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

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

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Holt McDougal Algebra 2 2-1 Solving Linear Equations and Inequalities Solve linear equations using a variety of methods. Solve linear inequalities. Objectives

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Holt McDougal Algebra 2 2-1 Solving Linear Equations and Inequalities equation solution set of an equation linear equation in one variable identify contradiction inequality Vocabulary

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Holt McDougal Algebra 2 2-1 Solving Linear Equations and Inequalities 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.

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Holt McDougal Algebra 2 2-1 Solving Linear Equations and Inequalities Linear Equations in One variable Nonlinear Equations 4x = 8 3x – = –9 2x – 5 = 0.1x +2 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. + 1 = 32 + 1 = 41 3 – 2 x = –5

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Holt McDougal Algebra 2 2-1 Solving Linear Equations and Inequalities

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Holt McDougal Algebra 2 2-1 Solving Linear Equations and Inequalities 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.

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Holt McDougal Algebra 2 2-1 Solving Linear Equations and Inequalities The local phone company charges $12.95 a month for the first 200 of air time, plus $0.07 for each additional minute. If Ninas bill for the month was $14.56, how many additional minutes did she use? Example 1: Consumer Application

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Holt McDougal Algebra 2 2-1 Solving Linear Equations and Inequalities Example 1 Continued monthly charge plus additional minute charge times 12.95 0.07 number of additional minutes total charge + = Let m represent the number of additional minutes that Nina used. m 14.56 * = Model

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Holt McDougal Algebra 2 2-1 Solving Linear Equations and Inequalities Solve. 12.95 + 0.07m = 14.56 0.07m = 1.61 0.07 m = 23 Subtract 12.95 from both sides. Divide both sides by 0.07. Nina used 23 additional minutes. Example 1 Continued –12.95

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Holt McDougal Algebra 2 2-1 Solving Linear Equations and Inequalities 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?

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Holt McDougal Algebra 2 2-1 Solving Linear Equations and Inequalities Check It Out! Example 1 Continued Let c represent the number of additional cups needed. one cup plus additional cup height times 3.25 0.25 number of additional cups total height + = c 14.00 * = Model

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Holt McDougal Algebra 2 2-1 Solving Linear Equations and Inequalities Check It Out! Example 1 Continued 3.25 + 0.25c = 14.00 0.25c = 10.75 0.25 c = 43 44 cups fit between the 14 in. shelves. Solve. Subtract 3.25 from both sides. Divide both sides by 0.25. –3.25

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

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Holt McDougal Algebra 2 2-1 Solving Linear Equations and Inequalities Check4(m + 12) = –36 4(–21 + 12) –36 4(–9) –36 –36 Example 2 Continued

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Holt McDougal Algebra 2 2-1 Solving Linear Equations and Inequalities Example 2 Continued Distribute 4. Distribute before solving. 4m + 48 = –36 4m = –84 –48 –48 Subtract 48 from both sides. Divide both sides by 4. = 4m –84 4 4 m = –21 Solve 4(m + 12) = –36 Method 2

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Holt McDougal Algebra 2 2-1 Solving Linear Equations and Inequalities Divide both sides by 3. Method 1 The quantity (2 – 3p) is multiplied by 3, so divide by 3 first. 3(2 – 3p) = 42 3 Check It Out! Example 2a Solve 3(2 –3p) = 42. Subtract 2 from both sides. –3p = 12 2 – 3p = 14 –2 –2 –3 Divide both sides by –3. p = –4

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Holt McDougal Algebra 2 2-1 Solving Linear Equations and Inequalities Check3(2 – 3p) = 42 3(2 + 12) 42 6 + 36 42 42 Check It Out! Example 2a Continued

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Holt McDougal Algebra 2 2-1 Solving Linear Equations and Inequalities Distribute 3. Method 2 Distribute before solving. 6 – 9p = 42 –9p = 36 –6 Subtract 6 from both sides. Divide both sides by –9. = –9p 36 –9 p = –4 Check It Out! Example 2a Continued Solve 3(2 – 3p) = 42.

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Holt McDougal Algebra 2 2-1 Solving Linear Equations and Inequalities Divide both sides by – 3. Method 1 The quantity (5 – 4r) is multiplied by –3, so divide by –3 first. –3(5 – 4r) –9 –3 = Check It Out! Example 2b Solve –3(5 – 4r) = –9. Subtract 5 from both sides. –4r = –2 5 – 4r = 3 –5 –5

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Holt McDougal Algebra 2 2-1 Solving Linear Equations and Inequalities Check It Out! Example 2b Continued Divide both sides by –4. = –4 –4r –2 r = –9 Check–3(5 –4r) = –9 –3(5 – 4 ) –9 –3(5 – 2) –9 –3(3) –9 Solve –3(5 – 4r) = –9. Method 1

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Holt McDougal Algebra 2 2-1 Solving Linear Equations and Inequalities Distribute 3. Distribute before solving. –15 + 12r = –9 12r = 6 +15 +15 Add 15 to both sides. Divide both sides by 12. = 12r 6 12 Check It Out! Example 2b Continued r = Solve –3(5 – 4r) = –9. Method 2

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Holt McDougal Algebra 2 2-1 Solving Linear Equations and Inequalities 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.

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Holt McDougal Algebra 2 2-1 Solving Linear Equations and Inequalities Example 3: Solving Equations with Variables on Both Sides Simplify each side by combining like terms. –11k + 25 = –6k – 10 Collect variables on the right side. Add. Collect constants on the left side. Isolate the variable. +11k 25 = 5k – 10 35 = 5k 5 7 = k +10 + 10 Solve 3k– 14k + 25 = 2 – 6k – 12.

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Holt McDougal Algebra 2 2-1 Solving Linear Equations and Inequalities Check It Out! Example 3 Solve 3(w + 7) – 5w = w + 12. Simplify each side by combining like terms. –2w + 21 = w + 12 Collect variables on the right side. Add. Collect constants on the left side. Isolate the variable. +2w 21 = 3w + 12 9 = 3w 3 3 = w –12

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Holt McDougal Algebra 2 2-1 Solving Linear Equations and Inequalities 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.

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Holt McDougal Algebra 2 2-1 Solving Linear Equations and Inequalities Solve 3v – 9 – 4v = –(5 + v). Example 4A: Identifying Identities and Contractions 3v – 9 – 4v = –(5 + v) Simplify. –9 – v = –5 – v + v + v –9 –5x Contradiction The equation has no solution. The solution set is the empty set, which is represented by the symbol.

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Holt McDougal Algebra 2 2-1 Solving Linear Equations and Inequalities Solve 2(x – 6) = –5x – 12 + 7x. Example 4B: Identifying Identities and Contractions 2(x – 6) = –5x – 12 + 7x Simplify. 2x – 12 = 2x – 12 –2x –12 = –12 Identity The solutions set is all real number, or.

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Holt McDougal Algebra 2 2-1 Solving Linear Equations and Inequalities Solve 5(x – 6) = 3x – 18 + 2x. The equation has no solution. The solution set is the empty set, which is represented by the symbol. Check It Out! Example 4a 5(x – 6) = 3x – 18 + 2x Simplify. 5x – 30 = 5x – 18 –5x –30 –18x Contradiction

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Holt McDougal Algebra 2 2-1 Solving Linear Equations and Inequalities Solve 3(2 –3x) = –7x – 2(x –3). 3(2 –3x) = –7x – 2(x –3) Simplify. 6 – 9x = –9x + 6 + 9x +9x 6 = 6 Identity The solutions set is all real numbers, or. Check It Out! Example 4b

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Holt McDougal Algebra 2 2-1 Solving Linear Equations and Inequalities 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.

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Holt McDougal Algebra 2 2-1 Solving Linear Equations and Inequalities These properties also apply to inequalities expressed with >,, and.

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Holt McDougal Algebra 2 2-1 Solving Linear Equations and Inequalities To check an inequality, test the value being compared with x a value less than that, and a value greater than that. Helpful Hint

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Holt McDougal Algebra 2 2-1 Solving Linear Equations and Inequalities Solve and graph 8a –2 13a + 8. Example 5: Solving Inequalities Subtract 13a from both sides. 8a – 2 13a + 8 –13a –5a – 2 8 Add 2 to both sides. +2 +2 –5a 10 Divide both sides by –5 and reverse the inequality. –5 –5 –5a 10 a –2

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Holt McDougal Algebra 2 2-1 Solving Linear Equations and Inequalities Example 5 Continued Check Test values in the original inequality. –10 –9 –8 –7 –6 –5 –4 –3 –2 –1 Test x = –4Test x = –2Test x = –1 8(–4) – 2 13(–4) + 88(–2) – 2 13(–2) + 88(–1) – 2 13(–1) + 8 –34 –44 So –4 is a solution. So –1 is not a solution. So –2 is a solution. –18 –10 –5 x Solve and graph 8a – 2 13a + 8.

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Holt McDougal Algebra 2 2-1 Solving Linear Equations and Inequalities Solve and graph x + 8 4x + 17. Subtract x from both sides. x + 8 4x + 17 –x 8 3x +17 Subtract 17 from both sides. –17 –9 3x Divide both sides by 3. 3 –9 3x –3 x or x –3 Check It Out! Example 5

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Holt McDougal Algebra 2 2-1 Solving Linear Equations and Inequalities Check Test values in the original inequality. Test x = –6Test x = –3Test x = 0 –6 + 8 4(–6) + 17 –3 +8 4(–3) + 17 0 +8 4(0) + 17 2 –7 So –6 is a solution. So 0 is not a solution. So –3 is a solution. 58 17 x Check It Out! Example 5 Continued Solve and graph x + 8 4x + 17. –6 –5 –4 –3 –2 –1 0 1 2 3

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Holt McDougal Algebra 2 2-1 Solving Linear Equations and Inequalities 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 companys pay-per-view service. If his bill last month was $32.49, how many movies did Alex order? 5 movies

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Holt McDougal Algebra 2 2-1 Solving Linear Equations and Inequalities Lesson Quiz: Part II y = –4 x = 6 all real numbers, or 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

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Holt McDougal Algebra 2 2-1 Solving Linear Equations and Inequalities Lesson Quiz: Part III 5. Solve and graph. 12 + 3q > 9q – 18q < 5 –2 –1 0 1 2 3 4 5 6 7 °

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