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© 2010 Pearson Prentice Hall. All rights reserved. CHAPTER 6 Algebra: Equations and Inequalities

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© 2010 Pearson Prentice Hall. All rights reserved. 2 6.2 Linear Equations in One Variable and Proportions

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© 2010 Pearson Prentice Hall. All rights reserved. 3 Objectives 1.Solve linear equations. 2.Solve linear equations containing fractions. 3.Solve proportions. 4. Solve problems using proportions. 5.Identify equations with no solution or infinitely many solutions.

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© 2010 Pearson Prentice Hall. All rights reserved. 4 Linear Equation A linear equation in one variable x is an equation that can be written in the form ax + b = 0, where a and b are real numbers, and a 0. Solving an equation in x involves determining all values of x that result in a true statement when substituted into the equation. Such values are solutions or roots. Equivalent equations have the same solution set. 4x + 12 = 0 and x = 3 are equivalent equations.

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© 2010 Pearson Prentice Hall. All rights reserved. 5 The Addition and Multiplication Properties of Equality The Addition Property of Equality The same real number or algebraic expression may be added to both sides of an equation without changing the equation’s solution set. a = b and a + c = b + c are equivalent equations. The Multiplication Property of Equality The same nonzero real number may multiply both sides of an equation without changing the equation’s solution set. a = b and ac = bc are equivalent equations.

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© 2010 Pearson Prentice Hall. All rights reserved. 6 Using Properties of Equality to Solve Equations

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© 2010 Pearson Prentice Hall. All rights reserved. 7 Solving a Linear Equation 1.Simplify the algebraic expression on each side by removing grouping symbols and combining like terms. 2.Collect all the variable terms on one side and all the constants, or numerical terms, on the other side. 3.Isolate the variable and solve. 4.Check the proposed solution in the original equation.

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© 2010 Pearson Prentice Hall. All rights reserved. 8 Example 2: Solving a Linear Equation Solve and check: 2(x – 4) – 5x = 5. Step 1. Simplify the algebraic expression on each side: 2(x – 4) – 5x = 5 This is the given equation. 2x – 8 – 5x = 5 Use the distributive property. 3x – 8 = 5 Combine like terms: 2x – 5x = 3x. Step 2. Collect variable terms on one side and constants on the other side. 3x – 8 + 8 = 5 + 8 Add 8 to both sides. 3x = 3 Simplify.

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© 2010 Pearson Prentice Hall. All rights reserved. 9 Example 1 continued Step 3. Isolate the variable and solve. 3x = 3 Divide both sides by 3. 3 3 Simplify. x = 1 Step 4. Check the proposed solution in the original equation by substituting 1 for x. 2(x – 4) – 5x = 5 2( 1 – 4) – 5( 1) = 5 10 – ( 5) = 5 5 = 5 This statement is true. Because the check results in a true statement, we conclude that the solution set of the given equation is { 1}.

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© 2010 Pearson Prentice Hall. All rights reserved. 10 Example 6: An Application: Responding to Negative Life Events These graphs indicate that persons with a low sense of humor have higher levels of depression. These graphs can be modeled by the following formulas:

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© 2010 Pearson Prentice Hall. All rights reserved. 11 Example 2: continued We are interested in the intensity of a negative life event with an average level of depression of 7/2 for the high humor group.

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© 2010 Pearson Prentice Hall. All rights reserved. 12 Example 10: Linear Equations with No Solution Solve: 2x + 6 = 2(x + 4) Solution: 2x + 6 = 2(x + 4) 2x + 6 = 2x + 8 2x + 6 – 2x = 2x + 8 – 2x 6 = 8 The original equation 2x + 6 = 2(x + 4) is equivalent to 6 = 8, which is false for every value of x. The equation has no solution. The solution set is Ø, the empty set.

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© 2010 Pearson Prentice Hall. All rights reserved. 13 Example 11 : Linear Equations with Infinitely Many Solutions Solve: 4x + 6 = 6(x + 1) – 2x Solution: 4x + 6 = 6(x + 1) – 2x 4x + 6 = 6x + 6 – 2x 4x + 6 = 4x + 6 The original statement is equivalent to the statement 6 = 6, which is true for every value of x. The solution set is the set of all real numbers, expressed as {x|x is a real number}.

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© 2010 Pearson Prentice Hall. All rights reserved. Proportions The property tax on a house with an assessed value of $480,000 is $5760. Determine the property tax no a house with an assessed value of $600,000, assuming the same tax rate. $5760 x ------------ = ------------ $480,000 $600,000 5760 x ------------ = ------------ 480,000 600,000 14

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© 2010 Pearson Prentice Hall. All rights reserved. 5760 x ------------ = ------------ 480,000 600,000 5760 x ----------- = -------- 48 60 5760 ∙ 60 ------------- = x 48 5 1440 15 5760 ∙ 60 ------------ = x 48 12 4 1 7200 = x $7200 = x 15

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© 2010 Pearson Prentice Hall. All rights reserved. 16 A chocolate-chip recipe for five dozen cookies requires ¾ cup of sugar. If you want to make eight dozen cookies, how much sugar is needed? Solution: 3 / 4 (cup) x (cup) ----------------- = -------------- 5 (dozen) 8 (dozen) 8 (3/4) 8 3 6 ---------- = x --- ∙ --- = --- = x 5 5 4 5 Example 15: Changing the Size of a Recipe

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© 2010 Pearson Prentice Hall. All rights reserved. Proportion 17 x 60” 240” 960” Find the height x of the tree, when a 60”-man casts a shadow 240” long.

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© 2010 Pearson Prentice Hall. All rights reserved. Proportion 18 x 60” 240” 960” 60 x ----- = ------- 240 960 60 ∙ 960 ----------- = x x = 240 240

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