CHAPTER 6
Linear
Equations and
Inequalities
6.1 – SOLVING
EQUATIONS BY USING
INVERSE OPERATIONS
Chapter 6
FIXING A FLAT TIRE
Listed are the steps to remove a flat tire. How can you use those
steps to figure out how to put on a spare?
EQUATIONS
TAKE OUT A SPARE SHEET OF PAPER
• Write x = ___. Pick any number you want to go
into the blank.
• Multiply or divide each side by another number
of your choice.
• Now add or subtract something from each side.
• This is your final equation.
• Trade equations with the person next to you, and
try to figure out what number the other person
started with.
What are some of the strategies that you used?
INVERSE OPERATIONS
• When we are solving linear equations, we use
inverse operations, which means that we do the
opposite of what’s in the equation.
• What’s the opposite of adding (+)?
• What’s the opposite of subtracting (–)?
• What’s the opposite of multiplication (×)?
• What’s the opposite of division (÷)?
• By using inverse operations we can cancel out what
is in the equation.
• When we do something to one side of the equation,
we need to do it to the other side as well. Think of it
as a scale, and you need to keep it balanced.
EXAMPLE
a) 3x = –3.6
 x = –3.6 ÷ 3
 x = –1.2
Verify: (3)(–1.2) = –3.6  it works
b)
x ÷ 4 = 1.5
 x = 4(1.5)
x=6
Verify: 6 ÷ 4 = 1.5  it works
EXAMPLE
a) How do we represent 7 percent?
 7 percent = 7/100 = 0.07
b) (0.07)(810) = 56.7
So, the solution is correct.
What does “of” mean in an equation
 it means multiply, so to find
7 percent of a number we need
to multiply.
 0.07x = 56.7
 0.07x ÷ 0.07 = 56.7 ÷ 0.07
 x = 810
TRY IT
L = 3.7
P = 13.2
W=?
a) P = L + L + W + W
 P = 2L + 2W
b) 13.2 = 7.4 + 2W
 2W + 7.4 – 7.4 = 13.2 – 7.4
 2W = 5.8
 2W ÷ 2 = 5.8 ÷ 2
 W = 2.9 cm
c) 7.4 + 2(2.9) = 7.4 + 5.8 = 13.2
 13.2 = 2(3.7) + 2W
 13.2 = 7.4 + 2W
The solution works.
TRANSLATING ENGLISH INTO ALGEBRA
In math, one of the most important skills is
understanding how to turn what we see in word
problems into equations. Write each phrase as an
algebraic expression to practice for word problems.
PG. 272-274, # 8, 9, 10,
13, 16, 18, 21, 24.
Independent
Practice
6.2 – SOLVING
EQUATIONS BY USING
BALANCE STRATEGIES
Chapter 6
EXAMPLE: BALANCE MODEL
Solve:
6x + 2 = 10 + 4x
Algebraically:
6x + 2 = 10 + 4x
 6x + 2 – 4x = 10 + 4x – 4x
 2x + 2 = 10
 2x + 2 – 2 = 10 – 2
 2x = 8
 2x ÷ 2 = 8 ÷ 2
x = 4
EXAMPLE
Solve:
122
=3
r
122
=3
r
122
Þ
´ r = 3´ r
r
Þ122 = 3r
3r 122
=
3
3
122
Þr =
= 40.6
3
Þ
A cell phone company offers two plans.
Plan A: 120 free minutes, $0.75 per additional minute
Plan B: 30 free minutes, $0.25 per additional minute
Which time for calls will result in the same cost for both plans?
a) Model the problem with an equation.
b) Solve the problem.
a) Let t be time, in minutes.
b) 0.75(t – 120) = 0.25(t – 30)
Plan A: You will pay for every minute
after 120 minutes. So, you will pay for
(t – 120) minutes.
Cost of Plan A = 0.75(t – 120)
 0.75(t – 120) ÷ 0.25 = 0.25(t – 30) ÷ 0.25
 3(t – 120) = t – 30
 3t – 360 = t – 30
 3t – 360 + 360 = t – 30 + 360
 3t = t + 330
 3t – t = t + 330 – t
They will be the
 2t = 330
same cost if
 2t ÷ 2 = 330 ÷ 2
there are 165
 t = 165 minutes
minutes of calls.
Similarly,
Cost of Plan B = 0.25(t – 30)
 0.75(t – 120) = 0.25(t – 30)
MAGIC SQUARES ACTIVIT Y
Answer all of the questions on the Magic
Squares Activity sheet to the best of your ability.
P. 280-283, # 4, 6, 8, 11,
13, 18, 19, 20, 21, 23
Independent
Practice
6.3 – INTRODUCTION TO
LINEAR INEQUALITIES
6.4 – SOLVING LINEAR
INEQUALITIES BY USING
ADDITION AND
SUBTRACTION
Chapter 6
INEQUALITIES
What is an inequality?
We use inequalities to model a situation that can be described by a range of
numbers instead of a single number.
<
Less than
>
Greater than
≤
Less than or equal to
≥
Greater than or equal to
t ≤ 30
TRY IT
EXAMPLE
We want to know the numbers that are greater than or equal to –4.
a)
b)
c)
d)
e)
–8 is lower than –4 (the more negative, the smaller a number is), so No.
–3.5 is a bit higher than –4, so Yes.
–4 is equal to –4, so Yes.
–4.5 is lower than –4, so No.
0 is higher than –4, so Yes.
INEQUALITIES AND NUMBER LINES
We can represent inequalities using number lines. It’s important to
differentiate between “greater/less than” and “greater/less than or equal to.”
Example: a < 3
Example: b ≤ –5
When it’s greater/less than, you leave the circle open (not
coloured in). When it’s greater/less than or equal to, you need
to remember to colour in the circle.
TRY IT
a)
b)
c)
d)
SOLVING INEQUALITIES
Inequalities are basically just equations, with a different symbol in
the place of the equals sign. We can represent inequalities with
balances, like we can regular equations.
How can represent this using an inequality?
7 grams < 9 grams
• What would happen if we added two grams to each side?
• What if we subtracted two grams from each side?
EXAMPLE
Solve: 6 ≤ x – 4.5
6 ≤ x – 4.5
 6 + 4.5 ≤ x – 4.5 + 4.5
 10.5 ≤ x
 x ≥ 10.5
Try it: 1 > r + 9
1>r+9
1–9>r+9–9
 –8 > r
 r < –8
EXAMPLE
Jake plans to board his dog while he is away on vacation.
• Boarding house A charges $90 plus $5 per day.
• Boarding house B charges $100 plus $4 per day.
For how many days must Jake board his dog for boarding house A to be less expensive
than boarding house B?
a) Choose a variable and write an equality that can be used to solve the problem.
b) Solve and graph the problem.
a) Let d be the number of days.
 Boarding house A = 90 + 5d
 Boarding house B = 100 + 4d
We want:
Boarding house A < Boarding house B
 90 + 5d < 100 + 4d
b) 90 + 5d < 100 + 4d
 90 + 5d – 90 < 100 + 4d – 90
 5d < 10 + 4d
 5d – 4d < 10 + 4d – 4d
 d < 10
For boarding house A to be less expensive
than B, he needs to board his dog for less
than 10 days.
PG. 292-293, # 3, 8, 12,
13, 14, 15.
PG. 298-300, # 9, 10,
13, 14, 15, 16.
Independent
Practice
6.5 – SOLVING LINEAR
INEQUALITIES BY USING
MULTIPLICATION AND
DIVISION
Chapter 6
PATTERN HANDOUT
Fill in the blanks with the appropriate inequality symbols.
From the patterns, what can we say about multiplying and dividing while solving
inequalities?
The properties of inequalities:
When each side of an inequalities is multiplied or divided
by a positive number, the resulting inequality is still true.
When each side of an inequality is multiplied or divided
by a negative number, the inequality sign must be
reversed for the inequality to remain true.
EXAMPLE
Solve:
a)
a)
b)
b)
7a -21
Þ
<
7
7
y
Þ
´ -4 < -3 ´ -4
-4
Þ a < -3
Þ y < 12
EXAMPLE
A super-slide charges $1.25 to rent a mat and $0.75 per ride. Haru as $10.25. How
many rides can Haru go on?
a) Choose a variable, then write an inequality to solve this problem.
b) Solve the problem and graph the solution.
a) Let r be the number of rides.
The total cost has to be less than or equal
to $10.25.
 1.25 + 0.75r ≤ 10.25
b) 1.25 + 0.75r ≤ 10.25
 1.25 + 0.75r – 1.25 ≤ 10.25 – 1.25
 0.75r ≤ 9
 0.75r ÷ 0.75 ≤ 9 ÷ 0.75
 r ≤ 12
Haru has to go on less than or equal to
12 rides.
Inequalities Bingo
PG. 304-306, # 9, 10,
11, 15, 16, 17, 18.
Independent
Practice