# So how do you use algebra to solve age problemsWhen it comes to age problems, like Jim is 3 times as old as Sue, I always guess and check. It's easy for me to look at numbers and see the relation....

So how do you use algebra to solve age problems

When it comes to age problems, like Jim is 3 times as old as Sue, I always guess and check. It's easy for me to look at numbers and see the relation. But my teacher noticed this and insists on using algebra instead. Can someone tell me the correct way to solve age probles using algebra? [*I tried but I failed*]

1. John is now 3 times as old as his brother Sam. In 5 years, John will be twice as old as Sam will be them. Find their present ages.

Answer: Currently: Jim- 15 ~ Sam: 5

2. A man is now 6 times as old as his son. In 6 years, the father will be 3 times as old as the son will be then. Find their present ages.

Answer: Currently: Man- 24 ~ Son: 4

*I solved these using guess and check, I need to solve it using algebra.*

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(1) Assign variables to the two boys' ages; let John's age be J, and Sam's age S. (It often helps to identify your variables in a way that represents the problem instead of blindly using x and y)

John is now 3 times as old as Sam -- translating to algebra we get J=3S.

In 5 years John will be twice Sam's new age: translated becomes J+5=2(S+5).

From the first equation, we can substitute 3S for J in the second equation:

3S+5=2(S+5) ==> 3S+5=2S+10 ==> S=5 so J=15.

**Thus John is 15 and Sam is 5.** (Check: 15 is 3x5, and in 5 years 20 is 2x10 as required.)

** We have a system of two equations in two unknowns; you can use any method you know to solve them.**

(2) Assign variables: M=man's age, S=son's age.

The man is 6 times as old as his son so M=6S.

In 6 years, the man's age will be 3 times the son's new age:

M+6=3(S+6).

Again we can use substitution, replacing M with 6S:

6S+6=3(S+6) ==> 6S+6=3S+18 ==> 3S=12 ==> S=4

**The son's age is 4, so the man's age is 24.**

(Check: 24=6x4 and 30=3x10)