#
class 10

lesson: pair of linear equation in two variable

## Answers

**Answer:**

The numerator of a fraction is 4 less than the denominator. If the numerator is decreased by 2 and denominator is increased by 1, then the denominator is eight times the numerator. Find the fraction.

Solution:

Let’s assume the numerator of the fraction to be x and the denominator of the fraction to be y.

So, the required fraction is x/y.

From the question it’s given as,

The numerator of the fraction is 4 less the denominator.

Thus, the equation so formed is,

x = y – 4

⇒ x – y = −4 …… (i)

And also it’s given in the question as,

If the numerator is decreased by 2 and denominator is increased by 1, then the denominator is 8 times the numerator.

Putting the above condition in an equation, we get

y + 1 = 8(x-2)

⇒ y + 1 = 8x–16

⇒ 8x – y = 1 + 16

⇒ 8x – y = 17 …… (ii)

Solving (i) and (ii),

Subtracting the equation (ii) from (i), we get

(x – y) – (8x – y) = – 4 – 17

⇒ x – y − 8x + y = −21

⇒ −7x = −21

⇒ x = 21/7

⇒ x = 3

Substituting the value of x =3 in the equation (i), we find y

3 – y = – 4

⇒ y = 3+4

⇒ y = 7

Therefore, the fraction is 3/7.