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This pdf includes the following topics:-

Review of Order of Operations

Review of Solving Equations

Inequalities Definitions

Solving Inequalities

Examples

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

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

you get the right answer.

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

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

you get the right answer.

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

Addition + − Subtraction

Subtraction − + Addition

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

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

Addition + − Subtraction

Subtraction − + Addition

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

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

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

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

|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

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

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

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

(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

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

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

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

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

(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