SOLUTION: 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 ages is 100. Find their age
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Question 815701: 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 ages is 100. Find their ages.
Answer by algebrahouse.com(1659) (Show Source): You can put this solution on YOUR website!
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
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