Question

In a certain fraction, the numerator is 4 less than the denominator. If 4 is added to both the numerator and the denominator , the resulting fraction is equal Vhat was the original 8/12 fraction (not necessarily written in lowest terms

Answer 1

Answer:

The original fraction was $\frac{4}{8}$

Step-by-step explanation:

The correct question is

In a certain fraction, the numerator is 4 less than the denominator. If 4 is added to both the numerator and the denominator , the resulting fraction is equal to 8/12. What was the original  fraction (not necessarily written in lowest terms)

Let

x ----> the numerator of the original fraction

y ----> the denominator of the original fraction

$\frac{x}{y}$

we know that

$x=y-4$

so

the original fraction is

$\frac{y-4}{y}$

If 4 is added to both the numerator and the denominator , the resulting fraction is equal to 8/12

so

$\frac{y-4+4}{y+4}=\frac{8}{12}$

$\frac{y}{y+4}=\frac{8}{12}$

Solve for y

$12y=8y+32\\12y-8y=32\\4y=32\\y=8$

Find the value of x

$x=y-4$

$x=8-4=4$

therefore

The original fraction was

$\frac{4}{8}$