Intermediate Algebra (11th edition)

(Marvins-Underground-K-12) #1
36.Concept Check The graph of the solution set of is given here.

Without actually doing the algebraic work, graph the solution set of the following, refer-
ring to the graph shown.
(a) (b)

Solve each inequality, and graph the solution set. See Example 3.(Hint: Compare your an-
swers with those in Exercises 23 – 34.)












































In Exercises 49 – 66, decide which method of solution applies, and find the solution set.
In Exercises 49 – 60, graph the solution set. See Examples 1–3.
































































Solve each equation or inequality. See Examples 4 and 5.




































77.|0.1x-2.5|+0.3Ú0.8 78.|0.5x-3.5|+0.2Ú0.6

`


2


3


x+

1


6


` +


1


2


=


5


2


`


1


2


x+

1


3


` +


1


4


=


3


4


|x+ 5 |- 6 ...- 1 |x- 2 |- 3 ... 4

|x+ 5 |- 2 = 12 | 2 x+ 1 |+ 378 | 6 x- 1 |- 276

|x|- 1 = 4 |x|+ 3 = 10 |x+ 4 |+ 1 = 2

|x- 4 |= 1 | 2 - 0.2x|= 2 | 5 - 0.5x|= 4

|x+ 2 |= 3 |x+ 3 |= 10 |x- 6 |= 3

|- 2 x- 6 |... 5 | 2 x- 1 |Ú 7 |- 4 +x|... 9

| 3 x- 1 |... 11 | 2 x- 6 |... 6 |- 6 x- 6 |... 1

| 2 x- 1 | 67 | 7 + 2 x|= 5 | 9 - 3 x|= 3

|- 4 +x| 79 |- 3 +x| 78 |x+ 5 | 720

| 5 - x|... 3 |- 5 x+ 3 | 612 |- 2 x- 4 | 65

|x+ 2 |... 10 | 4 x+ 1 | 621 | 3 - x|... 5

|x| 66 |r+ 5 | 620 | 3 r- 1 | 68

|x|... 3 |x|... 5 |x| 64

| 3 x- 4 |= 5 | 3 x- 4 | 75

| 3 x- 4 | 65

SECTION 2.7 Absolute Value Equations and Inequalities 119


Solve each equation. See Example 6.














82. 83. 84.


85. 86.


Solve each equation or inequality. See Examples 7 and 8.






















































102.|x- 4 |+ 5 Ú 4 103.| 10 x+ 7 |+ 361 104.| 4 x+ 1 |- 2 6- 5


| 5 x- 2 |= 0 | 7 x+ 4 |= 0 |x- 2 |+ 3 Ú 2

|x+ 9 |7- 3 | 7 x+ 3 |... 0 | 4 x- 1 |... 0

| 2 x- 1 |=- 6 | 8 x+ 4 |=- 4 |x+ 5 |7- 9

| 13 x+ 1 |=- 3 | 4 x+ 1 |= 0 | 6 x- 2 |= 0

|x|Ú- 10 |x|Ú- 15 | 12 t- 3 |=- 8

| 2 x- 6 |=| 2 x+ 11 | | 3 x- 1 |=| 3 x+ 9 |

` | 6 x|=| 9 x+ 1 | | 13 x|=| 2 x+ 1 |

2


3


x- 2 ` = `

1


3


x+ 3 `

` x-

1


2


=


1


2


| 3 x+ 1 |=| 2 x+ 4 | | 7 x+ 12 |=|x- 8 | x- 2 `

3

0
1


  • 3

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