Linear Equations in Two Variables

Topic : Draw the graphs of the equations:

Exercise 4.2 Solutions
1. Question:
Draw the graphs of the equations:
3x - 2y = 4 and x + y - 3 = 0 on the same graph.

Solution:
Graph of the equation 3x - 2y = 4
We have,
3x - 2y = 4

Now, x = 0

and x = 4

Thus, the abscissae and ordinates of two points on the lines represented by the equation 3x - 2y = 4
are as given in the following table:
x 0 4
y -2 4
(1/2 mark)
By plotting the points (0, -2) and (4, 4) on the graph paper and drawing a line passing through these
points, we obtain the graph of the equation 3x - 2y = 4 as shown in figure.

(2 marks)
Graph of the equation:
x+y-3=0
We have,
x+y-3=0
y=3-x
Now, x = 0
y = 3 and x = 4
y = -1.
Thus, the abscissae and ordinates of two points on the line represented by the given equation are as
shown in the following table:
x 0 4
y 3 -1
(1/2 mark)
By plotting the points (0, 3) and (4, -1) and joining them by a line, we obtain graph of the
equation x + y - 3 = 0 as shown in figure.

2. Question:
If the point (3, 3) lies on the graph of the equation 3x + ay - 12 = 0. Find the value of a.

Solution:
Given equation:
3x + ay - 12 = 0
Point (3, 3) lies on the graph
Therefore, point (3, 3) satisfies the equation 3x + ay - 12 = 0 (1 mark)
3(3) + a(3) - 12 = 0
9 + 3a - 12 = 0
3a - 3 = 0
3a = 3
a=1 (1 mark)

3. Question:
For each of the graphs given, select the equation whose graph it is from the choices given below:
(a) For figure (i)
(i) x + y = 0, (ii) y = 2x, (iii) y - x = 2, (iv) y - x = -2
(b) For figure (ii)
(i) x - 2y = 3, (ii) y = 2x + 4, (iii) y = x - 4, (iv) y = 2x
Figure (i)
 Figure (ii)

Solution:
(a) In figure (i), the points on the line are (1, 3), (3, 5). By inspection, y - x = 2 is the equation
corresponding to this graph. (1.5 marks)
(b) In figure (ii), the points on the line are (-5, -4), (-1,-2) and (3, 0).
By inspection, we find that the co-ordinates of the points of the graph (line) satisfy the equation x -
2y = 3.
So, x - 2y = 3 is the equation corresponding to the graph. (1.5 marks)

4. Question:
Draw the graph of the equation 2y - x = 7 and determine from the graph whether x = 3 and y = 2 is a
solution or not.
Solution:
The given equation is 2y - x = 7

(1/2 mark)
Table of values
x 1 3 -1
y 4 5 3
(1/2 mark)
Plotting the points (1, 4), (3, 5) and (-1, 3) and joining them, we get a straight line l. Also, plot the
point (3, 2). We see that the point (3, 2) does not lie on the graph of the equation 2y - x = 7.

(1 mark)
Hence, x = 3 and y = 2 is not a solution of the given equation. (1 mark)

5. Question:
Draw the graph of the line x - 2y = 3. From the graph, find the coordinate of the points when:
i) x = -5
ii) y = 0

Solution:
The given equation is x - 2y = 3
x = 2y + 3
Table of values
x 3 5 1
y 0 1 -1
(1 mark)
Plotting the points (3, 0), (5, 1) and (1, -1) and joining them, we get a straight line l. From the graph
we see that:
(i) When x = -5, y = -4 (1 mark)
(ii) When y = 0, x = 3 (1 mark)

(2 marks)

6. Question:
Solve the linear equation 2y + 3 = 9 and represent the solution
(i) in one variable
(ii) in two variables

Solution:
We have,
2y + 3 = 9
2y = 9 - 3
2y = 6
y=3
(i) The representation of the solution in one variable is as shown on the number line

(1 mark)
(ii) In the cartesian plane
The equation y = 3 does not contain x. So, its graph is a line parallel to x-axis passing through the
point (0, 3). Clearly, y = 3 means that for all values of the abscissa x, the ordinate y is 3.
Thus, we have the following table exhibiting the abscissae and ordinates of the points lying on the
line represented by the given equation.
x 0 -1 -2
y 3 3 3

Plotting the points (0, 3) and (-2, 3) on the graph paper and joining them by a line, we obtain the
graph of the line represented by the given equation as show in figure.

(2 marks)

7. Question:
Draw the graph of 2y = 4x - 6 and find the points where the line meets the y-axis.

Solution:
We have,
2y = 4x - 6

Table of values for 2y = 4x - 6

x 0 1 2
y -3 -1 1
(1 mark)
Plotting the points (0, -3), (1, -1), (2, 1) and joining them, we get the line , which is the graph of

2y = 4x - 6 (2
marks)
The line 2x = 4x - 6 meets the y-axis at (0,-3) (1 mark)

8. Question:
Express y in terms of x in the equation 2x + 3y = 11. Find the point where the line represented by the
equation 2x + 3y = 11 cuts the y-axis.

Solution:
We have,
2x + 3y = 11
or 3y = 11 - 2x

(1 mark)
The line represented by the equation 2x + 3y = 11 cuts the y-axis where
x = 0. (1/2 mark)
Putting x = 0 in 2x + 3y = 11, we get,
2(0) + 3y = 11
3y = 11

y= (1 mark)

Hence, the required point is (0, ) (1/2 mark)

9. Question:
Represent the solution of the equation 4x-15 = 5.
(i) on number line
(ii) in cartesian plane

Solution:
The equation 4x-15 = 5 can be written as 4x = 20
so, x = 5.
(i) Solution on number line can be represented as follows:

( 1 mark)
(ii) The given equation is x = 5.
This means that for all values of the ordinate y, the abscissa x is 5.
Thus, table of values is
x 5 5 5
y 0 1 2

Plotting the points (5, 0), (5, 1) and (5, 2) and joining them, we get a straight line .
(2 marks)

10. Question:
If the point (0, -2) lies on the graph of the equation 3x - 2y = a. Find the value of ''a''

Solution:
Given equation 3x - 2y = a
Point (0, -2) lies on the graph
Therefore, point (0, -2) satisfies the equation 3x - 2y = a (1 mark)
3(0) - 2(-2) = a
0+4=a
a=4 (1 mark)

11. Question:
Draw the graph of the equation 2x + 3y = 5. Check whether the points (-3, 4) and (7, -3) are the
solutions of the given equation.

Solution:
We have,
2x + 3y = 5
 3y = 5 - 2x

y= (1 mark)
x -2 1 4
y 3 1 -1

(1/2 mark)

(1 mark)

As the point (-3, 4) does not lie on the graph of the equation 2x + 3y = 5, it is not a solution. But the
point (7, -3) lies on the graph of the equation 2x + 3y = 5, hence it is a solution (1/2 mark)

12. Question:
Draw the graph of the line x - 2y = 3. From the graph, find the coordinates of the points when:
i) x = -5
ii) y = 0

Solution:
The given equation is x - 2y = 3
x = 2y + 3
Table of values
x 3 5 1
y 0 1 -1
(1 mark)
Plotting the points (3, 0), (5, 1) and (1, -1) and joining them, we get a straight line l. From the graph
we see that:
(i) When x = -5, y = -4 (1 mark)
(ii) When y = 0, x = 3 (1 mark)

(2 marks)
13. Question:
Find the value of p if the line represented by the equation 2x - py = 12 passes through the point (1, -
1).

Solution:
Since, the equation 2x - py = 12 passes through the point (1, -1),
the point (1, -1) must satisfy the given equation.
i.e. 2(1) - p(-1) = 12 (1 mark)
2 + p = 12
p = 10
Hence, p = 10 (1 mark)

14. Question:
Draw the graph of the equation 5y - x = 14.

Solution:
The given equation is 5y - x = 14
5y = x + 14

Table of values
x -4 1 6
y 2 3 4
(1 mark)
Plotting the points (-4, 2), (1, 3) and (6, 4) and joining them, we get a straight line , which is the
graph of 5y - x = 14.

(2 marks)

15. Question:
The sum of a two digit number and the number obtained by reversing the order of its digits is 165.
Write a linear equation which satisfied this data. Draw the graph of the same.

Solution:
Let the digit at unit's place be x
and the digit at ten's place be y.
Then, number = 10y + x (1/2 mark)
Number obtained by reversing the order of the digits = 10x + y
According to the given conditions, we have,
(10y + x) + (10x + y) = 165 (1/2 mark)
 10y + x + 10x + y = 165
11x + 11y = 165 (1/2 mark)
11(x + y) = 165
x + y = 15
x = 15 - y
when y = 0 x = 15
when y = 5 x = 10
when y = 10 x=5
when y = 15 x=0
x 15 10 5 0
y 0 5 10 15

16. Question:
Draw the graph of the equation 4 = 3x + y. From the graph, find the point at which the line is
intersecting the y-axis.

Solution:
To draw the graph we need at least two solutions of the equation.
4 = 3x + y
y = 4 - 3x

when x = 0 y = 4 - 3(0) = 4
when x = 1 y = 4 - 3(1) = 4 - 3 = 1
when x = -1 y = 4 - 3(-1) = 4 + 3 = 7
when x = 2 y = 4 - 3(2) = -2
x 0 1 -1 2
y 4 1 7 -2
(1 mark)

(2 marks)
Line 4 = 3x + y intersects y-axis at the point (0, 4). (1 mark)
 (2 marks)

17. Question:
Draw the graph of the line given by the equation: 3x + 2y = 0.

Solution:
When x = 0
We get, y = 0
When x = 4
We get y = -6
When x = 8
We get, y = -12

(1 mark)
Graph:

(2 marks)

18. Question:
Give the geometric representations of 2y + 5 = 0 as an equation
(i) in one variable
(ii) in two variable

Solution:
We solve
2y + 5 = 0
2y = -5

(1/2 mark)
(i) The representation of the solution on the number line, where y = -2.5 is treated as an equation in
one variable.

(1/2
mark)
(ii) We know that y = -2.5 can be written as 0 . x + 1 . y = -2.5, which is a linear equation in the
variables x and y.
This represented by a line.
Now, 0 . x + 1 . y = -2.5
Therefore, y must satisfies the equation y = -2.5 (1 mark)

(1
mark)

19. Question:
Draw the graph of the equation 3x - 2y = 7.
Or
Draw the graph of the equation 2y - x = 7 and determine from the graph whether x = 3 and y = 2 is a
solution or not.

Solution:
We have,
3x - 2y = 7
3x = 7 + 2y

x =
x 3 5 7
y 1 4 7

(1 mark)

(2 marks)
Or
The given equation is 2y - x = 7

Table of values
x 1 3 -1
y 4 5 3
(1/2 mark)
Plotting the points (1, 4), (3, 5) and (-1, 3) and joining them, we get a straight line .

(2 marks)
Now, plot the point (3, 2). We see that the point (3, 2) does not lie on the graph of the equation 2y -
x = 7.

Hence, x = 3 and y = 2 is not a solution of the given equation. (1/2 mark)

20. Question:
Draw the graph of the equation 2y - x = 8.

Solution:
The given equation is 2y - x = 8
2y = x + 8

(1/2 mark)
Table of values
x 0 2 -4
y 4 5 2
(1/2 mark)
Plotting the points (0, 4), (2, 5) and (-4, 2) and joining them, we get a straight line , which is the
graph of 2y - x = 8.

(2 marks)
21. Question:
Draw the graph of the equation 2x + 5y = 13 and determine from the graph whether
x = 2 and y = 5 is a solution or not.

Solution:
The given equation is 2x + 5y = 13
5y = 13 - 2x

Table of values
x 4 -1 9
y 1 3 -1
(1 mark)

Plotting the points (4, 1), (-1, 3) and (9, -1) and joining them, we get a straight line l. Also, plot the
point (2, 5). We see that the point (2, 5) does not lie on the graph of the equation 2x + 5y = 13.
Hence, x = 2 and y = 5 is not a solution of the given equation. (1/2 mark)

(1.5 marks)

22. Question:
Draw the graph of the equation 2x + 5y = 13. From the graph find whether x = 2 and y = 5 is a
solution or not. If x = 3, y = -2 is a solution of the given system of equations :
x + ay = -1
bx - 3y = 12
Find the values of a and b.

Solution:
The given equation is 2x + 5y = 13
5y = 13 - 2x

Table of values
x 4 -1 9
y 1 3 -1
(1.5 marks)
Plotting the points (4, 1), (-1, 3) and (9, -1) and joining them, we get a straight line .

(2 marks)
Also, plot the point (2, 5).
We see that the point (2, 5) does not lie on the graph of the equation 2x + 5y = 13.
Hence, x = 2 and y = 5 is not a solution of the given equation. (1/2 mark)
The given system of equations is
x + ay = -1 (1)
bx - 3y = 12 (2)
Since x = 3, y = -2 is a solution of the given system,
therefore, x = 3, y = -2 should satisfy both the equations (1 mark)
Putting x = 3, y = -2 in (1), we get ,
3 + a (-2) = -1

Again, putting x = 3, y = -2 in (2), we get

b 3 -3(-2) = 12

Hence a = 2 and b = 2. (1 mark)

 23. Question:
Draw the graph of the equation: 2y + 1 = 9.

Solution:
We have,
2y + 1 = 9
2y = 9 – 1 = 8
y=4 (1 mark)
The equation y = 4 does not contain x. So, its graph is a line parallel to x – axis passing through the
poont (0, 4) as shown:

( 1 mark)

24. Question:
Solve the equation 3x + 1 = x - 2 and represent the solutions (s) on
(i) the number line
(ii) the cartesian plane

Solution:
We have,
3x + 1 = x - 2
3x - x = -1 - 2
2x = -3

ie. (1/2 mark)

(i) The representation of the solution on the number line; x = -1.5 is treated as an equation in one
variable

(1
mark)
(ii) We know that x = -1.5 can be written as 2x + 0 . y = -3,
which is an equation in two variables x and y.
This is represented by a line 2x + 0 . y = -3

Therefore, x must satisfies the equation (1/2 mark)

(1 mark)
Question:
Solve the equation 2y + 1 = y - 3 and represent the solutions on
(i) the number line
(ii) the carterian plane

Solution:
We solve 2y + 1 = y - 3 to get
2y + 1 = y - 3
2y - y = -3 - 1
y = -4
i.e y = -4 (1 mark)
(i) Considering ''y'' as a variable only then it can be represented on the number line.
*****image (1 mark)
(ii) We know that y = -4 can be written as 0 . x + 1 . y = -4 which is a linear equation is the variables x
and y. This is represented by a line
Now, 0 . x + 1 . y = -4
As x = 0
Therefore, y must satisfies the equation y = -4
*****image (1 mark)

25. Question:
The sum of a two digit number and the number obtained by reversing the order of its digits is 121.
Write a linear equation in two variables to represent this statement. Draw its graph.

Solution:
Let the digit at the unit's place be x.
and the digit at the ten's place be y.
Then, number = 10y + x
The number obtained by reversing the order of the digits is 10x + y (1/2 mark)
According to the given condition, we have,
(10y + x) + (10x + y) = 121
10y + x + 10x + y = 121
11x + 11y = 121
11(x + y) = 11
x + y = 11 (1/2 mark)
Given equation x + y = 11
x = 11 - y
Table of values
x 0 11 7
y 11 0 4
(1 mark)
 (2 marks)

26. Question:
Draw the graph of equation 3x + y - 12 = 0. From the graph find the value of
(i) x when y = 3
(ii) y when x = 5

Solution:
3x + y - 12 = 0
y = 12 - 3x
x 2 4 3 2
y 6 0 3 6
(1 mark)

(2
marks)
Value of x when y = 3 is x = 3 (1/2 mark)
Value of y when x = 5 is y = -3 (1/2 mark)

27. Question:
Draw the graph of the equation 2x - 4y = 10.

Solution:
The given equation is 2x - 4y = 10
4y = 2x -10

Table of values
x -1 1 5
y -3 -2 0
(1 mark)
Plotting the points (-1, -3), (1, -2) and (5, 0) and joining them, we get a straight line .

(2 marks)

28. Question:
Draw a graph of the line x - 2y = 3. From the graph, find the co-ordinates of the point when (i) x = -5
(ii) y = 0.

Solution:
We have,
x - 2y = 3

(1 mark)
When x = 1, we have,

When x = -1, we have,

Thus, we have the following table:

x 1 -1 -3
y -1 -2 -3
(1 mark)
Plotting the points (1, -1), (-1, -2) and (-3, -3) on the graph paper and joining them, we get the
straight line. This line is the required graph of the equation x - 2y = 3.
(2 marks)
To find the co-ordinates of the point when x = 5, we draw a line parallel to y-axis and passing
through (-5, 0). This line meets the graph of x - 2y = 3 at a point from which we draw a line parallel
to x-axis which crosses y-axis at y = -4. So, the co-ordinates of the required point are (-5, -4).
Since, y = 0 is the x-axis. So, the required point is the point where the line meets x-axis. From the
graph the co-ordinates of such a point is (3, 0).
Hence, required points are (-5, -4) and (3, 0). (1 mark)

29. Question:
Draw the graph of x - 3y = 4. From the graph, find the coordinates of the point when x = -2.

Solution:
We have,
x - 3y = 4
x = 4 + 3y
x 4 7 10
y 0 1 2
 (1 mark)

(2 mark)
At x = -2 draw PQ X'OX meeting the line m in Q.
From Q, draw QR Y'O Y, we get,
OR = -2
Therefore, x = -2 gives y = -2. (1 mark)

30. Question:
Draw graphs of the equations
3x - 2y = 4 and x + y - 3 = 0

Solution:
Graph of the equation 3x - 2y = 4
We have,
3x - 2y = 4

Now, x = 0

and x = 4

Thus, the abscissae and ordinates of two points on the lines represented by the equation 3x - 2y = 4
are as given in the following table:
x 0 4
y -2 4

By plotting the points (0, -2) and (4, 4) on the graph paper and drawing a line passing through these
points, we obtain the graph of the equation 3x - 2y = 4 as shown in figure. (1 mark)
Graph of the equation: x + y - 3 = 0
We have,
x+y-3=0
y=3-x
Now, x = 0
y=3
and x = 4
y = -1
Thus, the abscissae and ordinates of two points on the line represented by the given equation are as
shown in the following table:
x 0 4
y 3 -1

By plotting the points (0, 3) and (4, -1) and joining them by a line, we obtain graph of the
equation x + y - 3 = 0 as shown in figure. (1 mark)

Clearly, line represented by the equations 3x - 2y = 4 and x + y - 3 = 0 are as shown above. (2

marks)

31. Question:
Draw the graph of the equation 2x + 3y = 5. Check whether the points (-3, 4) and (7, -3) are the
solutions of the given equation.
Solution:
We have,
2x + 3y = 5
3y = 5 - 2x

y= (1/2 mark)
x -2 1 4
y 3 1 -1

(1 mark)

(2 marks)

As the point (-3, 4) does not lie on the graph of the equation 2x + 3y = 5, it is not a solution. But the
point (7, -3) lies on the graph of the equation 2x + 3y = 5, hence it is a solution (1/2 mark)

32. Question:
A and B each have certain number of oranges. A says to B, "if you give me 10 of your oranges, I will
have twice the number of oranges left with you. Write a linear equation in two variables to represent
this statement. Draw its graph.

Solution:
Suppose A has x oranges and B has y oranges.
According to the given condition, we have,
x + 10 = 2(y - 10)
x + 10 = 2y - 20
x - 2y + 30 = 0 (1 mark)
Given equation
x - 2y + 30 = 0
x = 2y - 30
When y = 10 x = 2(10) - 30 = -10
when y = 12 x = 2(12) - 30 = 24 - 30 = -6
when y = 8 x = 2(8) - 30 = 16 - 30 = -14
x -10 -6 -14
y 10 12 8
(1.5 marks)

(1.5 marks)

33. Question:
In a ‘‘Cleanliness drive’’ residents of certain locality joined together to clean neighbourhood area.
Children and adults were participated in the drive and the total number of paticipants was 11. Taking
x as number of adults and y as number of children.
(a) Form a linear equation.
(b) Draw the graph of the linear equation obtained in (a).
(c) What values are depicted here? (V) (CBSE sample)

Solution:
(a) Here the number of adults is given as x and the number of children is given as y. Then, according
to the given question, we have:
x+ y = 11, which is the required linear equation. (1 mark)
(b) Given equation x + y = 11
x = 11 - y
Table of values
x 0 11 7
y 11 0 4

(2 marks)
(c) Co-operation, Sincerity, Environmental protection. (1 mark)
 34. Question:
Draw a graph for the linear equation 2x + 3y = 0.

Solution:
We have,
2x + 3y = 0

x=
Tables of values
x 0 3 -3
y 0 -2 2
(1 mark)
Plotting the points (0, 0), (3, -2) and (-3, 2) and joining them, we get a straight line, which is the
graph of 2x + 3y = 0.

(2 marks)

35. Question:
Draw agraph for the liner equation x + 2y = 3.

Solution:
We have,
x + 2y = 3
x = 3 - 2y
Tables of values
x 0 1 3
y 3/2 1 0
(1 mark)
Plotting the points (0, 3/2), (1, 1) and (3, 0) and joining them, we get a straight line, which is the
graph of x +2y =3.
(2 marks)

36. Question:
Draw the graph of line 3x - 2y = 7 and from the graph find the point at which the line is intersecting
the x-axis.

Solution:
Consider the equation 3x - 2y = 7
3x = 7 + 2y

(1/2 mark)
x 3 7 5
y 1 7 4
(1/2 mark)
Now, the points on the line are (3, 1), (5, 4), (7, 7) and on plotting these points on the line we obtain
the graph of linear equation 3x - 2y = 7
 (2 marks)
Line 3x - 2y = 7 intersects x-axis at the point

(1 mark)

37. Question:
Solve the equation 2y + 1 = y - 3, and represent the solution on
(i) the number lines,
(ii) the cartesian plane

Solution:
We solve
2y + 1 = y - 3, to get
2y - y = -3 - 1
i.e y = -4 (1/2 mark)
(i).

(1 mark)
(ii) We know that y = -4 can be written as 0 . x + 1 . y = -4, which is a linear equation in the
variables x and y.

Therefore, y must satisfies the equation y = -4 (1/2 mark)

(1
mark)

38. Question:
Draw the graph of 4x - 3y + 12 = 0.
Solution:

x 0 -3 3
y 4 0 8
(1 mark)
Graph:

(2 marks)

39. Question:
There are two examination rooms A and B. If 20 candidates are sent from B to A, the number of
students in A is doubled than the number of students in B. Express this statement in the form of an
equation in two variables and draw the graph of the same.

Solution:
Let the total number of students in room A be x
and the total number of student in room B be y
Therefore, according to given condition, we have,
2(y - 20) = x + 20
2y - 40 = x + 20
x - 2y = -60 (1 mark)
Now, given equation
x - 2y = -60
x = 2y - 60
(1 mark)
x -40 -20 0
Y 10 20 30

when y = 10 x = 2(10) - 60 = 20 - 50 = -40

when y = 20 x = 2(20) - 60 = 40 - 60 = -20
when y = 30 x = 2(30) - 60 = 60 - 60 = 0

40. Question:
The numerator of a fraction is 14 less than five times its denominator. Represent this in the form of a
linear equation in two variables. Also, draw the graph of the same .
From the graph find
(i) Find numerator when denominator is 5.
(ii) For what value of denominator the numerator is zero?

Solution:
Let the numerator be x and denominator be y then
According to given condition x = 5y - 14
So, x = 5y -14 represents the above condition. (1/2 mark)
Representing the given equation graphically
5y = x + 14

(1/2 mark)
Table of values
x -4 1 6
y 2 3 4
(1/2 mark)
Plotting the points (-4, 2), (1, 3) and (6, 4) and joining them, we get a straight line , which is the
graph of 5y - x = 14.
 (1.5 marks)
(i) Numerator = 11 when denominator is 5. (1/2 mark)

(ii) Denominator y = when numerator = 0. (1/2 mark)

 (2 marks)

41. Question:
Draw the graph of the equation 2y - x = 7 and determine from the graph whether x = 3 and y = 2 is a
solution or not.

Solution:
The given equation is 2y - x = 7

Table of values
x 1 3 -1
y 4 5 3
(1 mark)
Plotting the points (1, 4), (3, 5) and (-1, 3) and joining them, we get a straight line .

(2 marks)
Now, plot the point (3, 2). We see that the point (3, 2) does not lie on the graph of the equation 2y -
x = 7.

Hence, x = 3 and y = 2 is not a solution of the given equation. (1 mark)

42. Question:
Draw the graph of the equation 3x - 4y = 12. Check whether x = 4 and y = 2 is a solution.

Solution:
The given equation is 3x - 4y = 12

4y = 3x -12

Table of values
x 0 4 8
y -3 0 3
(1 mark)

Plotting the points (0, -3), (4, 0) and (8, 3) and joining them, we get a straight line .
Also, plot the point (4, 2). We see that the point (4, 2) does not lie on the graph of the equation 3x -
4y = 12.
Hence, x = 4 and y = 2 is not a solution of the given equation. (1 mark)

43. Question:

When 3 is added to the denominator and 2 is subtracted from the numerator, a fraction becomes .
Formulate a linear equation in two variables to represent this statement. Draw its graph.

Solution:
Let the numerator of the fraction be x
and the denominator of the fraction be y.

Then fraction = (1/2 mark)

According to the question, we have,

4(x - 2) = 1(y + 3)
4x - 8 = y + 3
4x - y = 11 (1/2 mark)
Now, given equation
4x - y = 11
y = 4x - 11
(1 mark)
when x = 1 y = 4(1) - 11 = 4 - 11 = -7
when x = 2 y = 4(2) - 11 = 8 - 11 = -3
when x = 3 y = 4(3) - 11 = 12 - 11 = 1
x 1 2 3
y -7 -3 1

(2
marks)
 44. Question:
Draw the graph of x + 6y = 12. From the graph, find the point at which the line is intersecting the y-
axis.

Solution:
To draw the graph, we need at least two solutions of the equations
x + 6y = 12
x = 12 - 6y
When y = 3
x = 12 - 6(3)
= 12 - 18
= -6
When y = 1
x = 12 - 6(1)
= 12 - 6
=6

x -6 6 0
y 3 1 2

When y = 2
x = 12 - 6(2)
= 12 - 12
=0

45. Question:
Draw the graph of the equation x - y = 4.

Solution:
The given equation is x - y = 4
x=y+4
Table of values
x 4 2 5
y 0 -2 1
(1 mark)
Plotting the point (4, 0), (2, -2) and (5, 1) and joining them, we get a straight line , which is the
graph of x - y = 4.
Question:
Find the point of intersection of the line represented by the equation -3x + 7y = 3 with the x-axis.

Solution:
The line represented by the equation -3x + 7y = 3 cuts the x-axis where y = 0. (1/2 mark)
Putting y = 0 in -3x + 7y = 3 we get,
-3x + 7(0) = 3
 -3x = 3

x= = -1 (1 mark)
Hence, the required point is (-1,0). (1/2 mark)

(2 marks)

46. Question:
Find the point of intersection of the line represented by the equation -3x + 7y = 3 with the x-axis.

Solution:
The line represented by the equation -3x + 7y = 3 cuts the x-axis where y = 0. (1/2 mark)
Putting y = 0 in -3x + 7y = 3 we get,
-3x + 7(0) = 3
-3x = 3

x= = -1 (1 mark)
Hence, the required point is (-1,0). (1/2 mark)
(1 mark)

(2 marks)
Line x + 6y = 12 intersects y-axis at the point (0, 2). (1 mark)

47. Question:
Find the points where the graph of the equation 3x + 4y = 12 cuts the x - axis and y - axis.

Solution:
If the graph of the equation 3x + 4y = 12 cuts the x-axis, then y = 0. (1/2 mark)
3x +4(0) = 12
 3x = 12
x=4
Hence, the graph of the equation 3x + 4y = 12 cuts the x-axis at the point (4, 0). (1 mark)
Now, If the graph of the equation 3x + 4y = 12 cuts the y-axis, then x = 0. (1/2 mark)
3(0) +4y = 12
4y = 12
y=3
Hence, the graph of the equation 3x + 4y = 12 cuts the y-axis at the point (0, 3). (1 mark)

48. Question:
From the choices given below. Choose the equation whose graphs are given in figure (i) and figure
(ii)
For figure (i)
(a) y = x (b) x + y = 0 (c) x + y = 3 (d) y = 2x
For Figure (ii)
(a) 3x - 2y = 4 (b) x + y = 3 (c) y = x - 2 (d) 2 + 3y = 7x

Figure (i)
Figure (ii)

Solution:
(a) In figure (i), the points on the line are (1, 2), (2, 1). The co-ordinates of the points of the graph
(line) satisfy the equation x + y = 3. So, x + y = 3 is the equation corresponding to the graph in figure
(i). (1.5 marks)
(b) In figure (ii), the points on the line are (0, -2), (4, 4). The co-ordinates of the points of the graph
(line) satisfy the equation 3x - 2y = 4. So, it is the equation corresponding to the graph in figure
(ii). (1.5 marks)

49. Question:
Ten years ago, father was twelve times as old as his son. Formulate a linear equation in two
variables to represent this statement. Draw its graph.

Solution:
Let the present age of father be x years
and the present age of son be y years.
Ten years ago, father's age was (x - 10) years. (1/2 mark)
Ten years ago, son's age was (y - 10) years.
Therefore, according to the question, we have,
x - 10 = 12(y - 10)
x - 12y + 110 = 0 (1/2 mark)
Now, given equation
x - 12y + 110 = 0
x = 12y - 110 (i)
Taking the points on line (i)
x 10 -50 -2
y 10 5 9
(1 mark)

(1
mark)

50. Question:
Draw the graph of the following equation x + y = 0

Solution:
We have, x + y = 0

Table of values
x 1 2 3
y -1 -2 -3
(1 mark)

(2 marks)

51. Question:
Draw the graph of the equation 3x - 2y = 0.

Solution:
The given equation is 3x - 2y = 0
2y = 3x
Table of values
x 2 0 -2
y 3 0 -3
(1 mark)
Plotting the points (2, 3), (0, 0) and (-2, -3) and joining them, we get a straight line , which is the
graph of 3x - 2y = 0.
 (2 marks)

52. Question:
Draw the graph of the equation x = y.

Solution:
The given equation is x = y.
Table of values
x 2 -3 0
y 2 -3 0
(1 mark)
Plotting the points (2, 2), (-3, -3) and (0, 0) and joining them, we get a straight line .
 (2 marks)

53. Question:
Draw the graph of line x - y = -1 and 3x + 2y - 12 = 0 on the same graph paper.

Solution:
Consider x - y = -1
x = -1 + y
x 0 1 2
y 1 2 3
(1 mark)
Plotting the points (0, 1), (1, 2) and (2, 3), we get, the line x - y = -1
Now, consider the equation 3x + 2y - 12 = 0
2y = 12 - 3x

x 0 2 4
y 6 3 0
(1 mark)
Plotting the points (0, 6), (2, 3), (4, 0), we get, of equation 3x + 2y - 12 = 0
 (2 marks)

54. Question:
Find the point of intersection of the line represented by the equation 5x + 7y = 10 with the x-axis.

Solution:
The line represented by the equation 5x + 7y = 10 cuts the x-axis where y = 0. (1/2 mark)
Putting y = 0 in 5x + 7y = 10, we get,
5x + 7 0 = 10 (1 mark)

5x = 10

Hence, the required point is (2, 0) (1/2 mark)

55. Question:
Draw a graph of the line x - 2y = 3. From the graph, find the co-ordinates of the point when
(i) x = -5 (ii) y = 0.

Solution:
We have,
x - 2y = 3

When x = 1, we have,

When x = -1, we have,

When x = -3, we have,

Thus, we have the following table:

x 1 -1 -3
y -1 -2 -3
(1 mark)
Plotting the points (1, -1), (-1, -2) and (-3, -3) on the graph paper and joining them, we get the
straight line. This line is the required graph of the equation x - 2y = 3.
(2 marks)
To find the co-ordinates of the point when x = 5, we draw a line parallel to y-axis and passing
through (-5, 0). This line meets the graph of x - 2y = 3 at a point from which we draw a line parallel
to x-axis which crosses y-axis at y = -4. So, the co-ordinates of the required point are (-5, -4).
Since, y = 0 is the x-axis. So, the required point is the point where the line meets x-axis. From the
graph the co-ordinates of such a point is (3, 0).
Hence, required points are (-5, -4) and (3, 0). (1 mark)