We already know that a big part of algebra is solving for an unknown value. Sometimes it takes more than one step to solve the equation. You have to be able to determine which step to do first. We still have to use our inverse operations. These are operations that “undo” each other. Addition and subtraction are inverse operations and so are multiplication and division. Let’s look at a few examples before we go any further.

Example 1:

 

Solve for x.

6x - 20 = 10

 

A way to look at these equations is to see which number is “closest” to x. This would be the number 6. So, we want to move this 6 last. Let’s cancel out the 20 by using the opposite operation and adding 20 to both sides.

6x - 20 = 10
    +20   +20
6x = 30

 

Now, we can cancel out the 6 by dividing both sides by 6.

6x = 30
6      6

 x = 5

 

You can check your answer by plugging the value back in for x. Let’s try.

 

6x - 20 = 10
6(5) - 20 = 10
30 - 20 = 10
10 = 10                      It works!

 

Example 2:

 

Solve for b.

\(\frac{b}{2} + 15 =  - 3\) 

This time the 2 is closest to the variable, so leave him and cancel out the 15 first. We have to subtract 15 from both sides.

 

 \(\frac{b}{2} + 15 =  - 3\)
                -15     -15

  \(\frac{b}{2} =  - 18\)

Be careful with this next step. The inverse operation is to multiply both sides by 2. This will cancel out the 2 in the denominator.

 \(\frac{b}{2} \bullet 2 =  - 18 \bullet 2\)

 \(b =  - 36\)

 A multi-step equation sometimes involves combining like terms. This is when you add together all the matching terms until there is only one of each. Here’s an example.

 

Example 3:

 

Solve the equation.

12 = 7m - 5 - 2m + 2

We have two sets of like terms in this equation. The 7m and the -2m will combine to be a 5m and the -5 and the +2 will combine to become -3.

12 = 5m - 3  Now, add 3 to both sides.
+3       +3

15 = 5m      Divide both sides by 5.
 5   5         

3 = m

These multi-step equations can get a bit long and difficult. Sometimes, there is an unknown on both sides of the equation. We can’t solve until it is on only one side. So, we must first move one of the terms to the other side by using an inverse operation.

Example 4:

 

 Solve the equation.
5p - 4 = 3p + 20

If the two variables were on the same side of the equation, you could combine them, but this is different. We have to “move” one to the other side. Let’s subtract 3p from each side of the equal sign.

 5p - 4 = 3p + 20
-3p      -3p

 2p - 4 = 20         Now, add 4 to each side.
    + 4  + 4

2p = 24             Divide each side by 2.
2      2

p = 12

Lastly, let’s take a look at an example that will use all of these steps.

Example 5:

 

Solve for x.
4(3x + 3) - 7 = 22 - 5x

This starts off with an old concept of the distributive property. We must distribure the 4 into that first parenthesis.

12x + 12 - 7 = 22 - 5x      Now, combine like terms on the left side.
12x + 5 = 22 - 5x           Since we have a -5x, we can add 5x to both sides.
+5x              +5x

17x + 5 = 22                Subtract 5 from each side.
       - 5   -5 

17x = 17            Divide both sides by 17.
17      17

x = 1

Below you can download some free math worksheets and practice.