Help on answering these Algebra age Problems

The sum of 1/3 of Debbie's age and Harry's age is 11 yrs. Six years from now, the sum of their ages is 40 years. How old are they now?


Billy is twice as old as his son. In 10 years, the ratio of their ages will be 3:5. How old are they now?

1)
d/3 + h/2 = 11
2d + 3h = 66 --------------- 1
d+6 + h+6 = 40
d + h = 28 --------------- 2
Those are the simultaneous equations you need to solve.

Eqn1 - 2Eqn2,
2d + 3h - 2d - 2h = 66-56
h = 10
Substitute into Eqn2
d + 10 = 28
d = 18

So Harry is 10, and Debbie is 18.

2)
b = 2s ---------- 1
(b+10)/(s+10) = 5/3
3(b+10) = 5(s+10)
3b + 30 - 5s - 50 = 0
3b - 5s - 20 = 0 -------- 2
Those are the simultaneous equations you need to solve.

Substituting Eqn1 into Eqn2,
6s - 5s - 20 = 0
s - 20 = 0
s = 20
Substituting into Eqn1,
b = 2*20 = 40

So Billy is 40, and his son is 20.

EDIT: Thanks for the additional information. I hope this helps you.