Algebra 1
Ch 3.3 – Solving Multi-Step Equations
Objective
 Students will solve multi-step
equations
Before we begin…
 In previous lessons we solved simple onestep linear equations…
 In this lesson we will look at solving multistep 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…
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
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.
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….
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
2x + 6 = 16
6 equals
The 6’s on
10
- 6 -6
the left
side cancel
2x
= 10
each other
out
What is left is a one step
equation 2x = 10
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:
2x
The 2’s on
the left
cancel out
leaving x
2
x
= 10
2
10 divided
by 2 is
equal to 5
=5
The solution to the equation 2x + 6 = 16 is x = 5
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
Example # 2
7x – 3x – 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
Example # 2 (continued)
4x – 8 = 24
Step 1: Add/Subtract
Since this equation has – 8, I will add 8 to both sides
4x – 8 = 24
+8 +8
The 8’s
on the
left
cancel
out
4x
= 32
I am left with a 1-step
equation
24 + 8 = 32
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
The 4’s on
the left
cancel out
leaving x
4x
= 32
4
4
x
=8
The solution that makes the
statement true is x = 8
32  4 = 8
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…
Example #3
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
Example # 3(continued)
8x + 12 = 28
I am now left with a 2-step equation
Step 1: Add/Subtract
The left side has +12. To undo the +12, I subtract 12 from both sides
The 12’s
on the left
cancel out
leaving 8x
8x + 12 = 28
-12
-12
8x
= 16
28 – 12 = 16
Example # 3 (continued)
8x
= 16
Step 2: Multiply/Divide
On the left side 8x means 8 times x. To undo the multiplication I
divide both sides by 8
8x
The 8’s on
the left
cancel out
leaving x
= 16
8
8
x
16  8 = 2
=2
The solution that makes the statement true is x = 2
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
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
Example # 4 (continued)
4x
– 3x + 6
x
+6
= 21
= 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
The 6’s
cancel
out
leaving x
+6
-6
x
= 21
-6
= 15
The solution is x = 15
21 – 6 = 15
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…
Example # 5
12 
3
( x  2)
10
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
10 I F
10 I 3
F
 GJ ( x  2)
G
J
H3 K H3 K10
12
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
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
40 = x + 2
40 – 2 = 38
-2
38 =
-2
x
The solution to the equation is x = 38
The 2’s on
the right
side cancel
out leaving
x
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…
Your Turn
1.
2.
3.
4.
5.
6x – 4(9 –x) = 106
2x + 7 = 15
6 = 14 – 2x
3(x – 2) = 18
12(2 – x) = 6
Your Turn
b g
6.
9
x  3  27
2
7.
4
 2 x  4  48
9
b g
8.
5m – (4m – 1) = -12
9.
55x – 3(9x + 12) = -64
10.
9x – 5(3x – 12) = 30
Your Turn Solutions
1.
2.
3.
4.
5.
14.2
4
4
8
1½
6. 3
7. -52
8. -13
9. -1
10.5
Summary
 A key tool in making learning effective is being
able to summarize what you learned in a
lesson in your own words…
 In this lesson we talked about Solving Multistep equations. Therefore, in your own
words summarize this lesson…be sure to
include key concepts that the lesson covered
as well as any points that are still not clear to
you…
 I will give you credit for doing this
lesson…please see the next slide…
Credit


I will add 25 points as an assignment grade for you
working on this lesson…
To receive the full 25 points you must do the following:
 Have your name, date and period as well a lesson
number as a heading.
 Do each of the your turn problems showing all work
 Have a 1 paragraph summary of the lesson in your own
words

Please be advised – I will not give any credit for work
submitted:
 Without a complete heading
 Without showing work for the your turn problems
 Without a summary in your own words…