SOLUTION: the sum of the present ages of Eric and his father is 58 years. In 10 years, his father will be twice as old as Eric will be at that time, Find their present ages
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Question 461537: the sum of the present ages of Eric and his father is 58 years. In 10 years, his father will be twice as old as Eric will be at that time, Find their present ages
Answer by algebrahouse.com(1659) (Show Source): You can put this solution on YOUR website!
the sum of the present ages of Eric and his father is 58 years. In 10 years, his father will be twice as old as Eric will be at that time, Find their present ages
x = Eric's age now
58 - x = father's age now {sum of present ages is 58}
x + 10 = Eric's age in 10 years {added 10 onto current age}
68 - x = father's age in 10 years {added 10 onto current age}
68 - x = 2(x + 10) {in ten years, the father will be twice Eric}
68 - x = 2x + 20 {used distributive property}
68 = 3x + 20 {added x to both sides}
3x = 48 {subtracted 20 from both sides}
x = 16 {divided both sides by 3}
58 - x = 42 {substituted 16, in for x, into 58 - x}
Eric is 16
father is 42
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