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Algebra's the easiest dear why ask for a simple one. What topic?

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}

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

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