SOLUTION: Five years ago, a mother was twice as old as her son. In 6 years time, the sum of the ages will be 82. Find their present ages

Question 985647: Five years ago, a mother was twice as old as her son. In 6 years time, the sum of the ages will be 82. Find their present ages
Answer by algebrahouse.com(1659) (Show Source): You can put this solution on YOUR website!
x = son's age 5 years ago
2x = mother's age 5 years ago {mother was twice as old as son 5 years ago}

x + 5 = son's age now
2x + 5 = mother's age now

x + 11 = son's age in 6 years
2x + 11 = mother's age in 6 years

x + 11 + 2x + 11 = 82 {in 6 years time, the sum of the ages will be 82}
3x + 22 = 82 {combined like terms}
3x = 60 {subtracted 22 from each side}
x = 20 {divided each side by 3}

x + 5 = 25 {substituted 20, in for x, into x + 5}
2x + 5 = 45 {substituted 20, in for x, into 2x + 5}

son is 25 now
mother is 45 now

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