SOLUTION: When Juanita is as old as her mother is now, she will be five times as old as her son is now. By then, Juanita's son will be 8 years older than Juanita is now. Juanitas age combine

Question 1120446: When Juanita is as old as her mother is now, she will be five times as old as her son is now. By then, Juanita's son will be 8 years older than Juanita is now. Juanitas age combined with her mothers age, equals 100 years. how old is Juanita's son?
Found 2 solutions by josgarithmetic, greenestamps:
Answer by josgarithmetic(39620) (Show Source): You can put this solution on YOUR website!
Assigning variables and carefully transcribing the description into equations:

PERSON VARIABLE IN x YEARS Juanita J J+x Son y y+x Mother m m+x

The question asks for J.

Steps can be different. First step could be, substitute for m wherever possible,...

May find that J is 35.
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!

The other tutor didn't show a complete solution, so I don't know how much work remained to reach the answer from where they left off. In any case, I get a different answer than they did....

There are many different ways the problem can be set up; the method I used shown below may not be the easiest or most efficient....

Let m be the mother's current age

Then 100-m is Juanita's current age, since the sum of their ages is 100.

Let s be the son's current age.

The number of years until Juanita is her mother's current age is the difference of their current ages: m-(100-m) = 2m-100.

According to the statement of the problem, m is Juanita's age "then".

The son's age "then" is s + (2m-100).

Juanita "then" will be 5 times as old as her son is now:
(1) m = 5s.

Her son's age "then" will be 8 more than Juanita's current age:
(2) s+2m-100 = 108-m --> s+3m = 208.

Substitute (1) in (2) to find s, the son's current age.

s+3(5s) = 208
s+15s = 208
16s = 208
s = 13

The son's current age is 13.
So the mother's current age is 5(13) = 65.
So Juanita's current age is 100-65 = 35.
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