Fred is 4 times as old as his niece, Selma. Ten years from now, he will be twice as old as she will be. How old is each now?

x = Selma's age now

4x = Fred's age now {Fred is 4 times as old as Selma}

x + 10 = Selma's age in 10 years

4x + 10 = Fred's age in 10 years

4x + 10 = 2(x + 10) {his age in 10 yrs is twice her age in 10 years}

4x + 10 = 2x + 20 {used distributive property}

2x + 10 = 20 {subtracted 2x from both sides}

2x = 10 {subtracted 10 from both sides}

x = 5 {divided both sides by 2}

4x = 20 {substituted 5, in for x, into 4x}

4x = Fred's age now {Fred is 4 times as old as Selma}

x + 10 = Selma's age in 10 years

4x + 10 = Fred's age in 10 years

4x + 10 = 2(x + 10) {his age in 10 yrs is twice her age in 10 years}

4x + 10 = 2x + 20 {used distributive property}

2x + 10 = 20 {subtracted 2x from both sides}

2x = 10 {subtracted 10 from both sides}

x = 5 {divided both sides by 2}

4x = 20 {substituted 5, in for x, into 4x}

**Selma is 5 now**

Fred is 20 nowFred is 20 now

*- Algebra House*