Download presentation

1
**Chapter 4 Inequalities and Problem Solving**

2
§ 4.1 Solving Linear Inequalities

3
**Intervals on the Real Number Line**

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

4
Linear Inequalities EXAMPLE Graph the solution of each inequality on a number line and then express the solutions in set-builder notation and interval notation. SOLUTION (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. {x|x > -4} = (-4, ) ( -4 Blitzer, Intermediate Algebra, 5e – Slide #4 Section 4.1

5
Linear Inequalities CONTINUED (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| x < 7} = [-2,7) [ ) -2 7 Blitzer, Intermediate Algebra, 5e – Slide #5 Section 4.1

6
**Properties of Inequalities The Addition Property of Inequality**

The Property in Words Example 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 Blitzer, Intermediate Algebra, 5e – Slide #6 Section 4.1

7
**Inequalities Properties of Inequalities CONTINUED Property**

The Property in Words Example 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 Blitzer, Intermediate Algebra, 5e – Slide #7 Section 4.1

8
**Properties of Inequalities The Negative Multiplication Property**

CONTINUED Properties of Inequalities Property The Property in Words Example The Negative Multiplication Property If a < b and c is negative, then ac > 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 Note that when you multiply both sides of an inequality by a negative, you must reverse the direction of the inequality symbol Blitzer, Intermediate Algebra, 5e – Slide #8 Section 4.1

9
**Solving a Linear Inequality**

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

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

11
**Linear Inequalities 3) Isolate the variable and solve.**

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

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

13
Linear Inequalities CONTINUED 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. Add x to both sides Subtract 10 from both sides Blitzer, Intermediate Algebra, 5e – Slide #13 Section 4.1

14
**Linear Inequalities 3) Isolate the variable and solve.**

CONTINUED 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, ) Blitzer, Intermediate Algebra, 5e – Slide #14 Section 4.1

15
**Linear Inequalities Solve the linear inequality. 5x < 5(x – 3)**

EXAMPLE Solve the linear inequality. 5x < 5(x – 3) SOLUTION 5x < 5(x – 3) 5x < 5x – 15 Distribute 0 < – 15 Subtract 5x from both sides Since the result is an obviously false statement, there is no solution. Blitzer, Intermediate Algebra, 5e – Slide #15 Section 4.1

16
**English Sentences and Inequalities**

Linear Inequalities English Sentences and Inequalities English Sentence Inequality 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. Blitzer, Intermediate Algebra, 5e – Slide #16 Section 4.1

17
Linear Inequalities EXAMPLE 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? SOLUTION 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. Blitzer, Intermediate Algebra, 5e – Slide #17 Section 4.1

18
Linear Inequalities CONTINUED 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. x < x Plan A is less than Plan B Blitzer, Intermediate Algebra, 5e – Slide #18 Section 4.1

19
**Linear Inequalities 4) Solve the inequality and answer the question.**

CONTINUED 4) Solve the inequality and answer the question. x < x x < 0.12x Subtract 3 from both sides 15 < 0.04x Subtract 0.08x from both sides 375 < x Divide both sides by 0.04 Therefore, Plan A costs less than Plan B when x > 375 long-distance minutes per month. Blitzer, Intermediate Algebra, 5e – Slide #19 Section 4.1

20
Linear Inequalities CONTINUED 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. Cost for Plan A = 15 + (450)(0.08) = $51 Cost for Plan B = 3 + (450)(0.12) = $57 Plan A has a lower monthly cost, making Plan A the better deal. Blitzer, Intermediate Algebra, 5e – Slide #20 Section 4.1

21
**POW (point the other way).**

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). Blitzer, Intermediate Algebra, 5e – Slide #21 Section 4.1

Similar presentations

OK

WARM UP Solve: 1. 3x – 5 = 24. -2(3x -5) = 2 1. 2x – 3 + 4x = 7 1. 3(2x – 4) = 6.

WARM UP Solve: 1. 3x – 5 = 24. -2(3x -5) = 2 1. 2x – 3 + 4x = 7 1. 3(2x – 4) = 6.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on eddy current probes Ppt on meeting skills Download ppt on indus valley civilization images Ppt on amplitude shift keying demodulation Ppt on any one mathematician byron Ppt on history and sport the story of cricket Ppt on channels of distribution in japan Ppt on manufacturing planning and control Ppt on heat treatment of steel Hrm ppt on recruitment process