If I times the second one by 3, I get 81 normal hours and 9 overtime hours for $189. This is 60 more normal hours and $120 more dollars, for $2 for each normal hour. Working backwards, I get 69 - 21*2 = 27, so each overtime hour is $3.

This answer works, but it seems too easy. Am I doing this wrong? Thanks!

You're actually doing quite a bit of calculation there, but the concise nature of your answer may make it appear simpler than it is. I see from a post in another thread that you're already doing algebra, so I would imagine that an algebraic solution may be expected. If so, here's one way of laying things out algebraically. I'll explain each step as I go, so that anyone not familiar with algebra can follow along. I've changed the figures in the question, but the method would be similar.

*A carpenter can make a cabinet in 10 hours. If he works for 8 hours at the normal rate of pay and 2 hours at the overtime rate, he will be paid $34. But if he works for 6 hours at the normal rate and 4 hours at the overtime rate, his earnings will be $38. Find his normal and overtime rates of pay.*

This example can be solved using 'simultaneous equations'. Basically, we are given enough information to form two separate equations. Neither equation is enough to give us the answer by itself, but together, they can.

To make things simpler to handle, we use letter variables as stand-ins for our unknown quantities. In this case, we don't know the number of normal hours or the number of overtime hours. We can choose any letters we like, but it's best to use something that's memorable. I'll choose

for 'normal hours' and

for 'extra' or 'overtime' hours. (I could have used

for 'overtime hours', but it would likely be confused with a zero at some point, so I decided against it.)

I've numbered the equations, so that we can refer to them later. As things stand, we can't solve either equation, since each has two variables - so, we need to eliminate one of the variables. Looking at the equations, I see that the simplest way would be to eliminate

.

Now,

could become

if we double it, or alternatively,

could become

if we halve it. In algebra, we can modify an equation in numerous ways and not change its equality, just so long as whatever we do to it, we do it equally to

*everything *on

*both *sides of the equals sign. The

and the

of equation 2 are both even, so they'll halve nicely, so we'll choose to halve all of equation 2. (We chose to halve rather than double, because it's usually easier to work with smaller rather than larger numbers.)

So, to eliminate the

, we substract one equation from the other, so the

term disappears. It doesn't matter which way round we do the subtraction (equation 2 - equation 3, or equation 3 - equation 2), but it's usually easier if we choose the one that results in the number attached to the remaining variable (

) being positive.

Rearranging equation 4 so that

is on its own, we get:

Great! We've got the value of one of the variables,

. Now, all we have to do is plug

back into either of our original equations that contain both

and

, and solve for

. We'll choose equation 2.

So, now we know

.

We should check that we have the correct values, so we'll plug

and

into the

*other original* equation (equation 1).

Now, all that remains to be done is to re-read the original question, check we've calculated what was asked of us, and write up (and underline) our final answer:

Your teacher / tutor / book may lay things out a little differently, but it'll probably be something along these lines.

As you can see, to write all that out is quite a bit more work than writing your explanation, but the actual method used in each case is actually quite similar. Here, I've written every little step to explain what I'm doing, but in practice, a few minor steps can be missed out if we are comfortable with manipulating equations.

For instance,

... could become...

But, as Eliz has said, as long as you show a logical progression from start to finish (and end up with the correct answer

), you should get full marks.

DAiv