# SOLUTION: Five years ago father's age was seven times his sons age. Ten years hence their ages will be in the ratio of 5 : 2. Find their present ages

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 Question 324481: Five years ago father's age was seven times his sons age. Ten years hence their ages will be in the ratio of 5 : 2. Find their present ages Answer by nerdybill(7003)   (Show Source): You can put this solution on YOUR website!Five years ago father's age was seven times his sons age. Ten years hence their ages will be in the ratio of 5 : 2. Find their present ages Let x = father's age and y = son's age . From:"Five years ago father's age was seven times his sons age." x-5 =7(y-5) x-5 =7y-35 x-7y = -30 equation 1 . From:"Ten years hence their ages will be in the ratio of 5 : 2." (x+10)/(y+10) = 5/2 2(x+10) = 5(y+10) 2x+20 = 5y+50 2x-5y = 30 equation 2 . x-7y = -30 2x-5y = 30 . Multiply the top equation by -2 and add to the second: -2x+14y = 60 2x-5y = 30 ------------- 9y = 90 y = 10 years old (son's age) . Father's age: x-7y = -30 x-7(10) = -30 x-70 = -30 x = 40 years old (father's age)