Question 82784
Let w=woman, h=husband, d=daughter

Set up the following equations:

{{{w=h+5}}} "A woman is 5 years older than her husband". Notice if h=30, w=40 which is 10 years older.Solving for h we get

{{{h=w-5}}}


{{{w=10d}}} "the woman is 10 times as old as their daughter". Notice if d=2, then w=20 which means the woman is 10 times older. Solving for d we get

{{{d=w/10}}}


So in 14 years the woman's age is 


{{{w=w+14}}}


In 14 years the husband's age is 


{{{h=h+14}}}


and in 14 years the daughter's age is 


{{{d=d+14}}}


Adding these together we get


{{{(w+14)+(h+14)+(d+14)=100}}}


{{{(w+14)+((w-5)+14)+((w/10)+14)=100}}}Plug in {{{h=w-5}}} and {{{d=w/10}}} (we solved for these earlier). This eliminates every variable but w, which means we can solve for w now.


{{{2w+37+w/10=100}}} Combine like terms


{{{10(2w+37+w/10)=10(100)}}}Multiply both sides by 10. This will eliminate the denominator.


{{{20w+370+w=1000)}}} Distribute


{{{21w+370+w=1000)}}} Combine like terms


{{{21w=630}}} Subtract 370 from both sides


{{{w=30}}} Divide both sides by 21


So the woman is 30 years old. In 14 years she will be 44 years old. The husband's age is:


{{{h=w-5}}}

{{{h=30-5=25}}} plug in w=30


So the husband is 25 now and in 14 years he will be 39. Finally the daughter's age is:


{{{d=w/10}}}


{{{d=30/10=3}}} plug in w=30


So the daughter is 3 years old now and in 14 years will be 17 years old. So if we add everyone's future age (ie their age in 14 years) together we get:

{{{future_age_woman+future_age_husband+future_age_daughter=100}}}

{{{44+17+39=100}}}

{{{100=100}}} works

This verifies our answer. 



note: you forgot to factor in the 14 year time lapse. That would explain the large age answer you got. Also, I'm not sure how you got Daughter=h+5/10