## Two Step Equations

Once you get the hang of solving simple equations like a + 4 = 7 and 9y = 72, it’s time to move on to some more difficult equations that take multiple steps to solve. Let’s start with an example:

2x + 5 = 17

To get x by itself, we’re going to have to get rid of the multiplication by 2 and the addition of 5. The question is, which should we get rid of first? Let’s attack the multiplication by 2 first and see what happens.

To get rid of the multiplication by 2, we have to divide both sides by 2. However, remember that we have to divide the entire left side by 2 as follows:

2x + 5 = 17
(2x + 5)/2 = 17/2
2x/2 + 5/2 = 17/2
x + 5/2 = 17/2

Now we can subtract 5/2 from both sides:

x + 5/2 = 17/2
x + 5/2 – 5/2 = 17/2 – 5/2
x = 12/2
x = 6

Once we get an answer, we check to make sure it’s the correct one:

2x + 5 = 17
2(6) + 5 = 17
12 + 5 = 17
17 = 17

And there we are. So now let’s solve the same equation again, but this time start by attacking the addition of 5 first.

2x + 5 = 17
2x + 5 – 5 = 17 – 5
2x = 12

Now we’ll divide both sides by 2:

2x/2 = 12/2
x = 6

And we get the same answer. The only difference between the two methods is that when we attack the addition and subtraction first, we can end up with messy fractions. While this is mathematically fine, dealing with more complicated fractions when we don’t have to opens us up to making mistakes that we wouldn’t make normally. However, you’re welcome to do it either way.

So now let’s look at a second example.

3c – 4 = 11

We need to c by itself, so we’ll need to attack the multiplication by 3 and the subtraction of 4. We saw earlier that attacking the addition and subtraction portion of the problem is easier, so we’ll attack the subtraction of 4 first by adding 4 to each side.

3c – 4 = 11
3c – 4 + 4 = 11 + 4
3c = 15

Now we have a single-step problem that’s pretty clear. Here we get rid of the multiplication by 3 by dividing each side by 3.

3c = 15
3c/3 = 15/3
c = 5