Set Builder Notation	Interval Notation	Graph
2	{ x | x ≥ - 7}	[-7, )
4	{ x | x < - 0.2 }	(- , - 0.2)
6
8	{ x | - 5 x - 1}	[- 5, - 1]
10	{ x | - 3 x - 7}	[- 7, - 3]

17.  Solve x - 7 - 9
Solution:
Begin with x - 7 - 9
                                     Add 7 to both sides of the inequality to obtain the inequality
x - 2                        equivalent to the original
Because this is a simplest inequality, the solution is all real numbers greater than or equal to - 2.
Using set builder notation we can write this solution set as {x | x - 2}
If we use interval notation we write this solution set as [- 2, )
The graph of this solution set is    

20.  Solve 11x < 10x + 5
Solution:
Begin with 11x < 10x + 5
                                     Add - 10x to both sides of the inequality to obtain the inequality
x < 5                            equivalent to the original
Because this is a simplest inequality, the solution is all real numbers less than 5.
Using set builder notation we can write this solution set as {x | x < 5}
If we use interval notation we write this solution set as (- , 5, )
The graph of this solution set is

22.  Solve 7x - 1 6x - 1
Solution:
Begin with 7x - 1 6x - 1
                                     Subtract 6x from both sides of the inequality to obtain the inequality
x - 1 - 1                     equivalent to the original
                                      Add 1 to both sides of the inequality to obtain the inequality
x 0
Because this is a simplest inequality, the solution is all real numbers greater than or equal to 0.
Using set builder notation we can write this solution set as {x | x 0}
If we use interval notation we write this solution set as [0, )
The graph of this solution set is

24.  Solve 7+ 8x > 9x + 12
Solution:
Begin with 7+ 8x > 9x + 12
                                     Subtract 9x from both sides of the inequality to obtain the inequality
7 - x > 12                     equivalent to the original
                                      Add - 7 to both sides of the inequality to obtain the inequality
- x > 5                           equivalent to the original
                          Divide both sides of the inequality by -1 and reverse the inequality symbol to obtain
x < - 5               equivalent to the original
Because this is a simplest inequality, the solution is all real numbers less than - 5.
Using set builder notation we can write this solution set as {x | x < - 5}
If we use interval notation we write this solution set as (- , - 5)
The graph of this solution set is

34	Solve 8 - 5x 23 Solution: 8 - 5x 23 -5x 15 x - 3 [- 3, ) = {x | x - 3} = { x | - 3 x}	Add - 8 to both sides D ivide both sides by - 5 and reverse the inequality. The solution set is written in interval notation and set builder notation. The graph is also shown.
36	Solve 10 + x < 6x - 10 Solution: 10 + x < 6x - 10 20 + x < 6x 20 < 5x 4 < x (4, ) = { x | x > 4 }	Add 10 to both sides Subtract x from both sides Divide both sides by 5. The solution set is written in interval notation and set builder notation. The graph is also shown.
38	Solve 5(x + 4) 4(2x + 3) Solution: 5(x + 4) 4(2x + 3) 5x + 20 8x + 12 -3x + 20 12 -3x -8	Use the Distributive Law to remove parenthsis. Subtract 8x from both sides Subtract 20 from both sides Divide both sides by - 3 and reverse the inequality. The solution set is written in interval notation and set builder notation. The graph is also shown.
40	Solve Solution:	Add the fractions Multiply both sides of the inequality by - 6 and reverse the inequality The solution set is written in interval notation and set builder notation. The graph is also shown.
42	Solve 3x + 1 < 3(x - 2) Solution: 3x + 1 < 3(x - 2) 3x + 1 < 3x - 6 1 < - 6 Because this is FALSE The solution set is therefore the empty set
44	Solve 8(x + 3) " 7(x + 5) + x Solution: 8(x + 3) " 7(x + 5) + x 8x + 24 7x +35 + x 8x 8x + 11   It is clear that this inequality is TRUE for all Real Numbers x. So the solution set will be the set of all Real Numbers R

75. Shureka has scores of 72, 67, 82, and 79 on her algebra tests. Use an inequality to find the minimum score she can make on the final exam to pass the course with an average of 60 or higher, given that the final exam counts as two tests.
Solution:
Let x be her score on the final exam.
Then her course average will be
Because this average must be greater than or equal to 60 we obtain the inequality:
     
Siimplyfy the left side to obtain:
     
Continue the solution process to obtain:
     

Conclusion:
If Shureka gets a score of 30 or higher on the final exam her course average will be at least 60.

78. A clerk must use the elevator to move boxes of paper. The elevator's weight limit is 1500 pounds. If each box of paper weighs 66 pounds and the clerk weighs 147 pounds, use an inequality to find the maximum number of boxes she can move on the elevator at one time.
Solution:
Let x be the number of boxes she can move.
This yields the inequality 66x + 147 1500
Subtract 147 from both sides to obtain 66x 1353
Divide both sides by 66 to obtain x 20.5
Conclusion:
Because she is moving whole (no partial boxes) boxes of paper, the maximum number of boxes she may move per trip is 20.

79. To mail an envelope first class, the U. S. Post office charges 37 cents for the first ounce and 23 cents per ounce for each additional ounce. Use an inequality to find the maximum number of whole ounces that can be mailed for $4.00.
Solution:
Let x be the number of ounces to be mailed.
Note that the first of these ounces costs 37 cents so x - 1 ounces cost 23 cents per ounce.
We can now write 0.37 + 0.23(x - 1) 4
Solve this inequality
0.37 + 0.23(x - 1) 4
0.37 + 0.23x - 0.23 4
0.23x 3.86
x 16.78
Conclusion:
16 ounces is the maximum number of ounces that may be mailed first class for $4.00

80. A shopping mall parking garage charges $2 for the first half hour and $1.20 for each additonal half hour or a portion of a half hour. Use an inequality to find how long you can park if you have $8.00 in cash.
Solution:
Let x be the number of half hours you park.
This yields the inequality 2 +1.2(x - 1) 8
Solve this inequality
2 +1.2(x - 1) 8
2 + 1.2x - 1.2 8
1.2x 7.2
x 6
Conclusion:
You may park for 6 half hours or 3 hours for a fee of $8.00

82. A car rental company offers two subcompact rental plans. Plan A charges $32 per day for unlimited mileage, and Plan B charges $24 per day plus 15 cents per mile. Use an inequality to find the number of daily miles for which Plan A is more economical than Plan B.
Solution:
Let x be the number of miles driven per day. The question being asked is what is the maximum value for x for which the cost of Plan B is less than or equal to the cost with Plan A.
The cost with Plan B will be 24 + 0.15x
This yields the inequality 24 + 0.15x 32
Solve the inequality
0.15x 8
x 53.33
Conclusion:
Plan B is the more economical if daily mileage is less than 54 miles.
Plan A is the more economical if daily mileage is 54 miles or more.