SOLUTION: yato is twice as old as his brother yukine. the quotient of their ages five years ago was seven less than yukine’s present age. Find their present ages.

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Question 1169566: yato is twice as old as his brother yukine. the quotient of their ages five years ago was seven less than yukine’s present age. Find their present ages.
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Yato, 2b
Brother Yukine, b

Second sentence, %282b-5%29%2F%28b-5%29=b-7
.
.
b%5E2-14b%2B40=0
.
.
b=%2814%2B-sqrt%2836%29%29%2F2
cross%28b=%2814%2B-+6%29%2F2%29
The one which works: b=4

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R E T R Y
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Returning to quadratic,
b%5E2-14b%2B40=0
Factorize,
%28b-4%29%28b-10%29=0
Possible solutions to check, b=4 or b=10.

Using the quotient description %282b-5%29%2F%28b-5%29=b-7
b if 4
%288-5%29%2F%284-5%29=4-7
arithmetic works but either age five years ago must not be negative.
b if 10
%2820-5%29%2F%2810-5%29=10-7
15%2F5=3
3=3
-
b=4 will not work.
b=10 will work.

Answer by ikleyn(52786) About Me  (Show Source):
You can put this solution on YOUR website!
.

My post is the  CORRECTION  to the answer by @josgarithetic.


The correct answer is :

            the present ages are  10 years  (the brother)  and  20  years (Yato).



The answer  b= 4  for the brother,  given by @josgarithmetic,  is INCORRECT  and does not work.