Brenda is 4 years older than Walter, and Carol is twice as old as Brenda. Three years ago, the sum of their ages was 35. How old is each now?
x = Walter's age now
x + 4 = Brenda's age now {Brenda is 4 yrs older than Walter}
2(x + 4) = 2x + 8 = Carol's age now {Carol is twice as old as Brenda, used distributive property}
x - 3 = Walter's age 3 years ago {subtracted 3 from x}
x + 1 = Brenda's age 3 years ago {subtracted 3 from x + 4}
2x + 5 = Carol's age 3 years ago {subtracted 3 from 2x + 8}
(x - 3) + (x + 1) + (2x + 5) = 35 {sum of ages, 3 years ago, was 35}
x - 3 + x + 1 + 2x + 5 = 35 {took out parentheses}
4x + 3 = 35 {combined like terms}
4x = 32 {subtracted 3 from both sides}
x = 8 = Walter now {divided both sides by 4}
x + 4 = 12 = Brenda now {substituted 8, in for x, into x + 4}
2(x + 4) = 24 = Carol now {substituted 8, in for x, into 2(x + 4)}
Walter is 8 now
Brenda is 12 now
Carol is 24 now
- Algebra House
x + 4 = Brenda's age now {Brenda is 4 yrs older than Walter}
2(x + 4) = 2x + 8 = Carol's age now {Carol is twice as old as Brenda, used distributive property}
x - 3 = Walter's age 3 years ago {subtracted 3 from x}
x + 1 = Brenda's age 3 years ago {subtracted 3 from x + 4}
2x + 5 = Carol's age 3 years ago {subtracted 3 from 2x + 8}
(x - 3) + (x + 1) + (2x + 5) = 35 {sum of ages, 3 years ago, was 35}
x - 3 + x + 1 + 2x + 5 = 35 {took out parentheses}
4x + 3 = 35 {combined like terms}
4x = 32 {subtracted 3 from both sides}
x = 8 = Walter now {divided both sides by 4}
x + 4 = 12 = Brenda now {substituted 8, in for x, into x + 4}
2(x + 4) = 24 = Carol now {substituted 8, in for x, into 2(x + 4)}
Walter is 8 now
Brenda is 12 now
Carol is 24 now
- Algebra House