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

1. 2-1 Objectives Solve linear equations using a variety of methods. Solve linear inequalities.

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

3. 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.

4. Linear Equations in One variable Nonlinear Equations 4x = 8 + 1 = 32 3x – = –9 + 1 = 41 2x – 5 = 0.1x +2 3 – 2x = –5

5. 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?

6. –12.95 –12.95 Example 1 Solve. 12.95 + 0.07m = 14.56 0.07m = 1.61 0.07 0.07 m = 23 Nina used 23 additional minutes.

7. 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?

8. –3.25 –3.25 0.25 0.25 Check It Out! Example 1 Continued Solve. 3.25 + 0.25c = 14.00 0.25c = 10.75 c = 43 44 cups fit between the 14 in. shelves.

9. 4 4 –12–12 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 m + 12 = –9 m = –21

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

11. –48 –48 4m –84 = 4 4 Example 2 Continued Solve 4(m + 12) = –36 Method 2 Distribute before solving. 4m + 48 = –36 4m = –84 m = –21

12. 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.

13. +11k +11k +10 + 10 5 5 Example 3: Solving Equations with Variables on Both Sides Solve 3k– 14k + 25 = 2 – 6k – 12. –11k + 25 = –6k– 10 25 = 5k – 10 35 = 5k 7 = k

14. 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.

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

16. –2x –2x Example 4B: Identifying Identities and Contractions Solve 2(x – 6) = –5x – 12 + 7x. 2(x – 6) = –5x – 12 + 7x 2x – 12 = 2x – 12 –12 = –12 Identity  The solutions set is all real number, or.

17. An inequalityis 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.

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

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

20. –10 –9 –8 –7 –6 –5 –4 –3 –2 –1 Example 5: Solving Inequalities Solve and graph 8a –2 ≥ 13a + 8. 8a – 2 ≥ 13a + 8 –13a –13a –5a – 2 ≥ 8 +2 +2 • –5a ≤ 10 –5–5 a ≤ –2

21. –10 –9 –8 –7 –6 –5 –4 –3 –2 –1 Example 5 Continued Solve and graph 8a –2 ≥ 13a + 8. Check Test values in the original inequality. • 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 ≥ –5x So –4 is a solution. So –2 is a solution. So –1 is not a solution.

22. 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

23. 2-2 Objective Apply proportional relationships to rates, similarity, and scale.

24. Vocabulary ratio proportion rate similar indirect measurement

25. 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.

26. 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.

27. 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.

28. 16 24 = 14c 1624 206.4 24p p 12.9 = = = = 88132 p12.9 2424 88c 1848 Example 1: Solving Proportions Solve each proportion. 14 c = A. B. 88132 206.4 = 24p 88c =1848 88 88 8.6 = p c = 21

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

30. 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).

31. 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 22.5(1800) = 100x 405 = x x = 405 So 405 voters are planning to vote for that candidate.

32. 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).

33. 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.

34. 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.

35. Write both ratios in the form . meters strides 600 m 482 strides x m 1 stride = Example 3: Fitness Application Ryan ran 600 meters and counted 482 strides. How long is Ryan’s stride in inches? (Hint: 1 m ≈ 39.37 in.) Use a proportion to find the length of his stride in meters. 600 = 482x x ≈ 1.24 m

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

37. Write both ratios in the form . meters strides 400 m 297 strides x m 1 stride = 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 = 297x x ≈1.35 m

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

39. Reading Math The ratio of the corresponding side lengths of similar figures is often called the scale factor. 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.

40. Step 1 Graph ∆XYZ. Then draw XB. 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. y z B x

41. = = height of ∆XAB width of ∆XAB height of ∆XYZ width of ∆XYZ 3x 9 6 Example 4 Continued Step 2To find the width of ∆XAB, use a proportion. 9x = 18 x = 2

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

43. = h ft 9 ft 6 ft 6 22 Shadow of tree Height of tree Shadow of house Height of house 22 ft = 9 h 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-foot 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. 6h = 198 h = 33 The house is 33 feet high.