Question

The age of a father is 5 more than twice of his son's age. The age of the mother is the sum of
her son's age and 1/3 of the father's age. The sum of all their ages is 100. Find their ages.

Answer 1

x = son's age

2x + 5 = father's age {age of father is 5 more than twice son's}

x + (1/3)(2x + 5) = mother's age {age of mother is sum of son and 1/3 father}

x + 2x + 5 + x + (1/3)(2x + 5) = 100 {sum of all ages is 100}

3[x + 2x + 5 + x + (1/3)(2x + 5) = 100] {multiplied entire equation by 3 to eliminate fraction}

3(x) + 3(2x) + 3(5) + 3(x) + 3[(1/3)(2x + 5)] = 3(100) {multiplied each term by 3}

3x + 6x + 15 + 3x + 2x + 5 = 300 {multiplied through}

14x + 20 = 300 {combined like terms}

14x = 280 {subtracted 20 from each side}

x = 20 {divided each side by 14}

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

x + (1/3)(2x + 5) = 35 {substituted 20 for x into mother's age}

son is 20, father is 45, and mother is 35