# Inequalities: Review of Order of Operations Contributed by: This pdf includes the following topics:-
Review of Order of Operations
Review of Solving Equations
Inequalities Definitions
Solving Inequalities
Examples
1. Faculty of Mathematics Centre for Education in
Waterloo, Ontario N2L 3G1 Mathematics and Computing
Grade 7 and 8 Math Circles
March 5th/6th/7th
Inequalities
See how many of the following equations you can solve.
1. 15 ÷ 3 − 1 6. 3x = 12
3x
=5−1 3
= 12
3
=4 x=4
7. 2x = −8
2. 7 × 3 + 4
2x
= −8
= 21 + 4 2 2
x = −4
= 25
8. −3x = 6
3. 48 ÷ 4 + 4 × 5 −3x 6
= −3
−3
= 12 + 20 x = −2
= 32
9. 6x = 18
6x
4. 48 ÷ (4 + 4) × 5 6
= 18
6
= 48 ÷ 8 × 5 x=3
=6×5
10. 2x + 5 = 15
= 30
2x + 5 − 5 = 15 − 5
2x = 10
5. 11 − 62 ÷ 12 + (4 + 5 × 2) ÷ 7 2x
= 10
= 11 − 62 ÷ 12 + (4 + 10) ÷ 7 2 2
x=5
= 11 − 62 ÷ 12 + 14 ÷ 7
= 11 − 36 ÷ 12 + 14 ÷ 7 11. 3x + 4 = 10
= 11 − 3 + 2 3x + 4 − 4 = 10 − 4
=8+2 3x = 6
3x
= 10 3
= 63
x=2
1
2. 12. 3x + 1 = 2x + 7 13. 2x + 9 = 6x + 1
3x + 1 − 2x = 2x + 7 − 2x 2x + 9 − 6x = 6x + 1 − 6x
x+1=7 −4x + 9 = 1
x+1−1=7−1 −4x + 9 − 9 = 1 − 9
x=6 −4x = −8
−4x
−4
= −8
−4
x=2
Review of Order of Operations
When working on more complicated mathematical expressions that involve multiple opera-
tions and many steps it is very important to do the math in the right order to ensure that
What is the right order?
1. Begin by simplifying everything inside of brackets.
2. Evaluate anything with exponents.
3. Starting from the left, work out all of the multiplication and division, whichever
comes first.
4. Starting from the left, finish by evaluating all of the addition and subtraction,
whichever comes first.
This can be summarized with the acronym BEDMAS:
Brackets Exponents Division Multiplication Addition Subtraction
Note: Remember that division and multiplication are evaluated in the same step, even
though D appears before M in the acronym; division is no more important than multiplica-
tion. Similarly, addition is no more important than subtraction.
2
3. Review of Solving Equations
Steps for Solving:
1. Determine what you are trying to isolate/solve for.
2. Simplify the equation as much as possible by adding and subtracting like terms.
Like terms are terms in a mathematical equation that have the exact same variables;
only their coefficients are different.
You can think of it like adding apples and oranges. If I have 3 apples plus 2 apples
plus 5 oranges plus 1 orange, I actually have 5 apples and 6 oranges. Another, more
mathematical, example:
5 + x + 3y − 2 − y + 2x = 3 + 3x + 2y
3. Isolate the desired variable on one side of the equal sign and everything else on the
other side by performing opposite operations in reverse BEDMAS order. The goal of
isolating a variable, say x, is to obtain the form x = ... or ... = x. Notice that x is
positive with a coefficient of 1. There should be no other xs on the other side of the
equal sign.
Original Operation Opposite Operation
Division ÷ × Multiplication
Multiplication × ÷ Division
Reverse BEDMAS (SAMDEB):
Subtraction Addition Multiplication Division Exponents Brackets
Note: Just as before, addition and subtraction have the same priority; as do multiplication
and division.
When performing opposite operations, what you do to one side of the equation
you must do to the other side of the equation.
3
4. Inequalities Definitions
To understand what equations are trying to tell us, we first must understand the symbols
being used in those equations.
= This symbol means that the first expression is equal to the second expression
< This symbol means that the first expression is less than the second expression
> This symbol means that the first expression is greater than the second expression
In order to know which of these symbols should be used in a specific equation, it can be
helpful to rember one of the following sayings:
• The smaller side goes towards the
smaller number, the bigger side goes
towards the bigger number
• The alligator always wants the
bigger meal
Determine which symbol belongs in the blank for the following equations to be correct.
1. 5 7 4. 3 + 4 6+1
< =
2. 7 7 5. 9 − 5 3+2
= <
3. 7 5 6. 1 + 3 12 − 7
> <
Solving Inequalities
When solving equations with equal signs the expression remains true as long as everything
you do to one side, you also do to the other side. When solving inequalities, there are cer-
tain operations that are safe to preform and have no effect on the direction of the inequality
(which symbol is being used). However, some operations, when preformed on inequalities,
cause the direction of the inequality to switch. If the direction of the inequality switches,
4
5. then > becomes <, or < becomes >. Below is a list of operations categorized by their effect
on the direction of the inequality.
Direction of the Inequality
Stays the Same Switches
Multiplying both sides by a positive number Multiplying both sides by a negative number
Dividing both sides by a positive number Dividing both sides by a negative number
Adding a number to both sides Switching left and right sides
Subtracting a number from both sides
Simplifying a side
Determine what the symbol will be after the operation is preformed.
1. x + 5 > 7 4. 12 > −3x
Subtracting 5 from both sides Switching left and right sides
> <
5. −3x < 12
2. 3x > 6
Dividing both sides by −3
Dividing both sides by 3
>
>
6. x + 5 = 7
3. x − 5 < 7 Subtracting 5 from both sides
Adding 5 to both sides =
< The equal sign never changes when an
operation is applied to both sides.
To solve an inequality you follow the same steps as when solving an equality expression,
with the additional step of paying attention to the direction of the inequality. It is normal
to have x be on the left side in the final expression. When x is on the left side it is easier to
read for meaning as it is of the form x is less than ... or x is greater than ... .
5
6. Solve the following inequalities for x.
1. x + 5 > 7 6. −3 > −3x
x+5−5>7−5 −3x < −3
−3x
x>2 −3
> −3
−3
x>1
2. x − 9 < 7
x−9+9<7+9
7. 1 − x < 7
x < 16
1−x−1<7−1
3. 3 < 2 + x −x < 6
−x 6
2+x>3 −1
> −1
2+x−2>3−2 x > −6
x>1
8. 2x + 5 > 7
4. 3x > 18
3x
2x + 5 − 5 > 7 − 5
3
> 18
3
2x > 2
x>6 2x
2
= 22
5. −4x < 16 x>1
−4x 16
−4
> −4
x > −4
Absolute Values
An absolute value is the distance between a number and zero on a number line. This
means that all absolute values are positive numbers or zero. Absolute values are represented
with the symbols | | surrounding a number. This means that | − 5| is the same as |5| as they
are both 5 units from zero when observed on a number line, just in different directions.
| − 5| = 5 = |5|
Absolute values play the biggest role when dealing with negative numbers and subtraction.
When subtracting two numbers normally, the order in which the numbers are placed matters
as it effects whether your answer is positive or negative. Because both the positive and
negative numbers have the same absolute value, the order of subtraction does not matter
when the absolute value is being considered.
6
7. |7 − 5| = |2| = 2
|5 − 7| = | − 2| = 2
When dealing with the multiplication and division by a negative number we see a similar
outcome. Normally, whether or not the number is negative has an effect on whether or
not the solution is negative, when taking the absolute value, it has no effect. This is again
because whether or not the number is negative the absolute value of the number will be the
positive equivalent.
|2 × 3| = |6| = 6
| − 2 × 3| = | − 6| = 6
|2 × −3| = | − 6| = 6
If the negative sign is outside the boundaries of the absolute value, then we treat the absolute
value as we would a set of brackets, evaluate the inside then apply the outside conditions.
−|7 − 5| = −|2| = −2
−|5 − 7| = −| − 2| = −2
Solving Absolute Value Equations
When solving an equation that contains an absolute value, we have to account for both the
positive and negative values that the absolute value could be. By considering both options,
absolute value equations will typically have two correct answers.
|2x − 1| = 5
2x − 1 = 5 2x − 1 = −5
2x = 6 2x = −4
x=3 x = −2
7
8. The same concept applies when there is an inequality. The expression containing the variable
must be considered with respect to both the negative and positive value. As we discussed
earlier with the direction of inequalities, when the sign has changed on the value we are
comparing the expression to, the direction of inequality also has to be switched.
Less Than
|2x − 1| < 5
−5 < 2x − 1 < 5
−5 + 1 < 2x < 5 + 1
−4 < 2x < 6
−2 < x < 3
Greater Than
|2x − 1| > 5
2x − 1 > 5 or 2x − 1 < −5
2x > 6 or 2x < −4
x > 3 or x < −2
8
9. Solve the following for x.
1. |3x| = 12 2x = −20
3x = 12 3x3
= 12
3
2x
2
= −20
2
x=4 x = −10
or
3. |x + 5| > 7
3x = −12
x+5>7
3x
3
= −12
3
x+5−5>7−5
x = −4
x>2
2. |2x + 5| = 15 or x + 5 > −7
2x + 5 = 15 x + 5 − 5 > −7 − 5
2x + 5 − 5 = 15 − 5 x < −12
2x = 10
2x 4. |x − 9| < 7
2
= 10
2
x=5
−7 < x − 9 < 7
or
−7 + 9 < x − 9 + 9 < 7 + 9
2x + 5 = −15
2 < x < 16
2x + 5 − 5 = −15 − 5
9
10. Problem Set
1. Determine if the following statements are true or false.
(a) 0 is less than −5 (e) −1 is less than | − 3|
False True
(b) 7 is less than −1 (f) |5| is less than |7|
False True
(c) −1 is less than −3 (g) | − 10| is less than | − 9|
False False
(d) −8 is less than −2 (h) 0 is less than | − 1|
True True
2. In a group of five friends:
• Amy is taller than Carla
• Dan is shorter than Eric but taller than Bob
• Eric is shorter than Carla
Who is the shortest?
Bob
Bob < Dan < Eric < Carla < Amy
2014 Cayley (Grade 10) # 9
3. Five children had dinner. Chris ate more than Max. Brandon ate less than Kayla.
Kayla ate less than Max but more than Tanya. Which child ate the second most?
Max
2011 Gauss (Grade 8) # 10
4. In downtown Gaussville, there are three buildings with different heights: The Euclid
(E), The Newton (N) and The Galileo (G). Only one of these statements below is true.
(a) The Newton is not the shortest.
(b) The Euclid is the tallest.
(c) The Galileo is not the tallest.
Order the buildings from shortest to tallest in height.
E, N, G
2012 Gauss (Grade 8) # 22
10
11. 5. Solve the following inequalities for x.
(a) 3x > 12 (g) 9 > 6 + x
3x 12
> 6+x<9
3 3
x>4 6+x−6<9−6
x<3
(b) 2x < −8
2x −8
<
2 2 (h) 2x + 5 > 15
x < −4
2x + 5 − 5 > 15 − 5
(c) 6 < −3x 2x > 10
−3x > 6 2x 10
>
−3x 6 2 2
>
−3 −3 x>5
x < −2
(i) 3x + 4 < 10
(d) 6x < 18
6x 18
< 3x + 4 − 4 < 10 − 4
6 6
x<3 3x < 6
3x 6
<
3 3
(e) x + 7 < 18
x<2
x + 7 − 7 < 18 − 7
x < 11 (j) 3 < 7 − 4x
7 − 4x > 3
(f) x − 1 > 5
7 − 4x − 7 > 3 − 7
x−1+1>5+1
−4x > −4
x>6 x<1
11
12. (k) 3x + 1 > 2x + 7 (l) 2x + 9 < 6x + 1
3x + 1 − 2x > 2x + 7 − 2x 2x + 9 − 6x < 6x + 1 − 6x
x+1>7 −4x + 9 < 1
x+1−1>7−1 −4x + 9 − 9 < 1 − 9
x>6 −4x < −8
−4x −8
<
−4 −4
x>2
6. Solve the following absolute value problems for x.
(a) |3x| = 12 (e) |x + 7| < 18
3x = 12 or 3x = −12
3x
3
= 12
3
or 3x
3
= −12
3 −18 < x + 7 < 18
x = 4 or x = −4 −18 − 7 < x + 7 − 7 < 18 − 7
−25 < x < 11
(b) |2x| = 8
2x = 8 or 2x = −8
2x
2
= 82 or 2x
2
= −8
2
(f) |x − 1| > 5
x = 4 or x = −4
(c) 6 < |3x| x − 1 < −5 or x − 1 > 5
x − 1 + 1 < −5 + 1 or x − 1 + 1 > 5 + 1
|3x| > 6
x < −4 or x > 6
3x < −6 or 3x > 6
3x
3
< −6
3
or 3x
3
> 63
x < −2 or x > 2
(g) 9 > |6 + x|
(d) |6x| > 18
6x < −18 or 6x > 18 |6 + x| < 9
6x
< −18 or 6x > 18 −9 < 6 + x < 9
6 6 6 6
x < −3 or x > 3 −9 − 6 < 6 + x − 6 < 9 − 6
−15 < x < 3
12
13. (h) |2x + 5| > 15 (j) 3 < |7 − 4x|
2x + 5 < −15 or 2x + 5 > 15 |7 − 4x| > 3
2x + 5 − 5 < −15 − 5 or 7 − 4x < −3 or 7 − 4x > 3
2x + 5 − 5 > 15 − 5 7 − 4x − 7 < −3 − 7 or
2x < −20 or 2x > 10 7 − 4x − 7 > 3 − 7
2x
2
< −20
2
or 2x
2
> 10
2 −4x < −10 or −4x > −4
x < −10 or x > 5 −4x
< −10 or −4x > −4
−4 −4 −4 −4
5
(i) |3x + 4| < 10 x > 2 or x < 1
−10 < 3x + 4 < 10
−10 − 4 < 3x + 4 − 4 < 10 − 4
−14 < 3x < 6
−14
3
< 3x
3
< 63
−14
3
7. Based on your knowledge of the inequality symbols what do you think is the meaning
of ≤ and ≥?
≤ means less than or equal to
≥ means greater than or equal to
8. Solve for x.
x x
(a) 3 < 2
(c) 2
+ 9 < 12
x x
2
>3 2
+ 9 − 9 < 12 − 9
x x
2
×2>3×2 2
<3
x
x>6 2
×2<3×2
x+6
x<6
(b) 4 < 3
x+6 (d) −2 < 6x − 2 < 4
3
>4
x+6
3
×3>4×3 −2 + 2 < 6x − 2 + 2 < 4 + 2
x + 6 > 12 0 < 6x < 6
0
x + 6 − 6 > 12 − 6 6
< 6x
6
< 66
x>6 0 13
14. x 2x−5
(e) 2 < 2
+3<5 (g) −1 < 3
<5
2 − 3 < x2 + 3 − 3 < 5 − 3 −1 × 3 < 2x−5 ×3<5×3
3
−1 < x2 < 2 −3 < 2x − 5 < 15
−1 × 2 < x2 × 2 < 2 × 2 −3 + 5 < 2x − 5 + 5 < 15 + 5
−2 < x < 4 2 < 2x < 20
2
(f) −2 < x−7
<1 2
< 2x
2
< 20
2
3
1 < x < 10
−2 × 3 < x−7
3
×3<1×3
−6 < x − 7 < 3 (h) −6 < 3x+6
<6
2
−6 + 7 < x − 7 + 7 < 3 + 7
−6 × 2 < 3x+6
2
×2<6×2
1 < x < 10
−12 < 3x + 6 < 12
−12 − 6 < 3x + 6 − 6 < 12 − 6
−18 < 3x < 6
−18
3
< 3x
3
< 63
−6 < x < 2
14