Chapter 1 Section 4: Compound Linear Inequalities

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Page 1 of 10

To solve the first inequality, first add 1 to both sides and then divide both sides by 2:

To solve the second, add 2 to both sides and then divide by 3:

Introduction

Compound linear inequalities involve finding the union or intersection of solution sets of two or more linear inequalities.

You’ve already encountered one type of compound inequality in the "Sets, Interval Notation, and Linear Inequalities" section—the three-way inequality. This section goes beyond that type of compound inequality.

When we write A< <x B, we mean that A< x and also that

x

<

B

. In other words, x is in between

A

and

B

. Three-way inequalities are best solved as we did previously.

For other types of compound inequalities, it’s best to first solve each of the inequalities. Next, graph their solutions on parallel number lines (with corresponding numbers lined up). Then, find the union or intersection of your graphed solutions visually, and graph the result on yet another copy of the number line. We illustrate this process in this section.

Example A

Solve, and graph the solution: 5≥2x −1 and 4≤3x−2.

Express your solution in interval notation and in set-builder notation. Note: The word “and” signifies that we are to find the intersection of the solution sets.

Step 1—Solve each of the inequalities separately.

1 1 5 2 1 6 2 6 2 2 x x ≥ ≥ + +− ≥ 2 x 3≥ x 2 2 4 3 2 6 3 6 3 3 x x ≤ ≤ + −+ ≤ 3 x 2 ≤ x

(This is the same as x ≤3.) (This is the same as x ≥ 2.)

Step 2—Graph each of these solutions on parallel number lines, and graph the intersection of the two solutions on a third number line (the intersection is the overlap of the two solution sets):

That last line is the graph of our solution—the intersection.

Note: View the animation in the course online to see the graphing process. Step 3—Write the intersection in interval notation and in set-builder notation:

Interval notation Set-builder notation

[ ]

2, 3

=

{

x

2 ≤ ≤

x

3 }

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Chapter 1 Section 4: Compound Linear Inequalities

To solve the first inequality, first subtract xand subtract3from both sides, and then divide both sides by 2:

To solve the second, add xand subtract 3

from both sides. Then divide both sides by 2: Note: Sometimes compound inequalities involving “and” have no solution.

Example B

Solve: 3+3x > +6 x and 5− > +x x 3.

2

2 x

3

2

3

2 x

>

Step 2—You don’t have to graph this to realize that x can’t be greater than 3

2 and less than 1 at

the same time. But if you do graph each solution, it’s also easy to see that there is nothing in the intersection of the two solution sets:

Step 3— The solution set for this compound inequality is the empty set, ∅.

Extended Example 1a

Solve, and graph the solution: 2x+ >5 6x−7 and 5− <x 7x−3. Express your solution in interval notation and in set-builder notation.

Hint: Solve 2x+ >5 6x−7. Subtract 2x from both sides, while adding

7

to both sides. Step 1:

2

5

6

7

12

4

2

7

2

7 x

x

−

+

>

−

>

+

Hint: Divide both sides by

4

. Step 2: 12 4 4 4

4

x > 3

4

⋅

4

>

4

x

3 >

x

which is

x

<

3

Hint: Now, solve 5− <x 7x−3. Add x and add 3 to both sides.

continued… 3 3 3 6 2 3 3 x x x x x − − + + > − > − 5 3 2 2 . 3 x x 3 x x x − + + − − > + > 2 2 2 > 2 x 1> x which is x <1

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Page 3 of 10

The last number line is the graph of the solution— the intersection. Extended Example 1a, continued…

Step 3:

5

7

3

8

3

3 x

x

−

<

+

−

<

+ +

+

Hint: Divide both sides by 8. Step 4: 8 8 8 8 8 x < 8 8 < 8 x 1< x which is x >1

Hint: Graph each solution on parallel number lines, and graph the intersection (overlap) on a third parallel number line.

Step 5:

Hint: Write this solution in interval notation and in set-builder notation. Answer:

( )

1, 3

=

{

x

1 < <

x

3 }

Extended Example 1b

Solve, and graph the solution: 2x+10>13x−12 and 6−3x< − −x 6. Express your solution in interval notation and in set-builder notation.

Hint: Solve 2x+10>13x−12. Subtract 2x from both sides, while adding 12 to both sides. Step 1: 2 10 13 1 2 2 22 11 12 2 12 x x x x x + > + >− + − −

Hint: Divide both sides by 11. Step 2: 22 11 11 11

11

x > 2

11

⋅

11

>

11

x

2 >

x

which is

x

<

2

Hint: Now, solve 6−3x < − −x 6. Add 3x and add 6 to both sides. Step 3: 6 3 6 12 2 6 3 3 6 x x x x x − < − − + < + + +

2

.

continued…

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Chapter 1 Section 4: Compound Linear Inequalities

Page 4 of 10

The last number line is the graph of the solution— this time there is nothing in the intersection!

The last number line is the graph of the solution—the intersection. Extended Example 1b, continued…

Step 4: 12 2 2

2

x < 6 2 ⋅ 2 < 2 x

6 <

x

which is

x

>

6

Step 5:

Hint: There is nothing in the intersection, so the solution set is empty. Answer:

In cases such as this, there’s no point in writing your answer in either interval notation or set-builder notation. The solution set is simply the empty set, ∅.

Extended Example 1c

Solve, and graph the solution: 5x − >8 4x−7 and 13−9x

≤

2x−9. Express your solution in interval notation and in set-builder notation.

Hint: Solve 5x− >8 4x−7. Subtract 4x from both sides, while adding 8 to both sides. Step 1: 5 8 4 7 1 4 8 4 8 x x x x x − > − > − + − +

Hint: Now, solve 13−9x ≤2x−9. Add 9x and add 9 to both sides. Step 2:

13

9

2

9

22

1

9

1

9 x

x

+ +

−

≤

+

−

+

≤

Hint: Divide both sides by 11. Step 3: 22 11 11

11

2

x

≤

11 x

2 ≤

x

Step 4:

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Page 5 of 10

The last number line is the graph of the solution—the intersection. Extended Example 1c, continued…

[

2,

∞ =

)

{

x x

≥

2 }

Extended Example 1d

Solve, and graph the solution: 11x + <1 5x+7 and 4x − ≤3 2x+7. Express your solution in interval notation and in set-builder notation.

Hint: Solve 11x+ <1 5x +7. Subtract

5 x

from both sides, while subtracting 1 from both sides. Step 1:

11

1

5

1

7

6

1

5

5 x

x

+

−

<

+

<

Hint: Divide both sides by 6. Step 2:

6

x

6

<

6

1 x

<

Hint: Now, solve 4x − ≤3 2x +7. Subtract 2x while adding 3 to both sides. Step 3: 4 3 2 7 2 10

2

3

2

3

x x x

x

− ≤ + ≤

−

+

−

+

2

. Step 4: 2 10 2 2 2 x ≤ 2 x 2 ≤ 5 2 ⋅ 5 x ≤

Step 5:

(

−∞

, 1

)

=

{

x

<

1 }

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To solve the first inequality, first add 3 to both sides, and then divide both sides by 5:

To solve the second, add 1 to both sides and then divide both sides by 2:

Example C

Solve, and graph the solution: 8<5x−3 or 3≥ 2x−1.

Express your solution in interval notation and in set-builder notation:

Note: The word “or” signifies that we are to find the union of the solution sets. Inclusive and Exclusive "or"

Ordinarily, in everyday English, “or” is interpreted as an exclusive or. For example, suppose a flight attendant asks if you would like the fish or chicken dinner. You can pick one or the other but NOT both. If it were the inclusive or used in logic and math, you could also choose both! The inclusive or is equivalent to the notion of union for sets.

11 5

8

5

3

5

3

11 x

x

<

−

+

<

+

<

5

x 11 11 5

<

x

which is

x

>

5 4 2

1

3

2

1

4

2

2 x

x

≥

+

≥

−

2

x

2 ≥

x

which is

x

≤

2

Step 2—Graph each of these solutions on parallel number lines, and graph the union of the two solutions on a third parallel number line:

The last line is the graph of our solution—the union.

Step 3—Write the union in interval notation and in set-builder notation. Interval notation:

(

, 2

]

11, .

5

−∞

⎛

_⎜

∞

⎞

_⎟

⎝

⎠

∪

There are two ways to write this in set-builder notation:

{

}

11 11 2 2 5 5 x x ≤ x > x x ≤ x x >

⎧

⎫

₌

⎧

⎫

⎨

⎬

⎨

⎬

⎩

or

⎭

∪

⎩

⎭

. Note that

(

, 2

]

11, 5 −∞

⎛

_⎜

∞

⎞

_⎟

⎝

⎠

∪ is not the same as

2,

11

5 ⎡

⎞

⎟

⎢⎣

⎠

; if the third number line in the graph of the solution set has two distinct parts, then the answer in notation form will also have two parts. After all, in such cases the solution consists of two distinct intervals!

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Page 7 of 10

To solve the first inequality, first subtract 1 from both sides, and then divide both sides by 2:

To solve the second, subtract 2xfrom both sides. Then divide both sides by 3: Note: Sometimes compound inequalities involving “or” have all real numbers as the solution.

Example D

Solve: 2x+ ≥ −1 3 or 5x <12+2x.

2

1

3

2

4

2

1

1 x

x

+ ≥ −

≥ −

−

2 x

4

2

2 x

−

≥

≥ −

2

5

12

2

3

12

3

2 x

x

−

<

+

<

−

3 x

12

3

4 x

<

Step 2—Graph these two inequalities’ solutions, and the union on a third line.

You can see that their union is all real numbers.

Step 3—The solution set is the set of all real numbers, expressed in set-builder notation and in interval notation below:

{

x

− ∞ < < ∞ = −∞ −∞

x

}

(

,

)

.

Extended Example 2a

Solve, and graph the solution: 6x− ≥1 5x +7 or 8x+ <5 2x−3. Express your solution in interval notation and in set-builder notation.

Hint: Solve 6x− ≥1 5x+7. Subtract

5 x

, while adding 1 to both sides. Step 1: 6 1 5 7 8

5

1

5

1

x x x

x

− ≥ + ≥

−

+

−

+

Hint: Solve 8x+ <5 2x−3. Subtract 2x and subtract

5

from both sides. Step 2: 8 5 2 3 6 8

2

5

2

5

x x x

x

+ < − < −

−

Hint: Divide both sides by 6.

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Extended Example 2a, continued… Step 3:

6

x 8 6 4 3 x − < < −

Hint: Graph each solution on parallel number lines, and graph the union on a third number line. Step 4:

[

)

{

}

4 4 , 8, 8 3 x x 3 x ⎛_{−∞ −} ⎞ _{∞ =} _{< −} _≥ ⎜ ⎟ ⎝ ⎠∪ or

Extended Example 2b

Solve, and graph the solution: 8x + ≥5 7x+9 or 7x−15<4x+9. Express your solution in interval notation and in set-builder notation.

Hint: Solve 8x+ ≥5 7x+9. Subtract

7 x

and subtract

5

from both sides. Step 1: 8 5 4 7 5 7 9 7 5 x x x x x − −+ ≥− +− ≥

Hint: Solve 7x −15< 4x+9. Subtract 4x, while adding 15 to both sides. Step 2: 4 15 4 15 7 15 4 9 3 24 x x x x x − +− <− ++ <

Hint: Divide both sides by 3. Step 3: 3 24

3

x <

3

x

3

< 8

3

⋅

8

x <

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Page 9 of 10

Extended Example 2c

Solve, and graph the solution: 14

y

+

13 ≥

11 y

+

1

or

17 y

− >

8

2 y

−

5

. Express your solution in interval notation and in set-builder notation.

Hint: Solve 14

y

+

13 ≥

11 y

+

1

. Subtract 11

y

and subtract 13 from both sides. Step 1: 14 13 1 11 13 11 1 1 2 3 1 3 1 y y y y y − + ≥ + − ≥ − − −

Hint: Divide both sides by 3. Step 2: 3 12

3

y − ≥

3

y

3

≥ − 4

3

⋅

4

y ≥ −

Hint: Solve 17

y

− >

8

2 y

−

5

. Subtract

2 y

, while adding 8 to both sides. Step 3: 2 17 8 2 5 1 8 8 5 2 3 y y y y y − > − > − + − +

Hint: Divide both sides by 15. Step 4: 1 15 15 5 3 15 y > 15 y 3 > 3 5 1 5 y ⋅ >

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Extended Example 2d

Solve, and graph the solution: 7z−23≥ 5z+5 or 12z− ≤2 3z−1. Express your solution in interval notation and in set-builder notation.

Hint: Solve 7z −23≥5z +5. Subtract

5 z

, while adding 23 to both sides. Step 1: 5 23 5 23 7 23 5 5 2 28 z z z z z − + − + − ≥ + ≥

2

. Step 2: 2 28

2

z ≥

2

z

2

≥ 14

2

⋅ 1

4

z ≥

Hint: Solve 12z− ≤2 3z−1. Subtract 3z, while adding

2

to both sides. Step 3: 12 2 3 3 2 1 3 9 1 2 z z z z z − − + − − ≤ ≤ +