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Solving and Graphing Linear Inequalities - Helpful Tips and Examples

Learn how to solve and graph linear inequalities with helpful tips and examples. Understand interval notation and properties of inequalities.

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Solving and Graphing Linear Inequalities - Helpful Tips and Examples

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  1. 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.

  2. 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.

  3. A set can be finite like the set of billiard ball numbers or infinite like the natural numbers {1, 2, 3, 4 …}. Afinite set has a definite, or finite, number of elements. An infinite set has an unlimited, or infinite number of elements. Helpful Hint The Density Property states that between any two numbers there is another real number. So any interval that includes more than one point contains infinitely many points.

  4. Many infinite sets, such as the real numbers, cannot be represented in roster notation. There are other methods of representing these sets. For example, the number line represents the sets of all real numbers. The set of real numbers between 3 and 5, which is also an infinite set, can be represented on a number line or by an inequality. -2 -1 0 1 2 3 4 5 6 7 8 3<x<5

  5. An interval is the set of all numbers between two endpoints, such as 3 and 5. In interval notation the symbols [ and ] are used to include an endpoint in an interval, and the symbols ( and ) are used to exclude an endpoint from an interval. -2 -1 0 1 2 3 4 5 6 7 8 (3,5) The set of real numbers between but not including 3 and 5. 3<x<5

  6. An interval that extends forever in the positive direction goes to infinity (∞), and an interval that extends forever in the negative direction goes to negative infinity (–∞). ∞ –∞ -5 0 5

  7. Because ∞ and –∞ are not numbers, they cannot be included in a set of numbers, so parentheses are used to enclose them in an interval. The table shows the relationship among some methods of representing intervals.

  8. Example 2A: Interval Notation Use interval notation to represent the set of numbers. 7 < x ≤ 12 (7, 12] 7 is not included, but 12 is.

  9. Example 2B: Interval Notation Use interval notation to represent the set of numbers. –6 –4 –2 0 2 4 6 There are two intervals graphed on the number line. [–6, –4] –6 and –4 are included. 5 is not included, and the interval continues forever in the positive direction. (5, ∞) The word “or” is used to indicate that a set includes more than one interval. [–6, –4] or (5, ∞)

  10. Check It Out! Example 2 Use interval notation to represent each set of numbers. a. -4 -3 -2 -1 0 1 2 3 4 –1 is included, and the interval continues forever in the negative direction. (–∞, –1] b. x ≤ 2 or 3 < x ≤ 11 (–∞, 2] 2 is included, and the interval continues forever in the negative direction. (3, 11] 3 is not included, but 11 is. (–∞, 2] or (3, 11]

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

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

  13. 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

  14. –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.

  15. 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

  16. –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.

  17. –2 –1 0 1 2 3 4 5 6 7 Lesson Quiz: Part III 5. Solve and graph. 12 + 3q > 9q – 18 q < 5 °

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

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

  20. 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.

  21. 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.

  22. 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?

  23. 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 *

  24. –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.

  25. 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

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

  27. –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

  28. 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

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

  30. –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

  31. –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

  32. –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.

  33. 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.

  34. 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.

  35. +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

  36. +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

  37. 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.

  38. + 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

  39. –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.

  40. Lesson Quiz: Part II x = 6 2. 2(3x – 1) = 34 3. 4y – 9 – 6y = 2(y + 5) – 3 4.r + 8 – 5r = 2(4 – 2r) y = –4 all real numbers, or 

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