Question

The numerator of a fraction is 6 less than the denominator. If 3is added to the numerator the fraction is equal to 2\3.what is the original fraction?

Answer 1

Answer:

The original fraction is $\dfrac{3}{9}$

Step-by-step explanation:

Solution :

Let,

• The Numerator of a fraction = x
• The Denominator of a fraction = x + 6

★ If 3 is added to the numerator the fraction is equal to 2\3

• The Numerator of a fraction = x + 3

★ According to the Question :

⇒ $\dfrac{x \: + \: 3}{ x\: + \: 6} \: = \: \dfrac{2}{3}$

⇒3 (x + 3) = 2 (x + 6)

⇒ 3x + 9 = 2x + 12

⇒ 3x - 2x = 12 - 9

⇒ x = 3

The Numerator of a fraction = 3

• The Denominator of a fraction = x + 6

⇒ x + 6

⇒ 3 + 6

⇒ 9

The Denominator of a fraction = 9

★ The original fraction = $\dfrac{3}{9}$

Therefore, the original fraction is

$\dfrac{3}{9}$

Answer 2

★ Solution :-

Let the numerator of the fraction be x-6.

Let the denominator of the fraction be x.

So, the fraction is

$\sf \leadsto \dfrac{x - 6}{x}$

As given in the question,

3 is added to numerator and the fraction is changed. So, the fraction is

$\sf \leadsto \dfrac{x - 6}{x}$

We are also given with another fraction that will be obtained while we add 3 to the numerator The given fraction is

$\sf \leadsto \dfrac{2}{3}$

According to the question,

$\sf \leadsto \dfrac{x - 6 + 3}{x} = \dfrac{2}{3}$

$\sf \leadsto 3(x - 6 + 3) = 2x$

$\sf \leadsto 3(x - 3) = 2x$

$\sf \leadsto 3x - 9 = 2x$

$\sf \leadsto 3x - 2x = 9$

$\sf \leadsto 1x = 9$

$\sf \leadsto x = 9$

As we know that x is denominator and the value of denominator will be 9. Now, we should find the value of numerator.

Numerator of fraction :

$\sf \leadsto 9 - 6$

$\sf \leadsto 3$

So, the fraction is 3/9. The simplest form is 1/3.

Therefore the original fraction is $\sf \dfrac{1}{3}$.