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Chapter 4 Inequalities and Problem Solving

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§ 4.1 Solving Linear Inequalities

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Blitzer, Intermediate Algebra, 5e – Slide #3 Section 4.1 Linear Inequalities Intervals on the Real Number Line Let a and b be real numbers such that a < b. Interval Notation Set-Builder NotationGraph (a,b)(a,b){x|a < x < b} [a,b][a,b]{x|a x b} [a,b)[a,b){x|a x < b} (a,b](a,b]{x|a < x b} (a, ){x|x > a} [a, ){x|x a} (-,b){x|x < b} (-,b]{x|x b} (-, ){x|x is a real number} ab () ab ab ab a b a b [] )[ (] ( [ ) ]

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Blitzer, Intermediate Algebra, 5e – Slide #4 Section 4.1 Linear InequalitiesEXAMPLE SOLUTION Graph the solution of each inequality on a number line and then express the solutions in set-builder notation and interval notation. (a) The solution to x > -4 is all real numbers that are greater than -4. They are graphed on a number line by shading all points to the right of -4. The parenthesis at -4 indicates that -4 is not a solution, but numbers such as and -3.3 are. The arrow shows that the graph extends indefinitely to the right. -4 ( {x|x > -4} = (-4, )

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Blitzer, Intermediate Algebra, 5e – Slide #5 Section 4.1 Linear InequalitiesCONTINUED ) (b) The solution to is all real numbers that are greater than -2 and less than 7. They are graphed on a number line by shading all points that are to the right of -2 and to the left of 7. The bracket at -2 indicates that -2 is part of the solution. The parenthesis at 7 indicates that 7 is not part of the solution. [ {x|-2 x < 7} = [-2,7) -27

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Blitzer, Intermediate Algebra, 5e – Slide #6 Section 4.1 Inequalities Properties of Inequalities PropertyThe Property in WordsExample The Addition Property of Inequality If a < b, then a + c < b + c If a < b, then a – c < b - c If the same quantity is added to or subtracted from both sides of an inequality, the resulting inequality is equivalent to the original one. 2x + 3 < 7 Subtract 3: 2x + 3 – 3 < 7 – 3 Simplify: 2x < 4

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Blitzer, Intermediate Algebra, 5e – Slide #7 Section 4.1 Inequalities Properties of Inequalities PropertyThe Property in WordsExample The Positive Multiplication Property of Inequality If a < b and c is positive, then ac < bc If a < b and c is positive, then If we multiply or divide both sides of an inequality by the same positive quantity, the resulting inequality is equivalent to the original one. 2x < 4 Divide by 2: Simplify: x < 2 CONTINUED

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Blitzer, Intermediate Algebra, 5e – Slide #8 Section 4.1 Inequalities Properties of Inequalities PropertyThe Property in WordsExample The Negative Multiplication Property If a bc If a < b and c is negative, then If we multiply or divide both sides of an inequality by the same negative quantity and reverse the direction of the inequality symbol, the resulting inequality is equivalent to the original one. -4x < 20 Divide by -4 and reverse the sense of the inequality: Simplify: x > -5 CONTINUED Note that when you multiply both sides of an inequality by a negative, you must reverse the direction of the inequality symbol

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Blitzer, Intermediate Algebra, 5e – Slide #9 Section 4.1 Inequalities Solving a Linear Inequality 1) Simplify the algebraic expression on each side. 2) Use the addition property of inequality to collect all the variable terms on one side and all the constant terms on the other side. 3) Use the multiplication property of inequality to isolate the variable and solve. Reverse the sense of the inequality when multiplying or dividing both sides by a negative number. 4) Express the solution set in set-builder or interval notation and graph the solution set on a number line.

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Blitzer, Intermediate Algebra, 5e – Slide #10 Section 4.1 Linear InequalitiesEXAMPLE SOLUTION Solve the linear inequality. Then graph the solution set on a number line. Distribute Add 5x to both sides Add 1 to both sides 1) Simplify each side. 2) Collect variable terms on one side and constant terms on the other side.

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Blitzer, Intermediate Algebra, 5e – Slide #11 Section 4.1 Linear Inequalities 3) Isolate the variable and solve [ Divide both sides by 8 CONTINUED 4) Express the solution set in set-builder or interval notation and graph the set on a number line. {x|x 2} = [2, )

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Blitzer, Intermediate Algebra, 5e – Slide #12 Section 4.1 Linear InequalitiesEXAMPLE SOLUTION Solve the linear inequality. Then graph the solution set on a number line. Distribute First we need to eliminate the denominators. Multiply by LCD = 10

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Blitzer, Intermediate Algebra, 5e – Slide #13 Section 4.1 Linear Inequalities Add x to both sides Subtract 10 from both sides 1) Simplify each side. Because each side is already simplified, we can skip this step. 2) Collect variable terms on one side and constant terms on the other side. CONTINUED 1 2 1

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Blitzer, Intermediate Algebra, 5e – Slide #14 Section 4.1 Linear InequalitiesCONTINUED 3) Isolate the variable and solve [ Divide both sides by 4 4) Express the solution set in set-builder or interval notation and graph the set on a number line. {x|x -2} = [-2, )

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Blitzer, Intermediate Algebra, 5e – Slide #15 Section 4.1 Linear InequalitiesEXAMPLE SOLUTION Solve the linear inequality. Distribute Since the result is an obviously false statement, there is no solution. 5x < 5(x – 3) 5x < 5x – 15 Subtract 5x from both sides0 < – 15

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Blitzer, Intermediate Algebra, 5e – Slide #16 Section 4.1 Linear Inequalities English Sentences and Inequalities English SentenceInequality x is at least 5. x is at most 5. x is between 5 and 7. x is no more than 5. x is no less than 5.

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Blitzer, Intermediate Algebra, 5e – Slide #17 Section 4.1 Linear InequalitiesEXAMPLE SOLUTION You are choosing between two long-distance telephone plans. Plan A has a monthly fee of $15 with a charge of $0.08 per minute for all long-distance calls. Plan B has a monthly fee of $3 with a charge of $0.12 per minute for all long-distance calls. How many minutes of long-distance calls in a month make plan A the better deal? 1) Let x represent one of the quantities. We are looking for the number of minutes of long-distance calls that make Plan A the better deal. Thus, Let x = the number of minutes of long-distance phone calls.

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Blitzer, Intermediate Algebra, 5e – Slide #18 Section 4.1 Linear Inequalities 2) Represent other quantities in terms of x. We are not asked to find another quantity, so we can skip this step. 3) Write an inequality in x that describes the conditions. Plan A is a better deal than Plan B if the monthly cost of Plan A is less than the monthly cost of Plan B. The inequality that represents this is the following, with the information for Plan A on the left side and the information for Plan B on the right side of the inequality. CONTINUED x < x Plan A is less than Plan B

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Blitzer, Intermediate Algebra, 5e – Slide #19 Section 4.1 Linear Inequalities 4) Solve the inequality and answer the question. Subtract 3 from both sides CONTINUED x < x x < 0.12x Subtract 0.08x from both sides15 < 0.04x Divide both sides by < x Therefore, Plan A costs less than Plan B when x > 375 long- distance minutes per month.

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Blitzer, Intermediate Algebra, 5e – Slide #20 Section 4.1 Linear Inequalities 5) Check the proposed solution in the original wording of the problem. One way to do this is to take a long-distance number of minutes greater than 375 per month to see if Plan A is the better deal. Suppose that the number of long-distance minutes we use is 450 in a month. CONTINUED Cost for Plan A = 15 + (450)(0.08) = $51 Plan A has a lower monthly cost, making Plan A the better deal. Cost for Plan B = 3 + (450)(0.12) = $57

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Blitzer, Intermediate Algebra, 5e – Slide #21 Section 4.1 Linear Inequalities In summary… Solving a linear inequality is similar to solving a linear equation except for two important differences: When you solve an inequality, your solution is usually an interval, not a point. and When you multiply both sides of an inequality by a negative number, you must remember to reverse the direction of the inequality sign… that is, POW (point the other way).

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