Study Guide and Review - Chapter 5
State whether each sentence is true or false . If false , replace the underlined term to make a true
sentence.
1. Set-builder notation is a less concise way of writing a solution set.
SOLUTION: The statement is false. Set-builder notation is a more concise way of writing a solution set because it fully describes
it in fewer words. 2. There are two types of compound inequalities.
SOLUTION: There are two types of compound inequalities: those that include and (graphs that include intersections) and those
that include or (graphs that include unions). So, the statement is true.
3. The graph of a compound inequality containing and shows the union of the individual graphs.
SOLUTION: The statement is false. The graph of a compound inequality containing and shows the intersection of two inequalities,
which is only where the two overlap. The graph of a union of two inequalities shows both inequalities, whether they
overlap or not.
4. A compound inequality containing or is true if one or both of the inequalities is true. Its graph is the union of the
graphs of the two inequalities.
SOLUTION: A compound inequality containing or is true if at least one of the inequalities is true. This union of the two inequalities
has a graph in which both inequalities are shown. So, the statement is true.
5. The graph of an inequality of the form y < ax + b is a region on the coordinate plane called a half-plane.
SOLUTION: This is the form of a linear inequality. The graph of a linear inequality involves shading one side of a boundary, which
divides the coordinate plane into two half-planes. So, the statement is true.
6. A point defines the boundary of an open half-plane.
SOLUTION: The statement is false. A line defines the boundary between the two half-planes. A point can be used as an open (circle) or closed (dot) endpoint in the solution set of an inequality.
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7. The boundary is the graph of the equation of the line that defines the edge of each half-plane.
SOLUTION: Page 1
5. The graph of an inequality of the form y < ax + b is a region on the coordinate plane called a half-plane.
SOLUTION: This
is theand
formReview
of a linear
inequality.
Study
Guide
- Chapter
5 The graph of a linear inequality involves shading one side of a boundary, which
divides the coordinate plane into two half-planes. So, the statement is true.
6. A point defines the boundary of an open half-plane.
SOLUTION: The statement is false. A line defines the boundary between the two half-planes. A point can be used as an open (circle) or closed (dot) endpoint in the solution set of an inequality.
7. The boundary is the graph of the equation of the line that defines the edge of each half-plane.
SOLUTION: The boundary is the graph of the equation of the line that defines the edge of each half-plane. If it is included in the
solution, the solution is a closed half-plane. If it is not included in the solution, then the solution is an open half-plane.
So, the statement is true.
8. The solution set to the inequality y ≥ x includes the boundary.
SOLUTION: The solution set to the inequality y ≥ x includes the boundary because y is greater than or equal to x. Its graph would
include a boundary with a solid line. So, the statement is true.
9. When solving an inequality, multiplying each side by a negative number reverses the inequality symbol.
SOLUTION: When solving an inequality, multiplying or dividing each side by a negative number reverses the inequality symbol.
The direction of the inequality sign is reversed to make the resulting inequality also true. So, the statement is true.
10. The graph of a compound inequality that contains and is the intersection of the graphs of the two inequalities.
SOLUTION: The graph of a compound inequality that contains and is the intersection of the graphs of the two inequalities. It is
where the two inequalities overlap. So, the statement is true.
Solve each inequality. Then graph it on a number line.
12. x + 8 ≤ 3
SOLUTION: The solution set is {x | x ≤ −5}.
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10. The graph of a compound inequality that contains and is the intersection of the graphs of the two inequalities.
SOLUTION: The
graphand
of aReview
compound
inequality
Study
Guide
- Chapter
5 that contains and is the intersection of the graphs of the two inequalities. It is
where the two inequalities overlap. So, the statement is true.
Solve each inequality. Then graph it on a number line.
12. x + 8 ≤ 3
SOLUTION: The solution set is {x | x ≤ −5}.
15. 13 − p ≥ 15
SOLUTION: The solution set is {p | p ≤ −2}.
Solve each inequality. Graph the solution on a number line.
18. SOLUTION: The solution set is {x | x > 18}.
21. −55 ≤ −5w
SOLUTION: The solution set is {w | w ≤ 11}.
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The solution set is {x | x > 18}.
Study Guide and Review - Chapter 5
21. −55 ≤ −5w
SOLUTION: The solution set is {w | w ≤ 11}.
24. MOVIE RENTAL Jack has no more than $24 to spend on DVDs for a party. Each DVD rents for $4. Find the
maximum number of DVDs Jack can rent for his party.
SOLUTION: The phrase no more than means the same as less than or equal to. Let d be the number of DVDs Jack rents. Write
an inequality.
The solution set is {d | d ≤ 6}. So, the maximum number of DVDs Jack can rent is 6.
Solve each inequality. Graph the solution on a number line.
27. 18 ≤ −2x + 8
SOLUTION: The solution set is {x | x ≤ −5}.
30. TICKET SALES The drama club collected $160 from ticket sales for the spring play. They need to collect at least
$400 to pay for new lighting for the stage. If tickets sell for $3 each, how many more tickets need to be sold?
SOLUTION: The phrase at least means the same as greater than or equal to. Let t be the number of tickets that still need to be
sold. Write an inequality.
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The solution set is {x | x ≤ −5}.
Study Guide and Review - Chapter 5
30. TICKET SALES The drama club collected $160 from ticket sales for the spring play. They need to collect at least
$400 to pay for new lighting for the stage. If tickets sell for $3 each, how many more tickets need to be sold?
SOLUTION: The phrase at least means the same as greater than or equal to. Let t be the number of tickets that still need to be
sold. Write an inequality.
The solution set is {t | t ≥ 80}. So, at least 80 more tickets need to be sold.
Solve each compound inequality. Then graph the solution set.
33. 3x + 2 ≤ 11 or 5x − 8 > 22
SOLUTION: or
The solution set is {x | x ≤ 3 or x > 6}.
Notice that the graphs do not intersect. To graph the solution set, graph x ≤ 3 and graph x > 6. Then find the union.
Solve each inequality. Then graph the solution set.
36. SOLUTION: Case 1 p + 2 is positive.
or
Case 2 p + 2 is negative.
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solution
set is {pby |Cognero
p < −9
or p > 5}.
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The solution set is {x | x ≤ 3 or x > 6}.
Notice that the graphs do not intersect. To graph the solution set, graph x ≤ 3 and graph x > 6. Then find the union.
Study Guide and Review - Chapter 5
Solve each inequality. Then graph the solution set.
36. SOLUTION: Case 1 p + 2 is positive.
or
Case 2 p + 2 is negative.
The solution set is {p | p < −9 or p > 5}.
39. SOLUTION: Case 1 3d − 1 is positive.
and
Case 2 3d − 1 is negative.
The solution set is
.
42. SOLUTION: Case 1 −4y − 3 is positive.
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The solution set is
.
Study Guide and Review - Chapter 5
42. SOLUTION: Case 1 −4y − 3 is positive.
and
Case 2 −4y − 3 is negative.
The solution set is {y |
}.
Graph each inequality.
45. y > x − 3
SOLUTION: y>x−3
Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point.
Since 0 is greater than −3, shade the half-plane that contains (0, 0).
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The solution set is {y |
}.
Study Guide and Review - Chapter 5
Graph each inequality.
45. y > x − 3
SOLUTION: y>x−3
Because the inequality involves >, graph the boundary using a dashed line. Choose (0, 0) as a test point.
Since 0 is greater than −3, shade the half-plane that contains (0, 0).
48. y ≥ −2x + 6
SOLUTION: y ≥ −2x + 6
Because the inequality involves ≥, graph the boundary using a solid line. Choose (0, 0) as a test point.
Since 0 is not greater than or equal to 6, shade the half-plane that does not contain (0, 0).
Graph each inequality. Determine which of the ordered pairs are part of the solution set for each
inequality.
51. y ≤ 4; {(3, 6), (1, 2), (−4, 8), (3, −2), (1, 7)}
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y ≤ 4
Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point.
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Study Guide and Review - Chapter 5
Graph each inequality. Determine which of the ordered pairs are part of the solution set for each
inequality.
51. y ≤ 4; {(3, 6), (1, 2), (−4, 8), (3, −2), (1, 7)}
SOLUTION: y ≤ 4
Because the inequality involves ≤, graph the boundary using a solid line. Choose (0, 0) as a test point.
0≤4
Since 0 is less than or equal to 4, shade the half-plane that contains (0, 0).
To determine which of the ordered pairs are part of the solution set, plot them on the graph. Because (1, 2) and (3, −2) are in the shaded half-plane, they are part of the solution set.
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