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