Question

the denominator of a fraction is grater than the numerator by 8 .if the numerator is increased by 1 and denominator is decreased by 1 the number obtained is 3/2 find the fraction ​

Answer 1

Let the numerator be x

So, denominator be x + 8

We know $\red{ \rm Fraction = \dfrac{Numerator}{Denominator}}$

So, $\rm \dfrac{x}{x+8}$

So, we get the fraction $\bf \dfrac{x}{x+8}$

Now according to question

→ $\sf \dfrac{x+1}{x+8-1} = \dfrac{3}{2}$

→ $\sf \dfrac{x+1}{x+7} = \dfrac{3}{2}$

→ $\sf 2(x+1) = 3(x+7)$

→ $\sf 2x+2 = 3x+21$

→ $\sf 2x - 3x = 21 - 2$

→ $\sf -x = 19$

→ $\sf x = -19$

So, we got the value of x which is -19

So, simply subsituting in our fraction $\rm \dfrac{x}{x+8}$

$\implies \sf \dfrac{-19}{-19+8}$

$\implies \sf \dfrac{-19}{-11}$

So, finally we got the fraction which is :-

$\green{\underbrace{\boxed{\overbrace{\dfrac{-19}{-11}}}}}$

Answer 2

Answer :-

• Let the numerator be x
• Denominator = x + 8

Adding 1 to numerator and subtracting 1 from denominator :-

• New numerator = x + 1
• New denominator = x + 8 - 1 = x + 7

New fraction = $\sf \dfrac{x + 1}{x + 7}$

According to the question :-

⟹ $\sf \dfrac{x + 1}{x + 7} = \dfrac{3}{2}$

⟹ $\sf 2(x + 1) = 3(x + 7)$

⟹ $\sf 2x + 2 = 3x + 21$

⟹ $\sf x = -19$

• Numerator = -19
• Denominator = -19 + 8 = -11

$\boxed{\sf Fraction = \dfrac{-19}{-11}}$