1. Chapter 4 Inequalities and Problem Solving

2. § 4.1 Solving Linear Inequalities

3. Blitzer, Intermediate Algebra, 5e – Slide #3 Section 4.1 Linear Inequalities p 241 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 [] )[ (] ( [ ) ] X is greater than a (a<x) and x is less than b (x<b) The parentheses in the graph and in interval notation indicate that a and b, the endpoints, are excluded from the interval.

4. Blitzer, Intermediate Algebra, 5e – Slide #4 Section 4.1 Linear Inequalities p 241 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 [] )[ (] ( [ ) ] X is greater than or equal to a [a x] and x is less than or equal to b (x b) The square brackets in the graph and in interval notation indicate that a and b, the endpoints, are included in the interval.

5. Blitzer, Intermediate Algebra, 5e – Slide #5 Section 4.1 Linear Inequalities p 241 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 [] )[ (] ( [ ) ]

6. Blitzer, Intermediate Algebra, 5e – Slide #6 Section 4.1 Linear Inequalities p 242EXAMPLE 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 -3.9999 and -3.3 are. The arrow shows that the graph extends indefinitely to the right. -4 ( {x|x > -4} = (-4, )

7. Blitzer, Intermediate Algebra, 5e – Slide #7 Section 4.1 Linear Inequalities p 242CONTINUED ) (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

8. Blitzer, Intermediate Algebra, 5e – Slide #8 Section 4.1 Linear Inequalities p 242 Check Point 1 ) [) [ )

9. Blitzer, Intermediate Algebra, 5e – Slide #9 Section 4.1 Inequalities p 243 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

10. Blitzer, Intermediate Algebra, 5e – Slide #10 Section 4.1 Inequalities p 243 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

11. Blitzer, Intermediate Algebra, 5e – Slide #11 Section 4.1 Inequalities p 243 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

12. Blitzer, Intermediate Algebra, 5e – Slide #12 Section 4.1 Inequalities p 243 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.

13. Blitzer, Intermediate Algebra, 5e – Slide #13 Section 4.1 Linear Inequalities p 244EXAMPLE 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.

14. Blitzer, Intermediate Algebra, 5e – Slide #14 Section 4.1 Linear Inequalities 3) Isolate the variable and solve. -1 0 1 2 3 4 5 6 [ 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, )

15. Blitzer, Intermediate Algebra, 5e – Slide #15 Section 4.1 Linear Inequalities p 244 - 245 Check Point 2 (complete on board) Check Point 3 (complete on board)

16. Blitzer, Intermediate Algebra, 5e – Slide #16 Section 4.1 Linear Inequalities p 246EXAMPLE 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

17. Blitzer, Intermediate Algebra, 5e – Slide #17 Section 4.1 Linear Inequalities Add x to both sides 1 1 1 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

18. Blitzer, Intermediate Algebra, 5e – Slide #18 Section 4.1 Linear InequalitiesCONTINUED 3) Isolate the variable and solve. -6 -5 -4 -3 -2 -1 0 1 2 [ 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, )

19. Blitzer, Intermediate Algebra, 5e – Slide #19 Section 4.1 Linear Inequalities p 247 Check Point 4 (Complete on board)

20. Blitzer, Intermediate Algebra, 5e – Slide #20 Section 4.1 Linear Inequalities EXAMPLE OF NO SOLUTION 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

21. Blitzer, Intermediate Algebra, 5e – Slide #21 Section 4.1 Linear Inequalities p 248 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.

22. Blitzer, Intermediate Algebra, 5e – Slide #22 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.

23. Blitzer, Intermediate Algebra, 5e – Slide #23 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 15 + 0.08x < 3 + 0.12x Plan A is less than Plan B

24. Blitzer, Intermediate Algebra, 5e – Slide #24 Section 4.1 Linear Inequalities 4) Solve the inequality and answer the question. Subtract 3 from both sides CONTINUED 15 + 0.08x < 3 + 0.12x 12 + 0.08x < 0.12x Subtract 0.08x from both sides12 < 0.04x Divide both sides by 0.04300 < x Therefore, Plan A costs less than Plan B when x > 300 long- distance minutes per month.

25. Blitzer, Intermediate Algebra, 5e – Slide #25 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 300 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

26. Blitzer, Intermediate Algebra, 5e – Slide #26 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).

27. DONE