Warm Up Compare. Write, or =. 1. −3 2 3. 2. 6.5 6.3 < > > 4. 0.25= Tell whether the inequality x < 5 is true or false for the following values of x. 5.

Warm UpCompare. Write <, >, or =. 1. −3 2 3.

2. 6.5 6.3 < >

> 4. 0.25 =

Tell whether the inequality x < 5 is true or false for the following values of x.

5. x = –10 T 6. x = 5 F

7. x = 4.99 T 8. x = T

An inequality is a statement that two quantities are not equal. The quantities are compared by using the following signs:

≤A ≤ B

A is less than or

equal to B.

<A < B

A is lessthan B.

>A > B

A is greaterthan B.

≥A ≥ B

A is greaterthan or

equal to B.

≠A ≠ B

A is notequal to B.

A solution of an inequality is any value that makes the inequality true.

An inequality like 3 + x < 9 has too many solutions to list. You can use a graph on a number line to show all the solutions.

The solutions are shaded and an arrow shows that the solutions continue past those shown on the graph. To show that an endpoint is a solution, draw a solid circle at the number. To show an endpoint is not a solution, draw an empty circle.

Example 2: Graphing Inequalities

Graph each inequality.

A. m ≥

0 1– 2 3 3

Draw a solid circle at .

Shade all the numbers

greater than and draw an

arrow pointing to the right.B. t < 5(–1 + 3)

t < 5(–1 + 3)

t < 5(2)

t < 10

–4 –2 0 2 4 6 8 10 12 –6 –8

Simplify.

Draw an empty circle at 10.Shade all the numbers less than 10 and draw an arrow pointing to the left.

Graph each inequality.Check It Out! Example 2

a. c > 2.5 Draw an empty circle at 2.5.

Shade in all the numbers greater than 2.5 and draw an arrow pointing to the right.

b. 22 – 4 ≥ w 22 – 4 ≥ w

4 – 4 ≥ w0 ≥ w

–4 –3 –2 –1 0 1 2 3 4 5 6

Draw a solid circle at 0.

Shade in all numbers less than 0 and draw an arrow pointing to the left.

c. m ≤ –3 Draw a solid circle at –3.

Shade in all numbers less than –3 and draw an arrow pointing to the left.

–4 –3 –2 –1 0 1 2 3 4 5 6

–4 –2 0 2 4 6 8 10 12 –6 –8

−3

2.5

Example 3: Writing an Inequality from a Graph

Write the inequality shown by each graph.

Use any variable. The arrow points to the left, so use either < or ≤. The empty circle at 2 means that 2 is not a solution, so use <.

x < 2

Use any variable. The arrow points to the right, so use either > or ≥. The solid circle at –0.5 means that –0.5 is a solution, so use ≥.

x ≥ –0.5

Write the inequality shown by the graph.

Check It Out! Example 3

Use any variable. The arrow points to the left, so use either < or ≤. The empty circle at 2.5 means that 2.5 is not a solution, so use so use <.

x < 2.5

Reading Math

“No more than” means “less than or equal to.”

“At least” means “greater than or equal to”.

Example 4: ApplicationRay’s dad told him not to turn on the air conditioner unless the temperature is at least 85°F. Define a variable and write an inequality for the temperatures at which Ray can turn on the air conditioner. Graph the solutions.Let t represent the temperatures at which Ray can turn on the air conditioner.

75 80 85 9070

Turn on the AC when temperature is at least 85°F

t ≥ 85

Draw a solid circle at 85. Shade all numbers greater than 85 and draw an arrow pointing to the right.

t 85

A store’s employees earn at least $8.50 per hour. Define a variable and write an inequality for the amount the employees may earn per hour. Graph the solutions.


Let w represent an employee’s wages.

An employee earns at least $8.50

w ≥ 8.50

4 6 8 10 12−2 0 2 14 16 18

8.5w ≥ 8.5

Lesson Quiz: Part I1. Describe the solutions of 7 < x + 4.

all real numbers greater than 3

2. Graph h ≥ –4.75

–5 –4.75 –4.5

Write the inequality shown by each graph.

3. x ≥ 3

4. x < –5.5

Lesson Quiz: Part II

5. A cell phone plan offers free minutes for no more than 250 minutes per month. Define a variable and write an inequality for the possible number of free minutes. Graph the solution.

0 ≤ m ≤ 250

0 250

Let m = number of minutes

Solving one-step inequalities is much like solving one-step equations. To solve an inequality, you need to isolate the variable using the properties of inequality and inverse operations.

Example 1A: Using Addition and Subtraction to SolveInequalities

Solve the inequality and graph the solutions.

x + 12 < 20

x + 12 < 20

–12 –12

x + 0 < 8

x < 8

Since 12 is added to x, subtract 12 from both sides to undo the addition.

–10 –8 –6 –4 –2 0 2 4 6 8 10

Draw an empty circle at 8.

Shade all numbers less than 8 and draw an arrow pointing to the left.

d – 5 > –7

Since 5 is subtracted from d, add 5 to both sides to undo the subtraction.

Draw an empty circle at –2.

Shade all numbers greater than –2 and draw an arrow pointing to the right.

+5 +5

d + 0 > –2

d > –2

d – 5 > –7

Example 1B: Using Addition and Subtraction to SolveInequalities


–10 –8 –6 –4 –2 0 2 4 6 8 10

Example 1C: Using Addition and Subtraction to SolveInequalities


0.9 ≥ n – 0.3

Since 0.3 is subtracted from n, add 0.3 to both sides to undo the subtraction.

Draw a solid circle at 1.2.

Shade all numbers less than 1.2 and draw an arrow pointing to the left.

0 1 2

+0.3 +0.3

1.2 ≥ n – 0

1.2 ≥ n

0.9 ≥ n – 0.3

1.2

a. s + 1 ≤ 10


–1– 1

s + 0 ≤ 9

s ≤ 9

Since 1 is added to s, subtract 1 from both sides to undo the addition.

b. > –3 + t

Since –3 is added to t, add 3 to both sides to undo the addition.

Solve each inequality and graph the solutions.

s + 1 ≤ 10

> –3 + t

+3 +3

> 0 + t

t <

9

–10 –8 –6 –4 –2 0 2 4 6 8 10

–10 –8 –6 –4 –2 0 2 4 6 8 10

q – 3.5 < 7.5

+3.5 +3.5

q – 0 < 11

q < 11

Since 3.5 is subtracted from q, add 3.5 to both sides to undo the subtraction.

Check It Out! Example 1c


q – 3.5 < 7.5

–7 –5 –3 –1 1 3 5 7 9 11 13

Since there can be an infinite number of solutions to an inequality, it is not possible to check all the solutions. You can check the endpoint and the direction of the inequality symbol.

The solutions of x + 9 < 15 are given by x < 6.

Lesson Quiz: Part I


1. 13 < x + 7

x > 6

2. –6 + h ≥ 15

h ≥ 21

3. 6.7 + y ≤ –2.1

y ≤ –8.8


4. A certain restaurant has room for 120 customers. On one night, there are 72 customers dining. Write and solve an inequality to show how many more people can eat at the restaurant.

x + 72 ≤ 120; x ≤ 48, where x is a natural number

Example 1A: Multiplying or Dividing by a Positive Number


7x > –42

7x > –42

>

1x > –6

Since x is multiplied by 7, divide both sides by 7 to undo the multiplication.

x > –6

–10 –8 –6 –4 –2 0 2 4 6 8 10

3(2.4) ≤ 3

7.2 ≤ m (or m ≥ 7.2)

Since m is divided by 3, multiply both sides by 3 to undo the division.

0 2 4 6 8 10 12 14 16 18 20

Example 1B: Multiplying or Dividing by a Positive Number


r < 16

0 2 4 6 8 10 12 14 16 18 20

Since r is multiplied by ,

multiply both sides by the

reciprocal of .

Example 1C: Multiplying or Dividing by a Positive Number


Check It Out! Example 1a


4k > 24

k > 6

0 2 4 6 8 10 12 16 18 2014

Since k is multiplied by 4, divide both sides by 4.

–50 ≥ 5q

–10 ≥ q

Since q is multiplied by 5, divide both sides by 5.

Check It Out! Example 1b


5–5 0–10–15 15

g > 36

Since g is multiplied by ,

multiply both sides by the

reciprocal of .

36

25 30 3520 4015



If you multiply or divide both sides of an inequality by a negative number, the resulting inequality is not a true statement. You need to reverse the inequality symbol to make the statement true.

This means there is another set of properties of inequality for multiplying or dividing by a negative number.

Caution!

Do not change the direction of the inequality symbol just because you see a negative sign. For example, you do not change the symbol when solving 4x < –24.

Example 2A: Multiplying or Dividing by a Negative Number


–12x > 84

x < –7

Since x is multiplied by –12, divide both sides by –12. Change > to <.

–10 –8 –6 –4 –2 0 2 4 6–12–14

–7

Since x is divided by –3, multiply both sides by –3. Change to .

16 18 20 22 2410 14 26 28 3012

Example 2B: Multiplying or Dividing by a Negative Number


24 x (or x 24)



a. 10 ≥ –x

–1(10) ≤ –1(–x)

–10 ≤ x

Multiply both sides by –1 to make x positive. Change to .

b. 4.25 > –0.25h

–17 < h

Since h is multiplied by –0.25, divide both sides by –0.25. Change > to <.

–20–16–12–8 –4 0 4 8 12 16 20

–17

–10 –8 –6 –4 –2 0 2 4 6 8 10

Lesson QuizSolve each inequality and graph the solutions.

1. 8x < –24 x < –3 2. –5x ≥ 30 x ≤ –6

3. x > 20 4. x ≥ 6

5. A soccer coach plans to order more shirts for her team. Each shirt costs $9.85. She has $77 left in her uniform budget. What are the possible number of shirts she can buy?0, 1, 2, 3, 4, 5, 6, or 7 shirts

Inequalities that contain more than one operation require more than one

step to solve.

Example 1A: Solving Multi-Step Inequalities


45 + 2b > 61

45 + 2b > 61–45 –45

2b > 16

b > 8

0 2 4 6 8 10 12 14 16 18 20

Since 45 is added to 2b, subtract 45 from both sides to undo the addition.

Since b is multiplied by 2, divide both sides by 2 to undo the multiplication.

8 – 3y ≥ 298 – 3y ≥ 29–8 –8

–3y ≥ 21

y ≤ –7

Since 8 is added to –3y, subtract 8 from both sides to undo the addition.

Since y is multiplied by –3, divide both sides by –3 to undo the multiplication. Change ≥ to ≤.

–10 –8 –6 –4 –2 0 2 4 6 8 10

–7

Example 1B: Solving Multi-Step Inequalities




–12 ≥ 3x + 6–12 ≥ 3x + 6– 6 – 6

–18 ≥ 3x

–6 ≥ x

Since 6 is added to 3x, subtract 6 from both sides to undo the addition.


–10 –8 –6 –4 –2 0 2 4 6 8 10



x < –11

–5 –5x + 5 < –6

–20 –12 –8 –4–16 0

–11

Since x is divided by –2, multiply both sides by –2 to undo the division. Change > to <.

Since 5 is added to x, subtract 5 from both sides to undo the addition.



1 – 2n ≥ 21–1 –1

–2n ≥ 20

n ≤ –10

Since 1 – 2n is divided by 3, multiply both sides by 3 to undo the division.

Since 1 is added to −2n, subtract 1 from both sides to undo the addition.

Since n is multiplied by −2, divide both sides by −2 to undo the multiplication. Change ≥ to ≤.

–10

–20 –12 –8 –4–16 0

To solve more complicated inequalities, you may first need to simplify the expressions on one or both sides by using the order of operations, combining like terms, or using the Distributive Property.

Example 2A: Simplifying Before Solving Inequalities


2 – (–10) > –4t

12 > –4t

–3 < t (or t > –3)

Combine like terms.

Since t is multiplied by –4, divide both sides by –4 to undo the multiplication. Change > to <.

–3

–10 –8 –6 –4 –2 0 2 4 6 8 10

Example 2B: Simplifying Before Solving Inequalities


–4(2 – x) ≤ 8

−4(2 – x) ≤ 8

−4(2) − 4(−x) ≤ 8 –8 + 4x ≤ 8

+8 +84x ≤ 16

x ≤ 4

Distribute –4 on the left side.

Since –8 is added to 4x, add 8 to both sides.


–10 –8 –6 –4 –2 0 2 4 6 8 10

Example 2C: Simplifying Before Solving Inequalities


4f + 3 > 2–3 –3

4f > –1

Multiply both sides by 6, the LCD of the fractions.

Distribute 6 on the left side.

Since 3 is added to 4f, subtract 3 from both sides to undo the addition.

4f > –1Since f is multiplied by 4, divide

both sides by 4 to undo the multiplication.

0

Example 2C Continued



– 5 > – 5

2m > 20

m > 10

Since 5 is added to 2m, subtract 5 from both sides to undo the addition.

Simplify 52.

Since m is multiplied by 2, divide both sides by 2 to undo the multiplication.

0 2 4 6 8 10 12 14 16 18 20

2m + 5 > 52

2m + 5 > 25



3 + 2(x + 4) > 3

3 + 2(x + 4) > 33 + 2x + 8 > 3

2x + 11 > 3– 11 – 11

2x > –8

x > –4


Combine like terms.Since 11 is added to 2x,

subtract 11 from both sides to undo the addition.


–10 –8 –6 –4 –2 0 2 4 6 8 10



5 < 3x – 2+2 + 2

7 < 3x

Multiply both sides by 8, the LCD of the fractions.

Distribute 8 on the right side.

Since 2 is subtracted from 3x, add 2 to both sides to undo the subtraction.

Check It Out! Example 2c Continued


7 < 3x

4 6 82 100


Lesson Quiz: Part ISolve each inequality and graph the solutions.

1. 13 – 2x ≥ 21 x ≤ –4

2. –11 + 2 < 3p p > –3

3. 23 < –2(3 – t) t > 7

4.


5. A video store has two movie rental plans. Plan A includes a $25 membership fee plus $1.25 for each movie rental. Plan B costs $40 for unlimited movie rentals. For what number of movie rentals is plan B less than plan A?

more than 12 movies

Some inequalities have variable terms on both sides of the inequality symbol. You can solve these inequalities like you solved equations with variables on both sides.

Use the properties of inequality to “collect” all the variable terms on one side and all the constant terms on the other side.

Example 1A: Solving Inequalities with Variables on Both Sides


y ≤ 4y + 18

y ≤ 4y + 18–y –y

0 ≤ 3y + 18

–18 – 18

–18 ≤ 3y

–6 ≤ y (or y –6)

To collect the variable terms on one side, subtract y from both sides.

Since 18 is added to 3y, subtract 18 from both sides to undo the addition.

Since y is multiplied by 3, divide both sides by 3 to undo the multiplication.

–10 –8 –6 –4 –2 0 2 4 6 8 10

4m – 3 < 2m + 6To collect the variable terms on one side,

subtract 2m from both sides.–2m – 2m

2m – 3 < + 6 Since 3 is subtracted from 2m, add 3 to

both sides to undo the subtraction + 3 + 3

2m < 9 Since m is multiplied by 2, divide both sides by 2 to undo the multiplication.

4 5 6

Example 1B: Solving Inequalities with Variables on Both Sides




4x ≥ 7x + 6

4x ≥ 7x + 6

–7x –7x

–3x ≥ 6

x ≤ –2

To collect the variable terms on one side, subtract 7x from both sides.

Since x is multiplied by –3, divide both sides by –3 to undo the multiplication. Change ≥ to ≤.

–10 –8 –6 –4 –2 0 2 4 6 8 10



5t + 1 < –2t – 6

5t + 1 < –2t – 6+2t +2t

7t + 1 < –6

– 1 < –1

7t < –7

7t < –7

7 7t < –1

–5 –4 –3 –2 –1 0 1 2 3 4 5

To collect the variable terms on one side, add 2t to both sides.

Since 1 is added to 7t, subtract 1 from both sides to undo the addition.

Since t is multiplied by 7, divide both sides by 7 to undo the multiplication.

You may need to simplify one or both sides of an inequality before solving it. Look for like terms to combine and places to use the

Distributive Property.

Example 3A: Simplify Each Side Before Solving


2(k – 3) > 6 + 3k – 3

2(k – 3) > 3 + 3k Distribute 2 on the left side of the

inequality.

2k + 2(–3) > 3 + 3k

2k – 6 > 3 + 3k

–2k – 2k

–6 > 3 + k

To collect the variable terms, subtract 2k from both sides.

–3 –3

–9 > k

Since 3 is added to k, subtract 3 from both sides to undo the addition.

Example 3A Continued

–9 > k

–12 –9 –6 –3 0 3

Example 3B: Simplify Each Side Before Solving

Solve the inequality and graph the solution.

0.9y ≥ 0.4y – 0.5

0.9y ≥ 0.4y – 0.5

–0.4y –0.4y

0.5y ≥ – 0.5

0.5y ≥ –0.5

0.5 0.5

y ≥ –1

To collect the variable terms, subtract 0.4y from both sides.

Since y is multiplied by 0.5, divide both sides by 0.5 to undo the multiplication.

–5 –4 –3 –2 –1 0 1 2 3 4 5



5(2 – r) ≥ 3(r – 2)

5(2 – r) ≥ 3(r – 2)

5(2) – 5(r) ≥ 3(r) + 3(–2)

10 – 5r ≥ 3r – 6

+6 +6

16 − 5r ≥ 3r

+ 5r +5r

16 ≥ 8r

Distribute 5 on the left side of the inequality and distribute 3 on the right side of the inequality.

Since 6 is subtracted from 3r, add 6 to both sides to undo the subtraction.

Since 5r is subtracted from 16 add 5r to both sides to undo the subtraction.

Check It Out! Example 3a Continued

–6 –2 0 2–4 4

16 ≥ 8r

Since r is multiplied by 8, divide both sides by 8 to undo the multiplication.

2 ≥ r



0.5x – 0.3 + 1.9x < 0.3x + 6

2.4x – 0.3 < 0.3x + 6

+ 0.3 + 0.3

2.4x < 0.3x + 6.3

–0.3x –0.3x

2.1x < 6.3

Since 0.3 is subtracted from 2.4x, add 0.3 to both sides.

Since 0.3x is added to 6.3, subtract 0.3x from both sides.

x < 3

Since x is multiplied by 2.1, divide both sides by 2.1.

Simplify.2.4x – 0.3 < 0.3x + 6

Check It Out! Example 3b Continued

x < 3

–5 –4 –3 –2 –1 0 1 2 3 4 5

There are special cases of inequalities called identities and contradictions.

Example 4A: Identities and Contradictions

Solve the inequality.

2x – 7 ≤ 5 + 2x

2x – 7 ≤ 5 + 2x

–2x –2x

–7 ≤ 5 Subtract 2x from both sides.

True statement.

The inequality 2x − 7 ≤ 5 + 2x is an identity. All values of x make the inequality true. Therefore, all real numbers are solutions.

2(3y – 2) – 4 ≥ 3(2y + 7)

2(3y – 2) – 4 ≥ 3(2y + 7)

2(3y) – 2(2) – 4 ≥ 3(2y) + 3(7)

6y – 4 – 4 ≥ 6y + 21

6y – 8 ≥ 6y + 21

Distribute 2 on the left side and 3 on the right side.

Example 4B: Identities and Contradictions


–6y –6y

–8 ≥ 21

Subtract 6y from both sides.

False statement.No values of y make the inequality true. There are no solutions.

4(y – 1) ≥ 4y + 2

4(y – 1) ≥ 4y + 2

4(y) + 4(–1) ≥ 4y + 2

4y – 4 ≥ 4y + 2




–4y –4y

–4 ≥ 2

Subtract 4y from both sides.

False statement.

No values of y make the inequality true. There are no solutions.


x – 2 < x + 1

x – 2 < x + 1 –x –x

–2 < 1 Subtract x from both sides.

True statement.

All values of x make the inequality true. All real numbers are solutions.


Lesson Quiz: Part I


1. t < 5t + 24 t > –6

2. 5x – 9 ≤ 4.1x – 81 x ≤ –80

b < 133. 4b + 4(1 – b) > b – 9


4. Rick bought a photo printer and supplies for $186.90, which will allow him to print photos for $0.29 each. A photo store charges $0.55 to print each photo. How many photos must Rick print before his total cost is less than getting prints made at the photo store?

Rick must print more than 718 photos.

Lesson Quiz: Part III

Solve each inequality.

5. 2y – 2 ≥ 2(y + 7)

contradiction, no solution

6. 2(–6r – 5) < –3(4r + 2)

identity, all real numbers

Solving Absolute Value Equations

Absolute Value (of x)

• Symbol lxl• The distance x is from 0 on the number line.• Always positive• Ex: l-3l=3

-4 -3 -2 -1 0 1 2

Example: x = 5

• What are the possible values of x?

x = 5 or x = -5

To solve an absolute value equation:

ax+b = c, where c>0

To solve, set up 2 new equations, then solve each equation.

ax+b = c or ax+b = -c

** make sure the absolute value is by itself before you split to solve.

Ex: Solve 6x-3 = 15

6x-3 = 15 or 6x-3 = -156x = 18 or 6x = -12

x = 3 or x = -2

* Plug in answers to check your solutions!

Ex: Solve 2x + 7 -3 = 8

Get the abs. value part by itself first!2x+7 = 11

Now split into 2 parts.2x+7 = 11 or 2x+7 = -11

2x = 4 or 2x = -18x = 2 or x = -9

Check the solutions.

Competition Problems

Solve the inequality:

5(k – 4) – 2(k + 6) ≥ 4(k + 1) – 1

Answer:

-35 ≥ k

Solve for x:

8 – 3x > x + 20

Answer:

-3 > x

Solve for x:

–7x +18 ≤ 9 + 5x

Answer:

x ≥ 3/4

Warm Up Compare. Write, or =. 1. −3 2 3. 2. 6.5 6.3 < > > 4. 0.25= Tell whether the inequality x < 5 is true or false for the following values of x. 5.

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