Answer to Question #202212 in Algebra for FZ Altamimi

Question

Nine years ago a mother was three times as old as her son in eight years from now the sum of their ages will be 78. How old are they today?

Answer 1

Let a mother be $x$ years old and her son be $y$ years today.

Given nine years ago a mother was three times as old as her son 

$$x-9=3(y-9)=>x=3y-18$$

Eight years from now the sum of their ages will be 78

$$x+8+y+8=78$$

Substitute

$$3y-18+y+16=78$$

Solve for $y$

$$4y=80$$

$$y=20$$

Then

$$x=3(20)-18=42$$

a mother is $42$ years old and her son is $20$ years today.