Hi, there--

The Problem:
A father is twice as old as his son and thrice as old as his daughter. If the father is eight years 
older than the sum of his children's ages, find the age of the daughter?

A Solution:
Let x be the age of the daughter.
Let y be the age of the son.
Let z be the age of the father.

We have three variables. We will write three equations using the information in the problem about 
this family.

"A father is twice as old as the son," can be expressed in algebra as z = 2y.

"A father is thrice as old as his daughter," can be written as z = 3x.

"The father is eight years older than the sum of his children's ages can be written as
z = x + y + 8

Now we solve this system of equations for x, y, and z. Rewrite the first equation in a "y=..." form.
z = 2y ------> y = (1/2)z

Rewrite the second equation in "x=..." form.
z = 3x  -------> x = (1/3)z

We see that x is equivalent to (1/3)z and y is equivalent to (1/2)z. Make these substitutions in the 
third equation.
z = x + y + 8
z = (1/3) z+ (1/2)z + 8

Solve for z. Combine like terms on the right hand side: (1/3)z + (1/2)z is (5/6)z.
z = 5z/6 + 8

Subtract (5/6)z from both sides of the equation: (1 - 5/6 = 1/6).
(1/6)z = 8

Multiply both sides of the equation by 6. (6 is the reciprocal of 1/6, so their product is 1.)
z = 8 * 6
z = 48

In the context of the problem, z = 48 means that the father is 48 years old.
The father is twice the age of the son, so the son is 24 years old.
The father thrice the age of the daughter so she is 16 years old.

We need to make sure that these ages satisfy the last requirement of the problem: The father 
is 8 years older than the sum of his children's ages.
The sum of the children's ages is 24+16=40. 40 is eight more than 48. Check!

That's it. Please email me if you have questions about the solution. I would appreciate the feedback.

Thanks!

Ms.Figgy
math.in.the.vortex@gmail.com