Download Ch 3.3 Solving Multi-Step Equations

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
12
10 I F
10 I 3
F
 GJ ( x  2)
G
J
H3 K H3 K10
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
Your Turn
Turn to pg. 148 in your book and we will
practice on some problems together.
#1,2,4,7,8
Assignment
Ch 3.3 pg. 148-149
#10-44 Even, #51