Section 6.5 Solving Linear Inequalities Math in Our World

Learning Objectives  Graph solution sets for simple inequalities.  Solve linear inequalities in one variable.  Solve three-part linear inequalities.  Solve real-world problems using inequalities.

Linear Inequalities A linear inequality in one variable is a statement that can be written in any of the following four forms: Ax + B 0, Ax + B ≤ 0, or Ax + B ≥ 0. In each case, B can be any real number, and A can be any real number except zero. To solve a linear inequality means to find the set of all numbers that make the inequality a true statement when substituted in for the variable. That set is called the solution set for the inequality.

Graphing Inequalities For an inequality of the form x ≥ 5, every real number that’s 5 or larger. To show that the number 5 is included in the solution set, a solid or closed circle (●) is used. On the other hand, the solution set for the inequality x > 5 includes all real numbers greater than 5; however, the actual number 5 is not in the solution set. To show that the number 5 is not included in the solution set, an open circle (○) is used.

Graphing Inequalities For an inequality of the form x ≤ 5, every real number that’s 5 or smaller. Again a closed circle indicates that 5 is included. When taking the intersection of two inequalities, such as x ≥ 2 and x ≤ 4, it’s more efficient to write it this way: 2 ≤ x ≤ 4. This is the way we’ll represent an interval between two numbers.

Graphing Inequalities

EXAMPLE 1 Graphing Solution Sets for Simple Inequalities Graph the solution set for each inequality. (a) x ≤ 10 (b) y > -4 (c) -30 < x ≤ 50 SOLUTION

Solving Linear Inequalities Addition and Subtraction Properties for Inequalities You can add or subtract the same real number or algebraic expression to both sides of an inequality without changing the solution set. In symbols, if a < b, then a + c, ≤, ≥.

Solving Linear Inequalities If you multiply or divide both sides of an inequality by the same negative real number, the direction of the inequality symbol is reversed. If a < b and c < 0 then ac > bc and ac bc, and likewise for >, ≤, ≥. Multiplication and Division Properties for Inequalities You can multiply or divide both sides of an inequality by the same positive real number without changing the solution set. In symbols, if a 0 then ac, ≤, ≥.

EXAMPLE 2 Solving a Linear Inequality Solve and graph the solution set for 5x – 9 ≥ 21. The solution set is {x | x ≥ 6}. The graph of the solution set is SOLUTION

EXAMPLE 3 Solving a Linear Inequality Solve and graph the solution set for 16 – 3x > 40. The solution set is {x | x < – 8}. The graph is SOLUTION

EXAMPLE 4 Solving a Linear Inequality Solve and graph the solution set for 4(x + 3) < 2x – 26.

EXAMPLE 4 Solving a Linear Inequality The solution set is {x | x < – 19}. The graph is SOLUTION

EXAMPLE 5 Solving a Three-Part Linear Inequality Solve and graph the solution set for – 4 < 3 – 2y ≤ 9. SOLUTION The solution set is {x | -3 < x < – 8}.

Common Phrases in Inequalities

EXAMPLE 6 Applying Inequalities to Vacation Planning With the stress of finals behind you, you decide to plan a vacation to relax a little bit. After poking around on the Internet, you find a room in the area you want to visit for $65 per night. Some quick estimating leads you to conclude that you’ll need at least $250 for gas, food, beverages, and entertainment expenses. Upon checking your bank balance, you decide that you can afford to spend at most $600 on the trip. How many nights can you stay?

EXAMPLE 6 Applying Inequalities to Vacation Planning SOLUTION Step 1 Relevant information: Lodging is $65 per night, other expenses are $250, maximum you can spend is $600. We’re asked to find the number of nights. Step 2 Use variable n to represent the number of nights. Step 3 Translate the relevant information into an inequality. $65 x number of nights + other expenses is no more than $ x n ≤ 600

EXAMPLE 6 Applying Inequalities to Vacation Planning SOLUTION Step 4 Solve the inequality. Step 5 Answer the question. Staying 5.4 nights doesn’t make sense, so you could stay at most 5 nights. Step 6 Check: Will 5 nights work?

Mike is planning to buy lunch for himself and some coworkers. He decides to buy cheeseburgers and fries from the value menu—the burgers are $1 each, and the fries cost $0.80. He also needs to pay 5% of the total in sales tax. What is the largest number of items he can buy if he wants to buy the same number of burgers as fries, and he only has $10 to spend? EXAMPLE 7 Applying Inequalities to the Cost of Buying Food

SOLUTION Step 1 Relevant information: Burgers are $1 each, fries are $0.80 each; 5% sales tax; same number of burgers and fries; maximum cost $10. Step 2 Use variable x to represent both the number of burgers and the number of fries because those numbers are equal. Step 3 $ 1 x is the total cost of burgers, and $0.80x is the total cost of fries. The tax is 5% of their sum, which is 0.05(x +.80x) dollars. So the total cost x x (x x), which must be less than or equal to 10. The inequality is

EXAMPLE 7 Applying Inequalities to the Cost of Buying Food SOLUTION Step 4 Solve the inequality. Step 5 Mike can’t buy 5.29 burgers, so we round down to 5. He can buy at most 5 burgers and 5 fries.