1. Ch 3.3 – Solving Multi-Step Equations
Algebra 1
Ch 3.3 – Solving Multi-Step Equations

2. Objective
Students will solve multi-step equations

3. Before we begin…
In previous lessons we solved simple one-step linear equations…
In this lesson we will look at solving multi-step equations…as the name suggests there are steps you must do before you can solve the equation…
To be successful here you have to be able to analyze the equation and decide what steps must be taken to solve the equation…

4. Multi-Step Equations This lesson will look at:
Solving linear equations with 2 operations
Combining like terms first
Using the distributive property
Distributing a negative
Multiplying by a reciprocal first
Your ability to be organized and lay out your problem will enable you to be successful with these concepts

5. Linear Equations with 2 Functions
Sometimes linear equations will have more than one operation
In this instance operations are defined as add, subtract, multiply or divide
The rule is
First you add or subtract
Second you multiply or divide.

6. Linear Equations with 2 Operations
You will have to analyze the equation first to see what it is saying…
Then, based upon the rules, undo each of the operations…
The best way to explain this is by looking at an example….

7. Example # 1
2x + 6 = 16
In this case the problem says 2 times a number plus 6 is equal to 16
Times means to multiply and plus means to add
The first step is to add or subtract.
To undo the addition of 6 I have to subtract 6 from both sides which looks like this:
16 minus 6 equals 10
The 6’s on the left side cancel each other out
2x + 6 = 16
2x = 10
What is left is a one step equation 2x = 10

8. Example #1 (Continued)
2x = 10
The second step is to multiply or divide
To undo the multiplication here you would divide both sides by 2 which looks like this:
10 divided by 2 is equal to 5
2x = 10
The 2’s on the left cancel out leaving x
x = 5
The solution to the equation 2x + 6 = 16 is x = 5

9. Combining Like Terms First
When solving multi-step equations, sometimes you have to combine like terms first.
The rule for combining like terms is that the terms must have the same variable and the same exponent.
Example:
You can combine x + 5x to get 6x
You cannot combine x + 2x2 because the terms do not have the same exponent

10. Example # 2 7x – 3x – 8 = 24 7x – 3x – 8 = 24 4x – 8 = 24
I begin working on the left side of the equation
On the left side I notice that I have two like terms (7x, -3x) since the terms are alike I can combine them to get 4x.
7x – 3x – 8 = 24
4x – 8 = 24
After I combine the terms I have a 2-step equation. To solve this equation add/subtract 1st and then multiply/divide

11. Example # 2 (continued) 4x – 8 = 24 4x – 8 = 24 +8 +8 4x = 32
Step 1: Add/Subtract
Since this equation has – 8, I will add 8 to both sides
4x – 8 = 24
= 32
The 8’s on the left cancel out
4x = 32
I am left with a 1-step equation

12. The solution that makes the statement true is x = 8
Example # 2 (continued)
4x = 32
Step 2: Multiply/Divide
In this instance 4x means 4 times x. To undo the multiplication divide both sides by 4
32  4 = 8
The 4’s on the left cancel out leaving x
4x = 32
x = 8
The solution that makes the statement true is x = 8

13. Solving equations using the Distributive Property
When solving equations, sometimes you will need to use the distributive property first.
At this level you are required to be able to recognize and know how to use the distributive property
Essentially, you multiply what’s on the outside of the parenthesis with EACH term on the inside of the parenthesis
Let’s see what that looks like…

14. Example #3 5x + 3(x +4) = 28 5x + 3(x +4) = 28
In this instance I begin on the left side of the equation
I recognize the distributive property as 3(x +4). I must simplify that before I can do anything else
5x + 3(x +4) = 28 5x
+3x
+12
= 28
After I do the distributive property I see that I have like terms (5x and 3x) I have to combine them to get 8x before I can solve this equation

15. I am now left with a 2-step equation
Example # 3(continued)
8x + 12 = 28
I am now left with a 2-step equation
Step 1: Add/Subtract
The left side has To undo the +12, I subtract 12 from both sides
8x + 12 = 28
28 – 12 = 16
The 12’s on the left cancel out leaving 8x
8x = 16

16. Example # 3 (continued) 8x = 16 8x = 16 8 8 x = 2
Step 2: Multiply/Divide
On the left side 8x means 8 times x. To undo the multiplication I divide both sides by 8
8x = 16
16  8 = 2
The 8’s on the left cancel out leaving x
x = 2
The solution that makes the statement true is x = 2

17. Distributing a Negative
Distributing a negative number is similar to using the distributive property.
However, students get this wrong because they forget to use the rules of integers
Quickly the rules are…when multiplying, if the signs are the same the answer is positive. If the signs are different the answer is negative

18. Example #4
4x – 3(x – 2) = 21
I begin by working on the left side of the equation.
In this problem I have to use the distributive property. However, the 3 in front of the parenthesis is a negative 3.
When multiplying here, multiply the -3 by both terms within the parenthesis. Use the rules of integers
4x – 3(x – 2) = 21 4x
– 3x
+ 6
= 21
After doing the distributive property, I see that I can combine the 4x and the -3x to get 1x or x

19. The 6’s cancel out leaving x
Example # 4 (continued) 4x
– 3x
+ 6
= 21
x = 21
After combining like terms you are left with a simple one step equation. To undo the +6 subtract 6 from both sides of the equation
x = 21
21 – 6 = 15
The 6’s cancel out leaving x
x = 15
The solution is x = 15

20. Multiplying by a Reciprocal First
Sometimes when doing the distributive property involving fractions you can multiply by the reciprocal first.
Recall that the reciprocal is the inverse of the fraction and when multiplied their product is equal to 1.
The thing about using the reciprocal is that you have to multiply both sides of the equation by the reciprocal.
Let’s see what that looks like…

21. Example # 5
In this example you could distribute the 3/10 to the x and the 2.
The quicker way to handle this is to use the reciprocal of 3/10 which is 10/3 and multiply both sides of the equation by 10/3
On the left side of the equation, after multiplying by the reciprocal 10/3 you are left with 120/3 which can be simplified to 40
On the right side of the equation the reciprocals cancel each other out leaving x + 2
The new equation is:
40 = x + 2

22. Example #5 (continued) 40 = x + 2
After using the reciprocals you are left with a simple one-step equation
To solve this equation begin by working on the right side and subtract 2 from both sides of the equation
The 2’s on the right side cancel out leaving x
40 = x + 2
40 – 2 = 38
38 = x
The solution to the equation is x = 38

23. Comments
On the next couple of slides are some practice problems…The answers are on the last slide…
Do the practice and then check your answers…If you do not get the same answer you must question what you did…go back and problem solve to find the error…
If you cannot find the error bring your work to me and I will help…

24. Your Turn 6x – 4(9 –x) = 106 2x + 7 = 15 6 = 14 – 2x 3(x – 2) = 18

25. Your Turn 5m – (4m – 1) = -12 55x – 3(9x + 12) = -646. 7. 8.
5m – (4m – 1) = -12 9.
55x – 3(9x + 12) = -64
10.
9x – 5(3x – 12) = 30

26. Your Turn Solutions
14.2 4 8
1 ½ 3
-52
-13 -1 5