Answer 1

Let the number of boys be x and the number of girls be y.

Then the equations formed are

x + y = 10 ….(1) and y = x + 4 ….(2)

Let us draw the graphs of equations (1) and (2) by finding two solutions for each of the equations.

The solutions of the equations are given.

x + y = 10 ⇒ y = 10 – x

x	0	8
y	10	2
Points	A	B

y = x + 4

x	0	1	3
y	4	5	7
Points	C	D	E

Plotting these points we draw the lines AB and CE passing through them to represent the equations. The two lines AB and Ce intersect at the point E (3, 7). So, x = 3 and y = 7 is the required solution of the pair of linear equations.

i.e. Number of boys = 3

Number of girls = 7.

Verification:

Putting x = 3 and y = 7 in (1), we get

L.H.S. = 3 + 7 = 10 = R.H.S., (1) is verified.

Putting x = 3 and y = 7 in (2), we get

7 = 3 + 4 = 7, (2) is verified.

Hence, both the equations are satisfied.

Answer 2

 Let the number of girls be x and the number of boys be y.

According to the question, the algebraic representation is

x + y = 10

x − y = 4

For x + y = 10,

x = 10 − y

Let the number of girls be x and the number of boys be y.

According to the question, the algebraic representation is

x + y = 10

x − y = 4

For x + y = 10,

x = 10 − y

x	5	4	6
y	5	6	4

For x − y = 4,

x = 4 + y

x	5	4	3
y	1	0	−1

Hence, the graphic representation is as follows.

From the figure, it can be observed that these lines intersect each other at point (7, 3).

Therefore, the number of girls and boys in the class are 7 and 3 respectively.