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?

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.

Karin

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