# Age Problem in Algebra

by Dianne Dagaas
(Manila)

Help me to solve this age problem in algebra:

Ana is ten years older than Belinda, while Belinda is 5 years younger than Celina. Twice Aana's age is six less than thrice Celina's age. How old are they?

## Karin from Algebra Class says:

In order to complete this age problem, we must first identify our variables:

Let a - Ana's age
Let b = Belinda's age
Let c = Celina's age

Now we need to write equations based on the information given:

Ana is ten years older than Belinda means that
a = b+10

Belinda is 5 years younger than Celina means that
b = c - 5

Twice Aana's age is six less than thrice Celina's age means:
2a = 3c - 6

So our three equations are:
a = b+10
b = c-5
2a = 3c - 6

Now we need to figure out how to substitute one equation into another so that we can solve for one of the missing variables.

First let's substitute c-5 for b into the equation: a =b+10.

a = b+10

a = c-5 +10

a = c +5

Now that we know that a = c+ 5, we can substitute this into the equation: 2a = 3c - 6

2a = 3c - 6
2(c+5) = 3c - 6     Substitute c+5 for a.

2c +10 = 3c - 6     Distribute 2 throughout the parenthesis.

Now solve for c.

2c - 2c +10 = 3c-2c - 6         Subtract 2c from both sides

10 = c - 6        3c - 2c = c

10+6 = c-6+6         Add 6 to both sides

16 = c

Now that we know that c = 16 we can substitute 16 for c.

b = c-6
b = 16-5         Substitute 16 for c

b = 11

Now substitute 11 for b into a= b+10

a = b+10
a = 11 +10
a = 21

Therefore, Ana is 21, Belinda is 11 and Celina is 16.

Hope this helps.

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