Question

The denominator of a fraction exceeds the numerator by 4. If 5 is taken away from each, the sum of the reciprocal of the new fraction and 4 times the original fraction is 5. Find the original fraction.​

Answer 1

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

Step-by-step explanation:

Let the numerator be x.

So denominator will be x + 4

Original fraction = $\sf\dfrac{x}{x+4}$

Now Let's understand what the question is asking.

If 5 is taken away from each,

$\sf\dfrac{x-5}{x+4-5}=\dfrac{x-5}{x-1}$

the sum of the reciprocal of the new fraction

$\sf New \ Fraction = \dfrac{x-5}{x-1}$

Reciprocal of New Fraction = $\sf\dfrac{x-1}{x-5}$

and 4 times the original fraction

Original fraction = $\sf\dfrac{x}{x+4}$

4 times of Original fraction = $\sf 4 \times \dfrac{x}{x+4}= \dfrac{4x}{x+4}$

is 5.

Hence,

$\sf\dfrac{x-1}{x-5}+\dfrac{4x}{x+4}=5$

Now, we can come to the solving part of the equation.

$\rm\dashrightarrow\dfrac{(x-1)(x+4)+4x(x-5)}{(x+4)(x-5)}=5$

$\rm\dashrightarrow\dfrac{x^2+3x-4+4x^2-20x}{x^2-x-20}=5$

$\rm\dashrightarrow\dfrac{5x^2-17x-4}{x^2-x-20}=5$

$\rm\dashrightarrow 5x^2-17x-4=5(x^2-x-20)$

$\rm\dashrightarrow 5x^2-17x-4=5x^2-5x-100$

5x^2 will be cancelled on both the sides of the equation

$\rm\dashrightarrow -17x-4=-5x-100$

$\rm\dashrightarrow -17x+5x=4-100$

$\rm\dashrightarrow -12x=-96$

$\rm\dashrightarrow x=\dfrac{-96}{-12}$

$\bf\dashrightarrow x=8$

Original fraction = $\sf\dfrac{x}{x+4}$

$\sf\dashrightarrow\dfrac{8}{8+4}=\dfrac{8}{12}=\bf \dfrac{2}{3}$