 Download Download Presentation 2-1

# 2-1

Download Presentation ## 2-1

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
##### Presentation Transcript

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

2. No; when 2 is substituted for x, the inequality is false: –4 < –6 Warm Up Simplify each expression. 2. –(w – 2) –w + 2 –x + 5 1. 2x + 5 – 3x 3. 6(2 – 3g) 12 – 18g Graph on a number line. 4.t> –2 –4 –3 –2 –1 0 1 2 3 4 5 5. Is 2 a solution of the inequality –2x < –6? Explain.

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.

7. 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.

8. 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?

9. 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 m + 0.07 *

10. –12.95 –12.95 Example 1 Continued Solve. 12.95 + 0.07m = 14.56 Subtract 12.95 from both sides. 0.07m = 1.61 Divide both sides by 0.07. 0.07 0.07 m = 23 Nina used 23 additional minutes.

11. 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?

12. 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 c 3.25 + 0.25 *

13. –3.25 –3.25 0.25 0.25 Check It Out! Example 1 Continued Solve. 3.25 + 0.25c = 14.00 Subtract 3.25 from both sides. 0.25c = 10.75 Divide both sides by 0.25. c = 43 44 cups fit between the 14 in. shelves.

14. 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 Divide both sides by 4. m + 12 = –9 Subtract 12 from both sides. m = –21

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

16. –48 –48 4m –84 = 4 4 Example 2 Continued Solve 4(m + 12) = –36 Method 2 Distribute before solving. 4m + 48 = –36 Distribute 4. Subtract 48 from both sides. 4m = –84 Divide both sides by 4. m = –21

17. 3 3 –2–2 –3 –3 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 Subtract 2 from both sides. –3p = 12 Divide both sides by –3. p = –4

18. Check 3(2 – 3p) = 42 3(2 + 12) 42 6 + 36 42 42 42  Check It Out! Example 2a Continued

19. –6 –6 –9 –9 –9p 36 = Check It Out! Example 2a Continued Solve 3(2 – 3p) = 42 . Method 2 Distribute before solving. 6 – 9p = 42 Distribute 3. Subtract 6 from both sides. –9p = 36 Divide both sides by –9. p = –4

20. –3(5 – 4r) –9 = –3 –3 –5–5 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. Divide both sides by –3. 5 – 4r = 3 Subtract 5 from both sides. –4r = –2

21. –4r –2 –4 –4 = r = Check –3(5 –4r) = –9 –3(5 – 4•) –9 –3(5 – 2) –9 –3(3) –9 –9 –9  Check It Out! Example 2b Continued Solve –3(5 – 4r) = –9. Method 1 Divide both sides by –4.

22. 12r 6 1212 = r = Check It Out! Example 2b Continued Solve –3(5 – 4r) = –9. Method 2 Distribute before solving. –15 + 12r = –9 Distribute 3. +15 +15 Add 15 to both sides. 12r = 6 Divide both sides by 12.

23. 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.

24. +11k +11k +10 + 10 5 5 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 Collect variables on the right side. 25 = 5k – 10 Add. Collect constants on the left side. 35 = 5k Isolate the variable. 7 = k

25. +2w +2w –12 –12 3 3 Check It Out! Example 3 Solve3(w + 7) – 5w = w + 12. Simplify each side by combining like terms. –2w+ 21 = w + 12 Collect variables on the right side. 21 = 3w + 12 Add. Collect constants on the left side. 9 = 3w Isolate the variable. 3 = w

26. 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.

27. + 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 Simplify. –9 ≠ –5 x Contradiction

28. –2x –2x Example 4B: Identifying Identities and Contractions Solve 2(x – 6) = –5x – 12 + 7x. 2(x – 6) = –5x – 12 + 7x Simplify. 2x – 12 = 2x – 12 –12 = –12 Identity  The solutions set is all real number, or.

29. –5x –5x The equation has no solution. The solution set is the empty set, which is represented by the symbol . Check It Out! Example 4a Solve 5(x – 6) = 3x – 18 + 2x. 5(x – 6) = 3x – 18 + 2x Simplify. 5x – 30 = 5x – 18 –30 ≠ –18 x Contradiction

30. + 9x +9x Check It Out! Example 4b Solve 3(2 –3x) = –7x – 2(x –3). 3(2 –3x) = –7x – 2(x –3) 6 – 9x = –9x + 6 Simplify. 6 = 6 Identity  The solutions set is all real numbers, or.

31. 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.

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

33. 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. +2 +2 –5a ≥ 10 Divide both sides by –5 and reverse the inequality. –5a ≤ 10 –5–5 a ≤ –2

34. –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.

35. 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 –17 –9 ≥ 3x Divide both sides by 3. –9 ≥ 3x 33 –3 ≥ x or x ≤ –3

36. –6 –5 –4 –3 –2 –1 0 1 2 3 Check It Out! Example 5 Continued Solve and graph x + 8 ≥ 4x + 17. Check Test values in the original inequality. • Test x = –6 Test x = –3 Test x = 0 –6+ 8 ≥ 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.

37. 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

38. 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) no solution, or Lesson Quiz: Part II x = 6 y = –4 all real numbers, or 

39. –2 –1 0 1 2 3 4 5 6 7 Lesson Quiz: Part III 5. Solve and graph. 12 + 3q > 9q – 18 q < 5 °