# R S AGGARWAL AND V AGGARWAL Solutions for Class 9 Maths Chapter 4 – Linear Equations in Two Variables

## Chapter 4 – Linear Equations in Two Variables Exercise Ex. 4A

Question 1

Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in
each case

3x + 5y = 7.5

Solution 1

We
have,

3x + 5y = 7.5

3x + 5y –
7.5 = 0

6x + 10y
– 15 = 0

On comparing this equation with ax
+ by + c = 0, we obtain

a
= 6, b = 10 and c =
-15

Question 2

Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in
each case

Solution 2

On comparing this equation with ax
+ by + c = 0, we obtain

a
= 10, b =
-1 and c = 30

Question 3

Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in
each case

3y – 2x = 6

Solution 3

We
have,

3y – 2x = 6

-2x + 3y – 6
= 0

On comparing this equation with ax
+ by + c = 0, we obtain

a
=
-2, b = 3 and c =
-6

Question 4

Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in
each case

4x = 5y

Solution 4

We
have,

4x = 5y

4x – 5y = 0

On comparing this equation with ax
+ by + c = 0, we obtain

a
=
4, b = -5 and c = 0

Question 5

Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in
each case

Solution 5

6x – 5y = 30

6x – 5y – 30 = 0

On comparing this equation with ax
+ by + c = 0, we obtain

a
= 6, b =
-5 and c = 30

Question 6

Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in
each case

Solution 6

On comparing this equation with ax
+ by + c = 0, we obtain

a
= , b =  and c =
-5

Question 7

Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in
each case

x = 6

Solution 7

We
have,

x
= 6

x – 6 = 0

1x + 0y – 6 =
0

x + 0y – 6 = 0

On comparing this equation with ax
+ by + c = 0, we obtain

a
= 1, b = 0 and c =
-6

Question 8

Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in
each case

3x – y = x – 1

Solution 8

We
have,

3x – y = x – 1

3x – x – y + 1
= 0

2x – y + 1 = 0

On comparing this equation with ax
+ by + c = 0, we obtain

a
= 2, b =
-1 and c = 1

Question 9

Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in
each case

2x + 9 = 0

Solution 9

We
have,

2x + 9 = 0

2x + 0y + 9 =
0

On comparing this equation with ax
+ by + c = 0, we obtain

a
= 2, b = 0 and c = 9

Question 10

Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in
each case

4y = 7

Solution 10

We
have,

4y = 7

0x + 4y – 7 = 0

On comparing this equation with ax
+ by + c = 0, we obtain

a
= 0, b = 4 and c =
-7

Question 11

Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in
each case

x + y = 4

Solution 11

We
have,

x + y = 4

x + y – 4 = 0

On comparing this equation with ax
+ by + c = 0, we obtain

a
= 1, b = 1 and c =
-4

Question 12

Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in
each case

Solution 12

We
have,

3x – 8y – 1 = 0

On comparing this equation with ax
+ by + c = 0, we obtain

a
= 3, b =
-8 and c = -1

Question 13

Check which of the following are the solutions of the
equation 5x – 4y = 20.

(4, 0)

Solution 13

Given
equation is 5x – 4y = 20

Substituting
x = 4 and y = 0 in L.H.S. of given equation, we get

L.H.S.
= 5x – 4y

= 5(4) – 4(0)

= 20 – 0

= 20

= R.H.S.

Hence,
(4, 0) is the solution of the given equation.

Question 14

Check which of the following are the solutions of the
equation 5x – 4y = 20.

(0, 5)

Solution 14

Given
equation is 5x – 4y = 20

Substituting
x = 0 and y = 5 in L.H.S. of given equation, we get

L.H.S.
= 5x – 4y

= 5(0) – 4(5)

= 0 – 20

= -20

R.H.S.

Hence,
(0, 5) is not the solution of the given equation.

Question 15

Check which of the following are the solutions of the
equation 5x – 4y = 20.

Solution 15

Given
equation is 5x – 4y = 20

Substituting
x = -2 and y =  in L.H.S. of given
equation, we get

L.H.S.
= 5x – 4y

= 5(-2) – 4

= -10 – 10

= -20

R.H.S.

Hence,
is not the
solution of the given equation.

Question 16

Check which of the following are the solutions of the
equation 5x – 4y = 20.

(0, -5)

Solution 16

Given
equation is 5x – 4y = 20

Substituting
x = 0 and y =
-5 in L.H.S. of given equation, we
get

L.H.S.
= 5x – 4y

= 5(0) – 4(-5)

= 0 + 20

= 20

= R.H.S.

Hence,
(0,
-5) is the solution of the given
equation.

Question 17

Check which of the following are the solutions of the
equation 5x – 4y = 20.

Solution 17

Given
equation is 5x – 4y = 20

Substituting
x = 2 and y =  in L.H.S. of given
equation, we get

L.H.S.
= 5x – 4y

= 5(2) – 4

= 10 + 10

= 20

= R.H.S.

Hence,
is the
solution of the given equation.

Question 18

Find five different solutions of each of the following
equations:

2x – 3y = 6

Solution 18

Given
equation is 2x – 3y = 6

Substituting
x = 0 in the given equation, we get

2(0)
– 3y = 6

0 – 3y = 6

3y = -6

y = -2

So,
(0, -2) is the solution of the given equation.

Substituting
y = 0 in the given equation, we get

2x
– 3(0) = 6

2x – 0 = 6

2x = 6

x = 3

So,
(3, 0) is the solution of the given equation.

Substituting
x = 6 in the given equation, we get

2(6)
– 3y = 6

12 – 3y = 6

3y = 6

y = 2

So,
(6, 2) is the solution of the given equation.

Substituting
y = 4 in the given equation, we get

2x
– 3(4) = 6

2x – 12 = 6

2x = 18

x = 9

So,
(9, 4) is the solution of the given equation.

Substituting
x = -3 in the given equation, we get

2(-3)
– 3y = 6

-6 – 3y = 6

3y = -12

y = -4

So,
(-3, -4) is the solution of the given equation.

Question 19

Find five different solutions of each of the following
equations:

Solution 19

Given
equation is

Substituting
x = 0 in (i), we get

4(0)
+ 3y = 30

3y = 30

y = 10

So,
(0, 10) is the solution of the given equation.

Substituting
x = 3 in (i), we get

4(3)
+ 3y = 30

12 + 3y = 30

3y = 18

y = 6

So,
(3, 6) is the solution of the given equation.

Substituting
x = -3 in (i), we get

4(-3)
+ 3y = 30

-12 + 3y = 30

3y = 42

y = 14

So,
(-3, 14) is the solution of the given equation.

Substituting
y = 2 in (i), we get

4x
+ 3(2) = 30

4x + 6 = 30

4x = 24

x = 6

So,
(6, 2) is the solution of the given equation.

Substituting
y = -2 in (i), we get

4x
+ 3(-2) = 30

4x – 6 = 30

4x = 36

x = 9

So,
(9, -2) is the solution of the given equation.

Question 20

Find five different solutions of each of the following
equations:

3y = 4x

Solution 20

Given
equation is 3y = 4x

Substituting
x = 3 in the given equation, we get

3y
= 4(3)

3y = 12

y = 4

So,
(3, 4) is the solution of the given equation.

Substituting
x = -3 in the given equation, we get

3y
= 4(-3)

3y = -12

y = -4

So,
(-3, -4) is the solution of the given equation.

Substituting
x = 9 in the given equation, we get

3y
= 4(9)

3y = 36

y = 12

So,
(9, 12) is the solution of the given equation.

Substituting
y = 8 in the given equation, we get

3(8)
= 4x

4x = 24

x = 6

So,
(6, 8) is the solution of the given equation.

Substituting
y = -8 in the given equation, we get

3(-8)
= 4x

4x = -24

x = -6

So,
(-6, -8) is the solution of the given equation.

Question 21

If x = 3 and y = 4 is a solution of the equation 5x – 3y =
k, find the value of k.

Solution 21

Since x = 3 and y = 4 is a
solution of the equation 5x – 3y = k, substituting x = 3 and y = 4 in
equation 5x – 3y = k, we get

5(3)
– 3(4) = k

15 – 12 = k

k = 3

Question 22

If x = 3k + 2 and y = 2k – 1 is a solution of the equation
4x – 3y + 1 = 0, find the value of k.

Solution 22

Since x = 3k + 2 and y = 2k – 1 is
a solution of the equation 4x – 3y + 1 = 0, substituting these values in
equation, we get

4(3k + 2) – 3(2k – 1) + 1 = 0

12k + 8 – 6k + 3 + 1 = 0

6k + 12 = 0

6k = -12

k = -2

Question 23

The cost of 5 pencils is equal to the cost of 2 ballpoints.
Write a linear equation in two variables to represent this statement. (Take
the cost of a pencil to be
Rs. x and that of a ballpoint to be Rs. y).

Solution 23

Let
the cost of one pencil be Rs. x and that of one
ballpoint be Rs. y.

Then,

Cost
of 5 pencils = Rs. 5x

Cost
of 2 ballpoints = Rs. 2y

According
to given statement, we have

5x
= 2y

5x – 2y = 0

## Chapter 4 – Linear Equations in Two Variables Exercise Ex. 4B

Question 1

Draw the graph of each of the following equation.

x = 4

Solution 1

x
= 4 is a line parallel to the Y-axis, at a distance of 4 units from it, to
its right.

Question 2

Draw the graph of each of the following equation.

x + 4 = 0

Solution 2

x
+ 4 = 0

x = -4, which
is a line parallel to the Y-axis, at a distance of 4 units from it, to its
left.

Question 3

Draw the graph of each of the following equation.

y = 3

Solution 3

y
= 3 is a line parallel to the X-axis, at a distance of 3 units from it, above
the X-axis.

Question 4

Draw the graph of each of the following equation.

y = -3

Solution 4

y
= -3 is a line parallel to the X-axis, at a distance of 3 units from it,
below the X-axis.

Question 5

Draw the graph of each of the following equation.

x = -2

Solution 5

x
= -2 is a line parallel to the Y-axis, at a distance of 2 units from it, to
its left.

Question 6

Draw the graph of each of the following equation.

x = 5

Solution 6

x
= 5 is a line parallel to the Y-axis, at a distance of 5 units from it, to
its right.

Question 7

Draw the graph of each of the following equation.

y + 5 = 0

Solution 7

y
+ 5 = 0

y = -5, which
is a line parallel to the X-axis, at a distance of 5 units from it, below the
X-axis.

Question 8

Draw the graph of each of the following equation.

y = 4

Solution 8

y
= 4 is a line parallel to the X-axis, at a distance of 4 units from it, above
the X-axis.

Question 9

Draw the graphs of the lines x – y = 1 and 2x + y = 8.
Shade the area formed by these two lines and the y-axis. Also, find this
area.

Solution 9

Graph of the equation x – y = 1

y = x – 1

When
x = 1, then y = 1 – 1 = 0

When
x = 2, then y = 2 – 1 = 1

Thus,
we have the following table:

 x 1 2 y 0 1

Now,
plot the points A(1, 0) and B(2, 1) on a graph
paper.

Join
AB and extend it in both the directions.

Then,
the line AB is the required graph of x – y = 1.

Graph of the equation 2x + y = 8

y = 8 – 2x

When
x = 2, then y = 8 – 2(2) = 8 – 4 = 4

When
x = 3, then y = 8 – 2(3) = 8 – 6 = 2

Thus,
we have the following table:

 x 2 3 y 4 2

Now,
plot the points C(2, 4) and D(3, 2) on a graph
paper.

Join
CD and extend it in both the directions.

Then,
the line CD is the required graph of 2x + y = 8.

The
two graph lines intersect at point D(3, 2).

The
area enclosed by the lines and Y-axis is shown in the graph.

Draw
DM perpendicular from D on Y-axis.

DM
= x-coordinate of point D(3, 2) = 3

And,
EF = 9

Area
of shaded region = Area of
ΔDEF

Question 10

Draw the graph for each of the equations x + y = 6 and x –
y = 2 on the same graph paper and find the coordinates of the point where the
two straight lines intersect.

Solution 10

Graph of the equation x + y = 6

y = 6 – x

When
x = 2, then y = 6 – 2 = 4

When
x = 3, then y = 6 – 3 = 3

Thus,
we have the following table:

 x 2 3 y 4 3

Now,
plot the points A(2, 4) and B(3, 3) on a graph
paper.

Join
AB and extend it in both the directions.

Then,
the line AB is the required graph of x + y = 6.

Graph of the equation x – y = 2

y = x – 2

When
x = 3, then y = 3 – 2 = 1

When
x = 4, then y = 4 – 2 = 2

Thus,
we have the following table:

 x 3 4 y 1 2

Now,
plot the points C(3, 1) and D(4, 2) on a graph
paper.

Join
CD and extend it in both the directions.

Then,
the line CD is the required graph of x – y = 2.

The
two graph lines intersect at point D(4, 2).

Question 11

Two students A and B contributed Rs. 100 towards the Prime Minister’s Relief Fund to help the
earthquake victims. Write a linear equation to satisfy the above data and
draw its graph.

Solution 11

Let
the amount contributed by students A and B be Rs. x
and Rs. y respectively.

Total
contribution = 100

x + y = 100

y = 100 – x

When
x = 25, then y = 100 – 25 = 75

When
x = 50, then y = 100 – 50 = 50

Thus,
we have the following table:

 x 25 50 y 75 50

Now,
plot the points A(25, 75) and B(50, 50) on a graph
paper.

Join
AB and extend it in both the directions.

Then,
the line AB is the required graph of x + y = 100.

Question 12

Draw the graph of the equation y = 3x.

From your graph, find the value of y when x = 2.

Solution 12

y
= 3x

When
x = 1, then y = 3(1) = 3

When
x = -1, then y = 3(-1) = -3

Thus,
we have the following table:

 x 1 -1 y 3 -3

Now,
plot the points A(1, 3) and B(-1, -3) on a graph
paper.

Join
AB and extend it in both the directions.

Then,
the line AB is the required graph of y = 3x.

the graph

Given:
x = 2. Take a point M on the X-axis such that OM = 2.

Draw
MP parallel to the Y-axis, cutting the line AB at P.

Clearly,
PM = 6

Thus,
when x = 2, then y = 6.

Question 13

Draw
the graph of the equation y = 3x. From your graph, find the value of y when x
= -2.

Solution 13

The
given equation is y = 3x.

Putting
x = 1, y = 3 1 = 3

Putting
x = 2, y = 3 2 = 6

Thus,
we have the following table:

 x 1 2 y 3 6

Plot
points (1,3) and (2,6) on a graph paper and join
them to get the required graph.

Take
a point P on the left of y-axis such that the distance of point P from the
y-axis is 2 units.

Draw
PQ parallel to y-axis cutting the line y = 3x at Q. Draw QN parallel to
x-axis meeting y-axis at N.

So,
y = ON = -6.

Question 14

Draw the graph of the equation x + 2y – 3 = 0. From your graph, find the value of y when x = 5.

Solution 14

The given equation is,

x + 2y – 3 = 0

x = 3 – 2y

Putting y = 1,x = 3 – (2 1) = 1

Putting y = 0,x = 3 – (2 0) = 3

Thus, we have the following table:

 x 1 3 y 1 0

Plot points (1,1) and (3,0) on a graph paper and join them to get the required graph.

Take a point Q on x-axis such that OQ = 5.

Draw QP parallel to y-axis meeting the line (x = 3 – 2y) at P.

Through P, draw PM parallel to x-axis cutting y-axis at M.

So, y = OM = -1.

Question 15

Draw the graph of the equation x + 2y – 3 = 0.

From your graph, find the value of y when x = -5

Solution 15

x
+ 2y – 3 = 0

2y = 3 – x

When
x = -1, then

When
x = 1, then

Thus,
we have the following table:

 x -1 1 y 2 1

Now,
plot the points A(-1, 2) and B(1, 1) on a graph
paper.

Join
AB and extend it in both the directions.

Then,
the line AB is the required graph of x + 2y – 3 = 0.

the graph

Given:
x = -5. Take a point M on the X-axis such that OM = -5.

Draw
MP parallel to the Y-axis, cutting the line AB at P.

Clearly,
PM = 4

Thus,
when x = -5, then y = 4.

Question 16

Draw
the graph of the equation 2x – 3y = 5. From the graph, find (i) the value of y when x = 4, and (ii) the value of x
when y = 3.

Solution 16

The given equation is, 2x – 3y = 5

Now, if x = 4, then

And, if x = -2, then

Thus, we have the following table:

 x 4 -2 y 1 -3

Plot points (4,1)
and (-2,-3) on a graph paper and join them to get the required graph.

(i) When
x = 4, draw a line parallel to y-axis at a distance of 4 units from y-axis to
its right cutting the line at Q and through Q draw a line parallel to x-axis
cutting y-axis which is found to be at a distance of 1 units above x-axis.

Thus, y = 1 when x = 4.

(ii) When y = 3, draw a line parallel to
x-axis at a distance of 3 units from x-axis and above it, cutting the line at
point P. Through P, draw a line parallel to y-axis meeting x-axis at a point
which is found be 7 units to the right of y axis.

Thus, when y = 3, x = 7.

Question 17

Draw the graph of the equation 2x + y = 6. Find the coordinates of the point, where the graph cuts the x-axis.

Solution 17

The given equation is 2x + y = 6

y = 6 – 2x

Now, if x = 1, then y = 6 – 2 1 = 4

And, if x = 2, then y = 6 – 2 2 = 2

Thus, we have the following table:

 x 1 2 y 4 2

Plot points (1,4) and (2,2) on a graph paper and join them to get the required graph.

We find that the line cuts the x-axis at a point P which is at a distance of 3 units to the right of y-axis.

So, the co-ordinates of P are (3,0).

Question 18

Draw
the graph of the equation 3x + 2y = 6. Find the coordinates of the point,
where the graph cuts the y-axis.

Solution 18

The given equation is 3x + 2y = 6

2y = 6 – 3x

Now, if x = 2, then

And, if x = 4, then

Thus, we have the following table:

 x 2 4 y 0 -3

Plot points (2, 0) and (4,-3) on a
graph paper and join them to get the required graph.

We find that the line 3x + 2y = 6
cuts the y-axis at a point P which is 3 units above the x-axis.

So, co-ordinates of P are (0,3).

Question 19

Draw the graphs of the equations 3x – 2y = 4 and x + y – 3
= 0. On the same graph paper, find the coordinates of the point where the two
graph lines intersect.

Solution 19

Graph of the equation 3x – 2y = 4

2y = 3x – 4

When
x = 2, then

When
x = -2, then

Thus,
we have the following table:

 x 2 -2 y 1 -5

Now,
plot the points A(2, 1) and B(-2, -5) on a graph
paper.

Join
AB and extend it in both the directions.

Then,
the line AB is the required graph of 3x – 2y = 4.

Graph of the equation x + y – 3 = 0

y = 3 – x

When
x = 1, then y = 3 – 1 = 2

When
x = -1, then y = 3 – (-1) = 4

Thus,
we have the following table:

 x 1 -1 y 2 4

Now,
plot the points C(1, 2) and D(-1, 4) on a graph
paper.

Join
CD and extend it in both the directions.

Then,
the line CD is the required graph of x + y – 3 = 0.

The
two graph lines intersect at point A(2, 1).

Question 20

Draw the graph of the line 4x + 3y = 24.

Write the coordinates of the points where this line
intersects the x-axis and the y-axis.

Solution 20

4x
+ 3y = 24

3y = 24 – 4x

When
x = 0, then

When
x = 3, then

Thus,
we have the following table:

 x 0 3 y 8 4

Now,
plot the points A(0, 8) and B(3, 4) on a graph
paper.

Join
AB and extend it in both the directions.

Then,
the line AB is the required graph of 4x + 3y = 24.

the graph

The
graph of line 4x + 3y = 24 intersects the X-axis at point C(6,
0) and the Y-axis at point A(0, 8).

Question 21

Draw the graph of the line 4x + 3y = 24.

Use this graph to find the area of the triangle formed by
the graph line and the coordinate axes.

Solution 21

4x
+ 3y = 24

3y = 24 – 4x

When
x = 0, then

When
x = 3, then

Thus,
we have the following table:

 x 0 3 y 8 4

Now,
plot the points A(0, 8) and B(3, 4) on a graph
paper.

Join
AB and extend it in both the directions.

Then,
the line AB is the required graph of 4x + 3y = 24.

the graph

Required
area = Area of
ΔAOC

Question 22

Draw the graphs of the lines 2x + y = 6 and 2x – y + 2 =
0. Shade the region bounded by these two lines and the x-axis. Find the area

Solution 22

Graph of the equation 2x + y = 6

y = 6 – 2x

When
x = 1, then y = 6 – 2(1) = 6 – 2 = 4

When
x = 2, then y = 6 – 2(2) = 6 – 4 = 2

Thus,
we have the following table:

 x 1 2 y 4 2

Now,
plot the points A(1, 4) and B(2, 2) on a graph
paper.

Join
AB and extend it in both the directions.

Then,
the line AB is the required graph of 2x + y = 6.

Graph of the equation 2x – y + 2 = 0

y = 2x + 2

When
x = -1, then y = 2(-1) + 2 = -2 + 2 = 0

When
x = 2, then y = 2(2) + 2 = 4 + 2 = 6

Thus,
we have the following table:

 x -1 2 y 0 6

Now,
plot the points C(-1, 0) and D(2, 6) on a graph
paper.

Join
CD and extend it in both the directions.

Then,
the line CD is the required graph of 2x – y + 2 = 0.

The
two graph lines intersect at point A(1, 4).

The
area enclosed by the lines and X-axis is shown in the graph.

Draw
AM perpendicular from A on X-axis.

PM
= y-coordinate of point A(1, 4) = 4

And,
CP = 4

Area
of shaded region = Area of
ΔACP

## Chapter 4 – Linear Equations in Two Variables Exercise MCQ

Question 1

The equation of the x-axis is

(a) x
= 0

(b) y
= 0

(c) x
= y

(d) x
+ y = 0

Solution 1

Correct
option: (b)

The
equation of the x-axis is y = 0.

Question 2

The graph of y + 2 = 0 is a line

(a) making
an intercept -2 on the x-axis

(b) making
an intercept -2 on the y-axis

(c) parallel
to the x-axis at a distance of 2 units below the x-axis

(d) parallel
to the y-axis at a distance of 2 units to the left of y-axis

Solution 2

Correct option: (c)

The graph of y + 2 = 0 is a line
parallel to the x-axis at a distance of 2 units below the x-axis.

Question 3

The graph of the linear equation 2x + 3y = 6 meets the
y-axis at the point

(a) (2,
0)

(b) (3,
0)

(c) (0,
2)

(d) (0, 3)

Solution 3

Correct
option: (c)

When
a graph meets the y-axis, the x coordinate is zero.

Thus,
substituting x = 0 in the given equation, we get

2(0)
+ 3y = 6

3y = 6

y = 2

Hence,
the required point is (0, 2).

Question 4

The graph of the linear equation 2x + 5y = 10 meets the
x-axis at the point

(a) (0,
2)

(b) (2,
0)

(c) (5,
0)

(d) (0, 5)

Solution 4

Correct
option: (c)

When
a graph meets the x-axis, the y coordinate is zero.

Thus,
substituting y = 0 in the given equation, we get

2x
+ 5(0) = 10

2x = 10

x = 5

Hence,
the required point is (5, 0).

Question 5

The graph of the line x
= 3 passes through the point

(a) (0,3)

(b) (2,3)

(c) (3,2)

(d) None of these

Solution 5

Question 6

The graph of the line y = 3 passes though the point

(a) (3,
0)

(b) (3,
2)

(c) (2,
3)

(d) none
of these

Solution 6

Correct
option: (c)

Since,
the y coordinate is 3, the graph of the line y = 3 passes through the point
(2, 3).

Question 7

The graph of the line y
= -3 does not pass through the point

(a) (2,-3)

(b) (3,-3)

(c) (0,-3)

(d) (-3,2)

Solution 7

Question 8

The graph of the linear
equation x-y=0 passes through the point

Solution 8

Question 9

If each of (-2,2), (0,0) and (2,-2) is a solution of a linear equation in x and y, then the equation is

(a) x-y=0

(b) x+y=0

(c) -x+2y=0

(d) x – 2y=0

Solution 9

Question 10

How many linear equations can be satisfied by x = 2 and y
= 3?

(a) only
one

(b) only
two

(c) only
three

(d) Infinitely
many

Solution 10

Correct option: (d)

Infinitely many linear equations can be satisfied by x = 2
and y = 3.

Question 11

A linear equation in two
variable x and y is of the form ax+by+c=0, where

(a) a≠0, b≠0

(b) a≠0, b=0

(c) a=0, b≠0

(d) a= 0, c=0

Solution 11

Question 12

The equation of the y-axis is

(a) x
= 0

(b) y
= 0

(c) x
= y

(d) x
+ y = 0

Solution 12

Correct
option: (a)

The
equation of the y-axis is x = 0.

Question 13

If (2, 0) is a solution of the linear equation 2x + 3y = k
then the value of k is

(a) 6

(b) 5

(c) 2

(d) 4

Solution 13

Correct option: (d)

Since, (2, 0) is a solution of the
linear equation 2x + 3y = k, substituting x = 2 and y = 0 in the given
equation, we have

2(2) + 3(0) = k

4 + 0 = k

k = 4

Question 14

Any point on x-axis is
of the form:

(a) (x,y), where x ≠0 and y ≠0

(b) (0,y), where y ≠0

(c) (x,0), where x ≠0

(d) (y,y), where y ≠0

Solution 14

Question 15

Any point on y-axis is
of the form

(a) (x,0), where x ≠ 0

(b) (0,y), where y ≠ 0

(c) (x,x), where x ≠ 0

(d) None of these

Solution 15

Question 16

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 16

Correct option: (c)

Substituting x = 5 and y = 2 in
L.H.S. of equation x + y = 7, we get

L.H.S. = 5 + 2 = 7 = R.H.S.

Hence, x = 5 and y = 2 is a
solution of the linear equation x + y = 7.

Question 17

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

(a)

(b)

(c)

(d)

Solution 17

Correct option: (b)

Since the point (3, 4) lies on the
graph of 3y = ax + 7, substituting x = 3 and y = 4 in the given equation, we
get

3(4) = a(3) + 7

12 = 3a + 7

3a = 5

Question 18

The point of the form (a,a), where a ≠ 0 lies on

(a) x-axis

(b) y-axis

(c) the line y = x

(d) the line x + y
= 0

Solution 18

Question 19

The point of the form
(a,-a), where a
≠ 0 lies on

(a) x-axis

(b) y-axis

(c) the line y-x=0

(d) the line x + y
= 0

Solution 19

Question 20

The linear equation 3x –
5y = 15 has

(a) a unique
solution

(b) two solutions

(c) infinitely many
solutions

(d) no solution

Solution 20

Question 21

The equation 2x + 5y = 7 has a unique solution, if x and y
are

(a) natural
numbers

(b) rational
numbers

(c) positive
real numbers

(d) real
numbers

Solution 21

Correct
option: (a)

The equation 2x + 5y = 7 has a unique solution, if x and y
are natural numbers.

If we take x = 1 and y = 1, the given equation is
satisfied.

Question 22

The graph of y = 5 is a line

(a) making
an intercept 5 on the x-axis

(b) making
an intercept 5 on the y-axis

(c) parallel
to the x-axis at a distance of 5 units from the origin

(d) parallel
to the y-axis at a distance of 5 units from the origin

Solution 22

Correct
option: (c)

The graph of y = 5 is a line
parallel to the x-axis at a distance of 5 units from the origin.

Question 23

The graph of x = 4 is a line

(a) making
an intercept 4 on the x-axis

(b) making
an intercept 4 on the y-axis

(c) parallel
to the x-axis at a distance of 4 units from the origin

(d) parallel
to the y-axis at a distance of 4 units from the origin

Solution 23

Correct
option: (d)

The graph of x = 4 is a line
parallel to the y-axis at a distance of 4 units from the origin.

Question 24

The graph of x + 3 = 0 is a line

(a) making
an intercept -3 on the x-axis

(b) making
an intercept -3 on the y-axis

(c) parallel
to the y-axis at a distance of 3 units to the left of y-axis

(d) parallel
to the x-axis at a distance of 3 units below the x-axis

Solution 24

Correct option: (c)

The graph of x + 3 = 0 is a line parallel to the y-axis at
a distance of 3 units to the left of y-axis.

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