In a certain​ fraction, the denominator is less than the numerator. If is added to both the numerator and​ denominator, the resulting fraction is equal to . Find the original fraction.

Question

In a certain​ fraction, the denominator is

7

less than the numerator. If

3

is added to both the numerator and​ denominator, the resulting fraction is equal to

15/8.

Find the original fraction pleaseee

Answer 1

If you start with a numerator N and a denominator that is 7 less then you have N / (N-7).

Now add 3 to each and you have (N+3) / (N-7+3) or (N+3) / (N-4) to simplify.

Now the result is 15/8 so you can set the numerators equal to each other (N+3)=15 and this results is N=12.

Same for the denominator (N-4)=8 also solves N=12.

Now to get the original fraction you plug 12 for N in the original formula of N / (N-7) and this gives you 12/5.

Good question!

Answer 2

In a certain​ fraction, the denominator is 7 less than the numerator.

^{d + 7} / _d

3 is added to both the numerator and​ denominator

^{d + 7 + 3} / _{d + 3}

the resulting fraction is equal to 15/8

d + 10 = 15

d + 3 = 8

=> d = 5

substitute in original fraction ^{5 + 7} / ₅ = ¹²/₅

Answer 3

D=N-7

N+3/D+3 = (N+3)(N-7+3) = 15/8

(N+3)/(N-4) = 15/8

8N+24 = 15N-60

7N = 84

N=84/7 = 12

D =5

original fraction = 12/5