The key to solving a problem like this is to be able to define all of the equations in terms of just one variable.

Let’s look:

If A’s age is 4 years more than 2x C’s age, we convert that to the following equation:

A – 4 = 2C

(we subtract 4 to equalize A’s age and 2 times C’s age)

We also know that B + 5 = C

(add 5 years to B’s age and he’s the same age as C)

since both equations are expressed in C, it’ll be easier to define both A and B in terms of C.

First equation: a – 4 = 2 c

a = 2 c + 4

Second equation: b + 5 = c

b = c – 5

Now it’s time to use that last equation we were given – the one that says that the average age is 41. If the average of 3 numbers is 41, we know that the total age of all three men is 41 x 3 or 123.

Or in other words, that:

A + B + C = 123

Now let’s get rid of A and B – we know how to define both of them in terms of C.

Every time we see A, write 2c+4

Every time we see B, write c-5

So A+B+C = (2c+4) + (c – 5) + c

simplified…

4c -1 = 123

4c = 124

c = 31

If c = 31, we can go back now and figure out what A and B are:

A is 2c + 4 (2)(31)+4 = 66

B is c – 5 (31-5) = 26

C is 31

Is the average of 66, 26 and 31 equal to 41? Yep.

There are our answers.

Hope this helps!

ok, so a is angelo, b is brandon, and c is carl. 4+2c=a ; b=c-5 ; (a+b+c)\3=41 . <- that means divided by 3... hope thats clear...

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