# SOLUTION: A father is six times as old as his son. Two years ago he was eight times as old. Find the son's age.

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 Question 457222: A father is six times as old as his son. Two years ago he was eight times as old. Find the son's age.Found 3 solutions by josmiceli, algebrahouse.com, karthiktcs:Answer by josmiceli(9817)   (Show Source): You can put this solution on YOUR website!Let = Father's age now Let = Son's age now given: (1) (2) -------------------- (2) (2) (2) Substitute (1) into (2) (2) (2) (2) The Son is 7 now check: (1) (1) (1) (2) (2) (2) (2) OK Answer by algebrahouse.com(1079)   (Show Source): You can put this solution on YOUR website!x = son's age now 6x = father's age now {father is six times as old as his son} x - 2 = son two years ago 6x - 2 = father two years ago 6x - 2 = 8(x - 2) {two years ago the father was eight times as old as his son} 6x - 2 = 8x - 16 {used distributive property} -2 = 2x - 16 {subtracted 6x from both sides} 14 = 2x {added 16 to both sides} x = 7 {divided both sides by 2} son is 7 years old now www.algebrahouse.com Answer by karthiktcs(3)   (Show Source): You can put this solution on YOUR website!let the present age of son be x and father age be 6x two years ago (6x-2)=8(x-2) 6x-2=8x-16 6x-2-8x+16 -2x+14=0 14=2x x=7