This is the solution to the algebra age problem asked by Joshua that read, "A man is 4 years older than his wife and 5 times as old as his son. When the son was born, the age of the wife was six-sevenths that of her husband's age. Find the age of each."

Ok so here we go...

"A man is 4 years older than his wife"

m = w + 4

Now we will subtract 4 from both sides to use this conveniently later ( you will see)

w = (m-4)

"and 5 times as old as his son."

m = 5s

"When the son was born, the age of the wife was six-sevenths that of her husband's age."

w - s = 6/7(m-s)

7(w-s) = 6(m-s) -->

7w - 7s = 6m - 6s --->

7w = 6m - 6s + 7s --->

7w = 6m + s

replace w with (m-4)

7(m-4) = 6m + s

7m - 28 = 6m + s

7m - 6m = s + 28

m = s + 28

replace m with 5s

5s = s + 28

5s - s = 28

4s = 28

s = 7 yrs is the son's age

then

5(7) = 35 yrs is Dad's age

and

35-4 = 31 yrs is Mom's age

Now check work...

"When the son was born, the age of the wife was six-sevenths that of her husband's age."

31 - 7 = 6/7(35 - 7)

24 = 6/7*28

24 = 24;

## Friday, October 21, 2011

### Age Problem 8 Solution

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I couldn't solve tht....nice one though

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