1. 2-1 Solving Linear Equations and Inequalities Warm Up
Lesson Presentation
Lesson Quiz
Holt Algebra 2

2. Warm Up Simplify each expression. 1. 2x + 5 – 3x –x + 5 2. –(w – 2)
3. 6(2 – 3g)
12 – 18g
Graph on a number line.
4. t > –2
–4 –3 –2 –
5. Is 2 a solution of the inequality –2x < –6? Explain.
No; when 2 is substituted for x, the inequality is false: –4 < –6

3. Objectives Solve linear equations using a variety of methods.
Solve linear inequalities.

4. Vocabulary equation solution set of an equation
linear equation in one variable
identify
contradiction
inequality

5. An equation is a mathematical statement that two expressions are equivalent. The solution set of an equation is the value or values of the variable that make the equation true. A linear equation in one variable can be written in the form ax = b, where a and b are constants and a ≠ 0.

6. Linear Equations in One variable
Nonlinear Equations
4x = 8
+ 1 = 32
3x – = –9
+ 1 = 41
2x – 5 = 0.1x +2
3 – 2x = –5
Notice that the variable in a linear equation is not under a radical sign and is not raised to a power other than 1. The variable is also not an exponent and is not in a denominator.
Solving a linear equation requires isolating the variable on one side of the equation by using the properties of equality.

8. To isolate the variable, perform the inverse or opposite of every operation in the equation on both sides of the equation. Do inverse operations in the reverse order of operations.

9. Example 1: Consumer Application
The local phone company charges $12.95 a month for the first 200 of air time, plus $0.07 for each additional minute. If Nina’s bill for the month was $14.56, how many additional minutes did she use?

10. Let m represent the number of additional minutes that Nina used.
Example 1 Continued
Let m represent the number of additional minutes that Nina used.
Model
additional minute charge
number of additional minutes
monthly charge
total charge
plus
times = =
14.56
12.95 +
0.07 * m

11. Example 1 Continued
Solve.
m =
Subtract from both sides.
– –12.95
0.07m =
Divide both sides by 0.07.
0.07
0.07
m = 23
Nina used 23 additional minutes.

12. Check It Out! Example 1
Stacked cups are to be placed in a pantry. One cup is 3.25 in. high and each additional cup raises the stack 0.25 in. How many cups fit between two shelves 14 in. apart?

13. Check It Out! Example 1 Continued
Let c represent the number of additional cups needed.
Model
additional cup height
number of additional cups
total height
plus
times
one cup = =
14.00
3.25 +
0.25 * c

14. Check It Out! Example 1 Continued
Solve.
c =
– –3.25
Subtract 3.25 from both sides.
0.25c = 10.75
Divide both sides by 0.25.
0.25
c = 43
44 cups fit between the 14 in. shelves.

15. Example 2: Solving Equations with the Distributive Property
Solve 4(m + 12) = –36
Method 1
The quantity (m + 12) is multiplied by 4, so divide by 4 first.
4(m + 12) = –36
Divide both sides by 4.
m + 12 = –9
–12 –12
Subtract 12 from both sides.
m = –21

16.  Example 2 Continued Check 4(m + 12) = –36 4(–21 + 12) –36 4(–9) –36
4(– ) –36
4(–9) –36
–36 –36 

17. Example 2 Continued
Solve 4(m + 12) = –36
Method 2
Distribute before solving.
4m + 48 = –36
Distribute 4.
–48 –48
Subtract 48 from both sides.
4m = –84 =
4m –84
Divide both sides by 4.
m = –21

18. Check It Out! Example 2a
Solve 3(2 –3p) = 42.
Method 1
The quantity (2 – 3p) is multiplied by 3, so divide by 3 first.
3(2 – 3p) = 42
Divide both sides by 3.
2 – 3p = 14
– –2
Subtract 2 from both sides.
–3p = 12
Divide both sides by –3.
– –3
p = –4

19. Check It Out! Example 2a Continued
3(2 + 12) 

20. Check It Out! Example 2a Continued
Solve 3(2 – 3p) = 42 .
Method 2
Distribute before solving.
6 – 9p = 42
Distribute 3.
– –6
Subtract 6 from both sides.
–9p = 36 =
–9p 36
–9 –9
Divide both sides by –9.
p = –4

21. Check It Out! Example 2b
Solve –3(5 – 4r) = –9.
Method 1
The quantity (5 – 4r) is multiplied by –3, so divide by –3 first.
–3(5 – 4r) –9
– –3 =
Divide both sides by –3.
5 – 4r = 3
– –5
Subtract 5 from both sides.
–4r = –2

22. Check It Out! Example 2b Continued
Solve –3(5 – 4r) = –9.
Method 1 =
–4 –4
–4r –2
Divide both sides by –4.
r =
– –9
Check
–3(5 –4r) = –9
–3(5 – 4• ) –9
–3(5 – 2) –9 
–3(3) –9

23. Check It Out! Example 2b Continued
Solve –3(5 – 4r) = –9.
Method 2
Distribute before solving.
– r = –9
Distribute 3.
Add 15 to both sides.
12r = 6 =
12r 6
Divide both sides by 12.
r =

24. If there are variables on both sides of the equation, (1) simplify each side. (2) collect all variable terms on one side and all constants terms on the other side. (3) isolate the variables as you did in the previous problems.

25. Example 3: Solving Equations with Variables on Both Sides
Solve 3k– 14k + 25 = 2 – 6k – 12.
Simplify each side by combining like terms.
–11k + 25 = –6k – 10
+11k k
Collect variables on the right side.
25 = 5k – 10
Add.
Collect constants on the left side.
35 = 5k
Isolate the variable.
7 = k

26. Check It Out! Example 3
Solve 3(w + 7) – 5w = w + 12.
Simplify each side by combining like terms.
–2w + 21 = w + 12
+2w w
Collect variables on the right side.
21 = 3w + 12
Add.
– –12
Collect constants on the left side.
9 = 3w
Isolate the variable.
3 = w

27. You have solved equations that have a single solution
You have solved equations that have a single solution. Equations may also have infinitely many solutions or no solution.
An equation that is true for all values of the variable, such as x = x, is an identity. An equation that has no solutions, such as 3 = 5, is a contradiction because there are no values that make it true.

28. Example 4A: Identifying Identities and Contractions
Solve 3v – 9 – 4v = –(5 + v).
3v – 9 – 4v = –(5 + v)
–9 – v = –5 – v
Simplify.
+ v v
–9 ≠ –5 x
Contradiction
The equation has no solution. The solution set is the empty set, which is represented by the symbol .

29. Example 4B: Identifying Identities and Contractions
Solve 2(x – 6) = –5x – x.
2(x – 6) = –5x – x
Simplify.
2x – 12 = 2x – 12
–2x –2x
–12 = –12 
Identity
The solutions set is all real number, or .

30. Check It Out! Example 4a
Solve 5(x – 6) = 3x – x.
5(x – 6) = 3x – x
Simplify.
5x – 30 = 5x – 18
–5x –5x
–30 ≠ –18 x
Contradiction
The equation has no solution. The solution set is the empty set, which is represented by the symbol .

31.  Check It Out! Example 4b Solve 3(2 –3x) = –7x – 2(x –3).
Simplify.
+ 9x x
6 = 6 
Identity
The solutions set is all real numbers, or .

32. An inequality is a statement that compares two expressions by using the symbols <, >, ≤, ≥, or ≠. The graph of an inequality is the solution set, the set of all points on the number line that satisfy the inequality.
The properties of equality are true for inequalities, with one important difference. If you multiply or divide both sides by a negative number, you must reverse the inequality symbol.

33. These properties also apply to inequalities expressed with >, ≥, and ≤.

34. To check an inequality, test
the value being compared with x
a value less than that, and
a value greater than that.
Helpful Hint

35. Example 5: Solving Inequalities
Solve and graph 8a –2 ≥ 13a + 8.
8a – 2 ≥ 13a + 8
–13a –13a
Subtract 13a from both sides.
–5a – 2 ≥ 8
Add 2 to both sides.
–5a ≥ 10
Divide both sides by –5 and reverse the inequality.
–5a ≤ 10
– –5
a ≤ –2

36.   Example 5 Continued Solve and graph 8a – 2 ≥ 13a + 8.
Check Test values in the original inequality. •
–10 –9 –8 –7 –6 –5 –4 –3 –2 –1
Test x = –4
Test x = –2
Test x = –1
8(–4) – 2 ≥ 13(–4) + 8
8(–2) – 2 ≥ 13(–2) + 8
8(–1) – 2 ≥ 13(–1) + 8
–34 ≥ –44 
–18 ≥ –18 
–10 ≥ –5 x
So –4 is a solution.
So –2 is a solution.
So –1 is not a solution.

37. Check It Out! Example 5
Solve and graph x + 8 ≥ 4x + 17.
x + 8 ≥ 4x + 17
–x –x
Subtract x from both sides.
8 ≥ 3x +17
Subtract 17 from both sides.
– –17
–9 ≥ 3x
–9 ≥ 3x
Divide both sides by 3.
–3 ≥ x or x ≤ –3

38. Check It Out! Example 5 Continued
Solve and graph x + 8 ≥ 4x + 17.
Check Test values in the original inequality. •
–6 –5 –4 –3 –2 –
Test x = –6
Test x = –3
Test x = 0
– ≥ 4(–6) + 17
–3 +8 ≥ 4(–3) + 17
0 +8 ≥ 4(0) + 17
2 ≥ –7 
5 ≥ 5 
8 ≥ 17 x
So –6 is a solution.
So –3 is a solution.
So 0 is not a solution.

39. Lesson Quiz: Part I
1. Alex pays $19.99 for cable service each month. He also pays $2.50 for each movie he orders through the cable company’s pay-per-view service. If his bill last month was $32.49, how many movies did Alex order?
5 movies

40. Lesson Quiz: Part II
Solve.
2. 2(3x – 1) = 34
3. 4y – 9 – 6y = 2(y + 5) – 3
4. r + 8 – 5r = 2(4 – 2r)
5. –4(2m + 7) = (6 – 16m)
x = 6
y = –4
all real numbers, or 
no solution, or

41. Lesson Quiz: Part III 5. Solve and graph. 12 + 3q > 9q – 18
–2 –

42. Proportional Reasoning
2-2
Proportional Reasoning
Warm Up
Lesson Presentation
Lesson Quiz
Holt Algebra 2

43. Warm Up
Write as a decimal and a percent. 1. 2.
0.4; 40%
1.875; 187.5%

44. Graph on a coordinate plane.
Warm Up Continued
Graph on a coordinate plane.
3. A(–1, 2)
4. B(0, –3)
A(–1, 2)
B(0, –3)

45. Warm Up Continued
5. The distance from Max’s house to the park is mi. What is the distance in feet? (1 mi = 5280 ft)
18,480 ft

46. Objective
Apply proportional relationships to rates, similarity, and scale.

47. Vocabulary
ratio
proportion
rate
similar
indirect measurement

48. Recall that a ratio is a comparison of two numbers by division and a proportion is an equation stating that two ratios are equal. In a proportion, the cross products are equal.

49. If a proportion contains a variable, you can cross multiply to solve for the variable. When you set the cross products equal, you create a linear equation that you can solve by using the skills that you learned in Lesson 2-1.

50. Reading Math
In a ÷ b = c ÷ d, b and c are the means, and a and d are the extremes. In a proportion, the product of the means is equal to the product of the extremes.

51. Example 1: Solving Proportions
Solve each proportion. c = p A. B. = p c = =
206.4 = 24p
Set cross products equal.
88c = 1848 p =
88c =
Divide both sides.
8.6 = p
c = 21

52. Check It Out! Example 1
Solve each proportion. y A. = B. = x y x = =
Set cross products equal.
924 = 84y
2.5x =105 y = =
2.5x
Divide both sides.
11 = y
x = 42

53. Because percents can be expressed as ratios, you can use the proportion to solve percent problems.
Percent is a ratio that means per hundred.
For example:
30% = 0.30 =
Remember! 30
100

54. Example 2: Solving Percent Problems
A poll taken one day before an election showed that 22.5% of voters planned to vote for a certain candidate. If 1800 voters participated in the poll, how many indicated that they planned to vote for that candidate?
You know the percent and the total number of voters, so you are trying to find the part of the whole (the number of voters who are planning to vote for that candidate).

55. Method 1 Use a proportion. Method 2 Use a percent equation.
Example 2 Continued
Method 1 Use a proportion.
Method 2 Use a percent equation.
Divide the percent by 100.
Percent (as decimal)  whole = part
0.225  1800 = x
Cross multiply.
22.5(1800) = 100x
405 = x
Solve for x.
x = 405
So 405 voters are planning to vote for that candidate.

56. Check It Out! Example 2
At Clay High School, 434 students, or 35% of the students, play a sport. How many students does Clay High School have?
You know the percent and the total number of students, so you are trying to find the part of the whole (the number of students that Clay High School has).

57. Check It Out! Example 2 Continued
Method 1 Use a proportion.
Method 2 Use a percent equation.
Divide the percent by 100.
35% = 0.35
Percent (as decimal)  whole = part
0.35x = 434
Cross multiply.
100(434) = 35x
x = 1240
Solve for x.
x = 1240
Clay High School has 1240 students.

58. A rate is a ratio that involves two different units
A rate is a ratio that involves two different units. You are familiar with many rates, such as miles per hour (mi/h), words per minute (wpm), or dollars per gallon of gasoline. Rates can be helpful in solving many problems.

59. Example 3: Fitness Application
Ryan ran 600 meters and counted 482 strides. How long is Ryan’s stride in inches? (Hint: 1 m ≈ in.)
Use a proportion to find the length of his stride in meters.
600 m
482 strides
x m
1 stride =
Write both ratios in the form
meters
strides
600 = 482x
Find the cross products.
x ≈ 1.24 m

60. Example 3: Fitness Application continued
Convert the stride length to inches.
is the conversion factor.
39.37 in.
1 m
 ≈
1.24 m
1 stride length
39.37 in.
1 m
49 in.
Ryan’s stride length is approximately 49 inches.

61. Luis ran 400 meters in 297 strides. Find his stride length in inches.
Check It Out! Example 3
Luis ran 400 meters in 297 strides. Find his stride length in inches.
Use a proportion to find the length of his stride in meters.
400 m
297 strides
x m
1 stride =
Write both ratios in the form
meters
strides
400 = 297x
Find the cross products.
x ≈ 1.35 m

62. Check It Out! Example 3 Continued
Convert the stride length to inches.
is the conversion factor.
39.37 in.
1 m
 ≈
1.35 m
1 stride length
39.37 in.
1 m
53 in.
Luis’s stride length is approximately 53 inches.

63. Similar figures have the same shape but not necessarily the same size
Similar figures have the same shape but not necessarily the same size. Two figures are similar if their corresponding angles are congruent and corresponding sides are proportional.
The ratio of the corresponding side lengths of similar figures is often called the scale factor.
Reading Math

64. Example 4: Scaling Geometric Figures in the Coordinate Plane
∆XYZ has vertices X(0, 0), Y(–6, 9) and Z(0, 9).
∆XAB is similar to ∆XYZ with a vertex at B(0, 3).
Graph ∆XYZ and ∆XAB on the same grid.
Step 1 Graph ∆XYZ. Then draw XB.

65. Example 4 Continued
Step 2 To find the width of ∆XAB, use a proportion. =
height of ∆XAB width of ∆XAB
height of ∆XYZ width of ∆XYZ = x
9x = 18, so x = 2

66. To graph ∆XAB, first find the coordinate of A.
Example 4 Continued
Step 3
To graph ∆XAB, first find the coordinate of A. Y Z
The width is 2 units, and the height is 3 units, so the coordinates of A are (–2, 3). B A X

67. Check It Out! Example 4
∆DEF has vertices D(0, 0), E(–6, 0) and F(0, –4).
∆DGH is similar to ∆DEF with a vertex at G(–3, 0).
Graph ∆DEF and ∆DGH on the same grid.
Step 1 Graph ∆DEF. Then draw DG.

68. Check It Out! Example 4 Continued
Step 2 To find the height of ∆DGH, use a proportion. =
width of ∆DGH height of ∆DGH
width of ∆DEF height of ∆DEF = 3 6 4 x
6x = 12, so x = 2

69. Check It Out! Example 4 Continued
Step 3
To graph ∆DGH, first find the coordinate of H. ●
E(–6, 0)
F(0,–4)
G(–3, 0)
D(0, 0)
H(0, –2) ●
The width is 3 units, and the height is 2 units, so the coordinates of H are (0, –2).

70. Example 5: Nature Application
The tree in front of Luka’s house casts a 6-foot shadow at the same time as the house casts a 22-fot shadow. If the tree is 9 feet tall, how tall is the house?
Sketch the situation. The triangles formed by using the shadows are similar, so Luka can use a proportion to find h the height of the house.
9 ft
6 ft = 6 9 h 22 =
Shadow of tree
Height of tree
Shadow of house
Height of house
h ft
22 ft
6h = 198
h = 33
The house is 33 feet high.

71. Check It Out! Example 5
A 6-foot-tall climber casts a 20-foot long shadow at the same time that a tree casts a 90-foot long shadow. How tall is the tree?
Sketch the situation. The triangles formed by using the shadows are similar, so the climber can use a proportion to find h the height of the tree.
6 ft
20 ft = 20 6 h 90 =
Shadow of climber
Height of climber
Shadow of tree
Height of tree
h ft
90 ft
20h = 540
h = 27
The tree is 27 feet high.

72. Lesson Quiz: Part I
Solve each proportion. 2.
3. The results of a recent survey showed that 61.5% of those surveyed had a pet. If 738 people had pets, how many were surveyed?
4. Gina earned $68.75 for 5 hours of tutoring. Approximately how much did she earn per minute?
g = 42
k = 8
1200
$0.23

73. Lesson Quiz: Part II
5. ∆XYZ has vertices, X(0, 0), Y(3, –6), and Z(0, –6). ∆XAB is similar to ∆XYZ, with a vertex at B(0, –4). Graph ∆XYZ and ∆XAB on the same grid. Y Z A B X

74. Lesson Quiz: Part III
6. A 12-foot flagpole casts a 10 foot-shadow. At the same time, a nearby building casts a 48-foot shadow. How tall is the building?
57.6 ft