7 …… (i)

Putting the value x = 3 and y = 4 in eq. (i),

3.4 = a.3 + 7

.3 12 7 a

5

3

a

Question 7

How many solution(s) of the equation 2x + 1 = x – 3

are there on the:

(a) Number line

(b) Cartesian plane

Solution:

(a) 2x + 1 = x – 3

2 3 1 x x

4 x

x = −4 can be represented on the number line as

shown.

(b) x = −4

x = −4 has infinitely many solutions on the

cartesian plane.

Question: 8

Find the solution of the linear equation x + 2y = 8

which represents a point on:

(a) x − axis

(b) y − axis

Solution:

(i) x + 2y = 8

When the point is on the x − axis, then putting y = 0,

we get,

x + 2 × 0 = 8

8 x

∴ The point on the x − axis is (8, 0)

When the point is on the y − axis, then putting x = 0,

we get,

0+ 2y = 8

8

4

2

y

∴ The point on the y − axis 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.

Solution:

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

Now, by the condition, and y − coordinate of given

linear equation are same i.e., x = y.

Putting x = y, we get

2x + cy = 8

8 2 cx x

8 2

, 0

x

c x

x

Thus, for every value of x, we have a corresponding

value of c.

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?

Solution:

It is given that y is directly proportional to x.

So,

y ∝ x

y = kx (k = constant)

Putting y =12 and x = 4, we get

.

12

3

4

k

3 y x

if x = 5 then

x = 3 × 5 = 15

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.

Solution:

y = 9x – 7

At the point A(1, 2), x = 1 and y = 2

we have,

2 = 9(1) – 7

2 2

Thus, the point A lies on the given line.

For the point B(–1, –16), x = –1, and y = –16

we have,

–16 = 9(1) – 7

16 16

At the point B lien on the given line.

For the point C(0, –7), we have

–7 = 9(0) – 7

7 7

Thus, the point C lies on the given line.

Hence, all the three points A, B and C lie on

y = 9x − 7.

Question: 2

The following values of x and y satisfy a linear

equation.

x

6

6

y

2

6

Write the linear equation.

Draw the graph using the values of x and 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:

(i) Given,

x

6

6

y

2

6

Let the linear equation be ax + by + c = 0.

Putting x = 6 and x = −2 in the above equation, we get

6a − 2b + c = 0 (A)

Putting x = −6 and y = 6 in the above equation, we get

−6a − 6b + c = 0 (B)

Adding equations (A) and (B), we get

4b + 2c = 0

2 c b

Putting the value of c in (A), we get

6a − 2b − 2b = 0

6 4 a b

2

3

b

a

Now, putting the values of and in ax + by + c = 0, we

get

2

2 0

3

b

x by b

2

2 0

3

x y

2

2

3

y x

Multiplying both the sides by 3, we get

3y = − 2x + 6

2 3 6 0 x y

The graph cuts the x − axis at (3, 0) and the 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.

Solution:

3x + 4y = 6

3 6 4 x y

6 4

3

y

x

x

0

2

y

3

2

0

The graph cuts the x − axis at (2, 0) and the y − axis at

.

3

0,

2

Question: 4

The linear equation that converts

Fahrenheit (F) to Celsius (C) is given by the relation

5 160

9

F

C

(a) If the temperature is 86°F, what is the temperature

in Celsius?

(b) If the temperature is 35°C, what is the temperature

in Fahrenheit?

(c) 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?

(d) What is the numerical value of the temperature

which is same in both the scales?

Solution:

(a)

5 160

9

F

C

Putting f = 86, we get,

5 86 160

9

C

430 160

9

270

30

9

The temperature is 30°C.

(b)

5 160

9

F

C

5 160

9

F

C

Putting c = 35,

5 160

35

9

F

5 160 315 F

5 315 160 F

475

5

F

∴ F = 95

The temperature is 95°F.

(c)

5 160

9

F

C

Putting C = 0,

5 160

0

9

F

5 160 0 F

5 160 F

160

5

F

∴ F = 32

The temperature is 32°F.

5 160

9

F

C

Putting F = 0,

5 160

9

F

C

160

9

C

(d)

5 160

9

F

C

Putting F = C

5 160

9

C

C

9 5 160 C C

9 5 160 C C

4 160 C

160

4

C

40 C

∴ The required value is 40°C.

Question: 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

9

273 32

5

y x

(a) Find the temperature of the liquid in Fahrenheit if

the temperature of the liquid is 313°K.

(b) If the temperature is 158° F, then find the

temperature in Kelvin.

Solution:

(a)

9

273 32

5

y x

Putting x = 313,

9

313 273 32

5

y

9

40 32

5

= 72 + 32

= 104

So the value is 140°F.

(b)

9

273 32

5

y x

Putting y = 158,

9

158 273 32

5

x

5

158 32 273

9

x

5

126 273

9

x

x = 343

∴ The required value is 343°K.

Question: 6

(i) 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

(a) 5 m/sec

2

(b) 6 m/sec

2

Solution:

y = mx, where y denotes the force, x denotes the

acceleration and m denotes the constant mass.

Let the force exerted to pull a cart be F and the

acceleration be a.

F ∝ a

F ma

6 F a

The required linear equation in two variables is

F = 6a.

a

1

2

f

6

12

Representing the force on the x − axis and the

acceleration on the y − axis, we get the following

graph

From the graph, we find that

(a) When acceleration produced = 5m/sec

2

,

the forced required = 30 N

(b) When acceleration produced = 6m/sec

2

,

the forced required = 36 N.