Algebra 1
Ch 3.4 – Solving Equations
with Variables on Both Sides
Objective
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Students will solve equations with
variables on both sides of the equation
Before we begin…
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In the last lesson we solved multi-step
equations…
In this lesson we will continue that theme
with another variation of equations…that is,
equations that have variables on both sides
of the equals sign…
In this lesson we will also look at equations
that have many solution or no solutions…
General Rule
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It really doesn’t matter which side you
collect the variable to….however, the
general rule for solving equations with
variables on both sides of the equals sign is
to collect the variables to the side with
the largest variable…
The reason we use the general rule is to
minimize the amount of work with negative
numbers…students are not always proficient
with working with negative numbers….
Example 1
7x + 19 = - 2x + 55
Let’s analyze this equation first…
Deciding where to start….I see that on the left side is 7x and on the
right side is – 2x.
In this instance I will use the general rule and collect the x’s to the left
side of the equation…I do that by working on the right side of the
equation and by adding 2x to both sides of the equation…which looks
like this:
7x + 2x = 9x
7x + 19 = - 2x + 55
+2x
9x + 19 =
+2x
The 2x’s on the
right side cancel
out leaving +55
+ 55
After collecting the variables to the right side I am left with a 2-step
equation. Add/Subtract first and then multiply or divide…
Example #1 (continued)
9x + 19 =
+ 55
In this 2-step equation to undo the +19, subtract 19 from both sides of
the equation
9x + 19
=
- 19
9x
+ 55
-19
=
36
To undo the multiplication, divide both sides by 9
9x
The 9’s
cancel out
leaving x
=
9
x
36
9
=
4
36  9 = 4
Example #2
90 – 9y = 6y
Let’s analyze this equation…
I see that there is a – 9y on the left and a +6y on the right…I’m going to
make the decision to collect the variables to the right side of the equation.
To undo the – 9y, add 9y to both sides of the
equation…it looks like this:
The 9y’s
on the left
cancel
out
leaving
90
90 – 9y = 6y
6y + 9y = 15y
+ 9y +9y
90
= 15y
You are left with a 1 step equation…
Example #2 (continued)
90
= 15y
To undo the multiplication of 15y…divide both sides by 15
90  15 = 6
90
= 15y
15
15
6
The 15’s cancel
out leaving y
= y
The solution that will make the equation true is y = 6
Many or No Solutions
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When working with linear equations you will
come across situations in which there are
many or no solutions to the equation…
Later on in the course when we plot systems
of equations you will find that the many or no
solution equations mean something…more
about that later in the course….
Let’s look at some examples…
Equations with many solutions
3(x + 2) = 3x + 6
First, we have to analyze this equation and decide what we want to do…
Notice, that the distributive property is illustrated on the left side of the
equation…we must take care of that before we do anything else…
3(x + 2) = 3x + 6
Distribute the 3 on the left side of the equation to get:
3x + 6 = 3x + 6
Now you can collect the x’s to either side of the equation…I choose the left
3x + 6 = 3x + 6
The x’s on the
left cancel each
other out
- 3x
-3x
6=
6
The x’s on the
right cancel
each other out
also
Equations with many solutions
6
=
6
What you are left with is a true statement 6 does equal 6
What does the solution mean?
Any value of x will make this a true statement
If you choose any number and
substitute it for x you will always
get a true statement
This type of linear equation is called an identity
Equations with no solutions
x+2=x+4
Again, we should analyze the equation first…
I notice that there is an x on both sides of the equation…It doesn’t
matter which side you collect the variables to….I will collect them
to the left side like this:
x+2=x+4
-x
-x
2=
4
I am left with the statement 2 = 4. This is not a true statement. To
write this correctly you would use the symbol ≠ as follows:
2≠
4
Equations with no solutions
2≠
4
What does it mean?
In this equation, no value of x will make this a true statement
The solution is written as no solution
Solving Complex Equations
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Ok…you have all the information you need to
solve any equation…
The key is to analyze the equation first and make
some decisions…
It’s ok if you don’t get it right the first time…go
back and problem solve to see where you made
your error…
I will walk you through a more complex equation
so that you can see my thought process..
Example #3
4(1 –x) + 3x = -2(x – 1)
Ok…I notice right off that I have the distributive property on both the left
and right side of the equation….
Don’t forget…at the level you are required to be able to recognize
and know how to work with the distributive property….
I must take care of the distributive property first before I do anything else
4(1 –x) + 3x = -2(x – 1)
Don’t forget that
-2 times -1 = +2
4 – 4x + 3x = -2x + 2
Ok…now I see that I have multiple x’s on the left side and -2x on
the right…I’m going to make the decision to combine the x’s on the
left before I collect all the x’s to the same side… My new equation
will look like this…..
Example #3 (continued)
4 – 4x + 3x = -2x + 2
4 – x = -2x + 2
-4x + 3x = -1x.
We write -1x as
just -x
Ok…now I have x’s on both sides of the equation…
I need to get the x’s on the same side…I am going to use the
general rule and collect the x’s to the left side….that’s because
–x is bigger than -2x and I don’t want to have to work with
negative numbers…to undo the -2x add 2x to both sides…
-x + 2x = +1x.
This is written
as just +x
4 – x = -2x + 2
+2x
4+x =
+2x
+2
The 2x’s cancel
out leaving 2
Example #3 (continued)
4+x =
+2
Ok…now all I have to do is isolate the x on the left and I have the
solution….to undo the +4, subtract 4 from both sides of the
equation….like this…
The 4’s on the left
cancel out leaving
just x
4+x =
+2
-4
-4
x =
+2–4=-2
-2
The solution that makes the equation true is x = -2
Reminder: Your answer must always be stated as a positive variable, If you
have a negative variable you have to do something to make it positive!
Comments
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On the next couple of slides are some practice
problems…The answers are on the last slide…
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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.
4x + 27 = 3x
-2m = 16m – 9
12c – 4 = 12c
12p – 7 = - 3p + 8
-7 + 4m = 6m - 5
Your Turn
6.
7.
8.
9.
10.
24 – 6r = (4 – r)
-4(x – 3) = -x
-2(6 – 10n) = 10(2n – 6)
¼ (60 + 16s) = 15 + 4s
¾ (24 – 8b) = 2(5b + 1)
Your Turn Solutions
1.
2.
3.
4.
5.
-27
½
No solution
1
-1
6.
7.
8.
9.
10.
Identity; all real
numbers
4
No solution
Identity; all real
numbers
1
Summary
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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 Equations
with Variables on Both Sides. 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
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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…