Section 2.3
Pre-Activity
Preparation
Isolating a Variable
How long? How fast? How far?
Base equation: d = rt (d = distance, r = rate, t = time)
You are on a road trip with friends. It is your turn to drive the next 3
hours. You have your cruise control set at 70 mph. The state line is
only 220 miles away. Will you still be behind the wheel when you see
the “Welcome!” signs?
d = rt
d = (3)(70) = 210 miles (sorry, not this time)
The next day, you and your friends all agree that stopping for dinner
at a famous steak house on the way to the hotel sounds great. It is
4:00 p.m. and the reservations you have made are for 7:30 p.m. The restaurant is 200 miles away. How fast
will you have to drive to make it there on time?
r=
d
t
r=
(200)
. 57 miles per hour (good thing the speed limit is 60 miles per hour!)
(3.5)
Dinner was great, but you only have 2 hours left to make it to your hotel before midnight. Because of
congestion and road construction, the speed of traffic has dropped to 40 miles per hour. The hotel is 85 miles
away. Will you make it there in time, or should you phone the hotel to let them know you will be late?
t=
d
r
t=
(85)
= 2.125 hours (you will need to pull over and make a phone call)
(40)
The same base equation was used in each case, but it was manipulated so that the variable we needed to solve
for was isolated on one side of the equals sign.
Learning Objectives
• Expanding the tool set for manipulating and transforming equations
• Validate that transformations are correct
• Learn to isolate a variable
Terminology
Previously Used
coefficient
New Terms
to
Learn
isolate a variable
exponent
like terms
terms
variable
127
Chapter 2 — Solving Equations
128
Building Mathematical Language
Simplifying Expressions
Expressions are in their simplest form when there are no parentheses to remove and when all like terms
have been combined. Simplify expressions by applying the Distributive Property to remove parentheses
and then use the Methodology for Combining Like Terms from Chapter 1.
???
= 3x + 7y – 1( x + 3y) Why can we do this?
= 3x + 7y – x – 3y Distributive Property
= 3x – x + 7y – 3y
Commutative Property
3x + 7y – (x + 3y) = (3x – x) + (7y – 3y)
Associative Property
= (3 – 1)x + (7 – 3)y
Distributive Property
= 2x + 4y
???
A negative sign in front of parentheses within an expression
indicates subtraction. Subtraction is addition of the opposite of
a number. Inserting the coefficient of one helps highlight the distribution of the negative
sign to each term in the expression. Therefore, when subtracting an expression within
parentheses, multiply the expression by negative one.
Why can we do this?
Isolating a Variable
Variables can appear on both sides of the equal sign such as in the equation:
3x + 5 = x – 7
Each side of the equation is simplified, but the variable is present on both sides. Isolating the variable
is the process of applying the Addition Property of Equality to get the variable on just one side of the equal
sign so that the equation can be solved.
Solving equations is a process of applying the principles and properties of equations to rewrite the equation
in successive equivalent forms until the solution is reached. The methodology for solving equations is
presented in Section 2.4. Isolating the variable is step 3 of the methodology. The methodology for isolating
a variable has some of the same steps used in solving an equation.
Section 2.3 — Isolating a Variable
Properties
129
Principles
and
In order to successfully work with the equations in this section, you should remember
and be able to use the Distributive Property, which has already been introduced.
Distributive Property Review
1. Distribute multiplication over addition (remove parentheses):
5(x + y) = 5x + 5y
–2(x + y) = –2x – 2y
–3(x – y) = –3x + 3y
2. Remove common factors from two or more terms:
–x – y = –1(x + y) = – (x + y)
3x + 6y = 3(x + 2y)
–2a + 6b = –2(a – 3b)
3. Combine like terms:
4x + 7x = x(4 + 7) = (4 + 7)x = 11x
5y3 – 9y3 = (5 – 9)y3 = –4y3
Models
Model 1
Simplify the expression: 3(x – 5y) – 4(2x – 3y)
3(x – 5y) – 4(2x – 3y)
= 3x – 15y – 8x + 12y
Distributive Property
= 3x – 8x – 15y + 12y
Commutative Property
= (3x – 8x) + (–15y + 12y)
Associative Property
= (3 – 8)x + (–15 + 12)y
Distributive Property
= –5x + (–3)y
Answer: = –5x – 3y
To validate an expression, you can select a value for the variable and substitute it into both the
original expression and your answer (the simplified version). Compare those values. Note that
this method will only work when validating an expression, not an equation. To validate the work
above, let x = 2 and y = 3:
original expression
3( x - 5 y ) - 4(2 x - 3 y )
3((2) - 5(3)) - 4(2(2) - 3(3))
3(2 -15) - 4(4 - 9)
3(-13) - 4(-5)
-39 + 20 = -19
simplified expression
-5 x - 3 y
-5(2) - 3(3)
-10 - 9 = -19 
Chapter 2 — Solving Equations
130
Model 2
Simplify the expression: 3x2 – 5x + 4 – (2x2 + 7x – 9)
3x2 – 5x + 4 – (2x2 + 7x – 9)
= 3x2 – 5x + 4 – 1(2x2 + 7x – 9)
(Remember we can insert a
coefficient of 1 and distribute
the –1 in the next step.)
= 3x2 – 5x + 4 – 2x2 – 7x + 9
Distributive Property
= 3x2 – 2x2 – 5x – 7x + 4 + 9
Associative Property
Answer: = x2 – 12x + 13
Let x = 3:
Distributive Property
original expression
simplified expression
3 x 2 - 5 x + 4 - (2 x 2 + 7 x - 9)
3(3) 2 - 5(3) + 4 - (2(3) 2 + 7(3) - 9)
3(9) - 5(3) + 4 - (2(9) + 7(3) - 9)
27 -15 + 4 - (18 + 21 - 9)
27 -15 + 4 -18 - 21 + 9
= -14
x 2 -12 x +13
(3) 2 -12(3) +13
9 - 36 +13
= -14 
Methodologies
Isolating a Variable
►
►
Example 1: Isolate the variable in the equation: –3(4x – 6) = 2(x + 2)
Example 2: Isolate the variable in the equation: 2(3x + 7) = 3(x + 5) + 3
Steps in the Methodology
Step 1
Observe the
equation,
noting where
the variables
and constants
occur.
The equation has variables
on both sides of the equal
sign. Each side must be
simplified by applying the
Distributive Property.
Special
Case:
For equations
that contain only
variables, treat all
variables except the
one being isolated
as constants.
See Model 4.
Example 1
The variables are
highlighted and the
constant factors to be
distributed are shown.
variables
–3(4x – 6) = 2(x + 2)
constant factors
Try It!
Example 2
Section 2.3 — Isolating a Variable
131
Steps in the Methodology
Step 2
Choose the
variable to
isolate.
Step 3
Remove
parentheses
and combine
like terms.
Step 4
Move or
collect the
variables to
one side of
the equal
sign; combine
like terms.
Step 5
Move or
collect the
constants to
one side of
the equal sign
and combine
like terms.
Step 6
Validate.
Example 1
In this case, there is only one variable, x, so there is
no choice. If the equation has more than one variable,
after choosing the variable, treat all other variables like
constants.
In simplifying expressions
with the Order of Operations,
parentheses are evaluated
first; for equations, first
remove parentheses by
applying the Distributive
Property.
Use the Addition Property
of Equality to add like
quantities to both sides
so that the term you want
to move adds to zero or
“cancels out” on one side
and adds to the other side.
Recall that adding the
opposite is the same as
subtracting.
–3(4x – 6) = 2(x + 2)
–12x + 18 = 2x + 4
–12x – 2x + 18 = 2x – 2x + 4
–14x + 18 = 4
You might choose, instead,
to align the terms like this:
–12x + 18 = 2x + 4
= –2x
–2x
–14x + 18 = 4
Do the same for the
constants.
–14x + 18 –18 = 4 –18
The equation is ready to
solve.
You can align the terms like
this:
–14x = – 14
–14x + 18 = 4
–18 = –18
= –14
–14x
Make sure the equation looks
like:
(coefficient)(chosen variable)
=constant (or simplified
expression).
Keep track of the arithmetic,
especially the arithmetic of
signed numbers.
Work backwards through
Steps 5, 4, 3, and 2.
Alternatively, rework the
problem after a short break.
–14x
+18
–14x + 18
2x
–12x + 18
–3(4x – 6)
=
=
=
=
=
=
–14
+18
4
2x
2x + 4
2(x + 2)
–3(4x – 6) = 2(x + 2) 
Example 2
Chapter 2 — Solving Equations
132
Model 3
Isolate the variable in the equation: 3x – (x + 5) = 7 – 2(x – 4)
Step 1
Step 2
Step 3
Step 4
Step 5
Step 6
Variables are on both sides of the equation, the Distributive Property is needed to remove
parentheses, and constants are on both sides.
Variable is x.
Remove parentheses and combine like terms. 3x – 1(x + 5) = 7 – 2(x – 4)
3x – x – 5 = 7 – 2x + 8
2x – 5 = 15 – 2x
Move variables to one side and combine like 2x – 5 + 2x = 15 – 2x + 2x
terms.
4x – 5 = 15
Collect constants and combine like terms.
4x – 5 + 5 = 15 + 5
Answer: 4x = 20
Validate.
4x
=
20
–2x – 5 = –2x – 5
2x – 5 = –2x + 15
Model 4
Isolate the variable x: y = mx + b
Step 1
Step 2
Step 3
Step 4
Step 5
The equation has all variables and no constants; therefore, treat all but the chosen
variable like constants.
Choose variable x.
No parentheses, omit this step.
x is on one side of the equation.
Collect constants.
y – b = mx + b – b
Answer: y – b = mx or mx = y – b
???
Why can we do this?
Step 6
y – b = mx
+b=+b
y = mx + b
Validate.
Alternatively, validate by choosing values for y, m, and b: Let y = 7, m = 3, b = 4
???
Why can we do this?
Substitute values into
the original equation:
Substitute values in our final
equation (from Step 5):
7 = 3x + 4
7 - 4 = 3x
3x = 3
x =1
3x = 7 - 4
3x = 3
x =1 
compare
Remember that the Symmetric Property of
Equations states that if a = b, then b = a.
Section 2.3 — Isolating a Variable
133
Addressing Common Errors
Issue
Multiplying
instead of
adding to
combine like
terms
Subtracting
only the first
term of an
expression
within
parentheses
Leaving off
a term that
does not
combine
with any
other term
(forgetting a
term)
Incorrect
Process
x + 2x = 3x2
5 – (2x + 3) =
=5 – 2x + 3
= –2x + 8
Resolution
Understand that
the Distributive
Property is used
for combining
like terms. Use
parentheses to
factor out the like
variable or constants
and then combine
the like terms within
parentheses.
Insert a –1 in place
of the negative
sign to remind you
that both terms
are subtracted.
Distribute the –1
to each term within
parentheses.
Scan through the
3x + 2y – 1 + x – 3y = expression and
determine how
= 4x – y
many different
terms there are,
including variables
and constants.
Highlight, underline,
circle or mark off
each term as it is
combined. Every
term has to be
accounted for, even
if it adds up to zero.
Correct
Process
x + 2x =
= (1 + 2)x
= 3x
Validation
Another way
to validate an
expression is to
substitute a value
and compare the
answer with the
original expression:
Let x = 5.
?
(5) + (2 × 5) = 3 × 52
?
5 + 10 = 3 × 25
15 ≠ 75
But:
?
5 + (2 × 5) = 3 × 5
?
5 + 10 = 15
15 = 15 
5 – (2x + 3) =
= 5 – 1(2x + 3)
= 5 + (–1)2x + (–1)3
= 5 – 2x – 3
= –2x + 2
Let x = 10
is –2(10) +2
the same as
5 – [(2 × 10) + 3)]?
–20 + 2 = –18 and
5 – 23 = –18 
Check the terms of
3x + 2y – 1 + x – 3y = your solution against
x variables: 3x + x the original problem.
y variables: 2y –3y Make sure that all
variables are present
constants: –1
(in this case, x and y),
= 3x + x + 2y – 3y – 1 as well as constants
(terms without a
= 4x – y – 1
variable).
x, y, constant 
Chapter 2 — Solving Equations
134
Issue
Adding a
term when
it should
have been
subtracted
or the
reverse
(sign error)
Incorrect
Process
2x – 3 = 12 + x
3x – 3 = 12
3x = 9
Correct
Process
Resolution
Write in the step of
showing the addition
process. The term to
move must add out
to zero (cancel out)
on one side.
Validation
2x – 3 = 12 + x
?
2 × 15 – 3 = 12 + 15
2x – x – 3 = 12 + x – x
?
30 – 3 = 27
x – 3 = 12
17 = 17 
x – 3 + 3 = 12 + 3
x = 15
Not
distributing
to all terms
–2(x + 1 – y) = 3
Dropping
a negative
sign
5 – 2x = 7
–2x + 1 + 2y = 3
–5 + 5 – 2x = 7 – 5
Understand that the
process distributes
the multiplication
over addition; each
addition term must
be multiplied by
the factor being
distributed.
Use parentheses to
keep your signs and
terms together.
2x = 2
–2 (x + 1 – y) = 3
–2x –2 + 2y = 3
Even though the
sign in front of the
2x does indicate
subtraction, once
the 5 and –5 add
out, that subtraction
sign still stands.
Changing it to
addition of –2x will
help you remember.
–2x
= 2
+5
+5
5 + (–2x) = 7
5 – 2x = 7 
5 – 2x = 7
–5 + 5 + (–2x) = 7 – 5
–2x = 2
Preparation Inventory
Before proceeding, you should have an understanding of each of the following:
Use of the Distributive Property to combine like terms
Use of the Distributive Property to remove parentheses
Use of the Addition Property of Equality to isolate a variable
Section 2.3
Activity
Isolating a Variable
Performance Criteria
• Application of the Distributive Property
– each term is included or accounted for
– terms are combined as needed
– accuracy of calculations
• Isolate a variable in preparation for solving an
equation
– appropriate variable is isolated
– steps are validated
Critical Thinking Questions
1. What does it mean to isolate a variable?
2. If you are given the formula, A = lw where A = area, l = length, and w = width, provide an example of three
ways in which you can make use of isolating a variable. (Hint: review the scenarios in the introduction to
this section.)
3. How do you ensure that you have isolated a variable correctly?
135
136
Chapter 2 — Solving Equations
4. What can you do to ensure that your signs are correct?
5. When validating by substituting a value for the variables, how do you know what value to choose?
6. When you distribute across parentheses, how many terms should you end up with?
7. What properties are used for isolating a variable?
Section 2.3 — Isolating a Variable
Tips
for
137
Success
• Experts often do more than one step at a time when isolating a variable. Work until you can combine the
two steps of adding variables and constants into the same step.
• Combine like terms whenever you can. The result is a simpler equation with fewer symbols and thus fewer
chances for error.
Demonstrate Your Understanding
1. Use the Distributive Property to simplify the following:
Problem
a)
3(x + y)
b)
–2(x + y)
c)
–2(x – y)
d)
5(x2 – 2y)
Worked Solution
Validation
2. Combine like terms to simplify the following expressions:
Problem
a)
7x2 + 5x2
b)
3x2 + 2 + 4x2 + 1
c)
6x2 + 7x + 3x + x2
Worked Solution
Validation
Chapter 2 — Solving Equations
138
Problem
d)
x + 2 + 4 + x + 3x + 2 + x + 7
e)
–3x + 5x – 2 – x – 4x + x
Worked Solution
Validation
3. Use the Distributive Property first, then combine like terms to simplify the following:
Problem
a)
4(n + 3) – 2(n – 5)
b)
–2(3x + 6y – 1) + 5(x – 3y)
c)
2(x – 1) + 4(x + 7)
d)
–4(x2 + 2) – 2(x – 7)
e)
–2(6x – 1) + 5(x – 2)
f)
(x + 2) – (x – 2)
Worked Solution
Validation
Section 2.3 — Isolating a Variable
139
4. Isolate the variable for the following equations. Do not solve.
Problem
a)
9 – 5x = –12
b)
x + 4 = 2x + 19
c)
7x – 3 = x
d)
2(3x + 5) = 3x – 35
e)
–(4x – 17) = 3(x + 3)
f)
–2(x + 1) – 4 = 7x – 3(x + 4)
g)
2 – (3x + 7) = 4(x + 3) + 3
h)
2(2x + 5) = x – 15
i)
–3(4x – 7) = 3(x + 6)
Answer
Validation
Chapter 2 — Solving Equations
140
5. Isolate for the given variable:
Problem
a)
Isolate for x:
5x + 2y = 0
b)
Isolate for y:
5x + 2y = 0
c)
Isolate for C:
F = 1.8C + 32
d)
Isolate for x:
2x + 3y = x – 7y
e)
Isolate for y:
2x + 3y = x – 7y
Worked Solution
Validation
Section 2.3 — Isolating a Variable
Identify
and
Correct
141
the
Errors
In the second column, identify the error(s) in the worked solution or validate its answer. If the worked solution
is incorrect, solve the problem correctly in the third column and validate your answer.
Worked Solution
1)
Combine like terms:
2x2 – 5y + 7 – y + 3x2 – 3y
2x2 + 3x2 – 5y – y – 3y
5x2 – 9y
2)
Isolate x:
3x – 2(x + 1) = 7
3x – 2x + 1 = 7
x+1=7
x=6
3)
Isolate t:
–5(t – 2) = 3(t – 1)
5t + 10 = 3t – 3
–3t
–3t
2t + 10 = –3
–10 = –10
2t = –13
4)
Isolate x:
y = –m(x + 7)
y = –mx – 7m
y + 7m = –mx
Identify Errors
or Validate
Correct Process
Validation
Chapter 2 — Solving Equations
142
Worked Solution
5)
Simplify and combine like
terms:
2(x –1) + 5x
2x – 2 + 5x
7x2 – 2
Identify Errors
or Validate
Correct Process
Validation