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Solving Linear Inequalities 2.9 Solving Linear Inequalities

Linear Inequalities An inequality is a statement that contains of the symbols: < , >, ≤ or ≥. Equations Inequalities x = 3 x > 3 12 = 7 – 3y 12 ≤ 7 – 3y

Graphing Solutions Graphing solutions to linear inequalities in one variable Use a number line Use a bracket at the endpoint of a interval if you want to include the point Use a parenthesis at the endpoint if you DO NOT want to include the point Represents the set {xx  7} Represents the set {xx > – 4}

Solutions to Linear Inequalities x < 3 Interval notation: (–∞, 3) NOT included –2 < x < 0 Interval notation: (–2, 0) Included –1.5  x  3 Interval notation: [–1.5, 3)

Example Graph:

Addition Property of Inequality If a, b, and c are real numbers, then a < b and a + c < b + c are equivalent inequalities.

Multiplication Property of Inequality 1. If a, b, and c are real numbers, and c is positive, then a < b and ac < bc are equivalent inequalities. 2. If a, b, and c are real numbers, and c is negative, then a < b and ac > bc

Solving Linear Inequalities Step 1: Clear the inequality of fractions by multiplying both sides of the inequality by the LCD of all fractions in the inequality. Step 2: Remove grouping symbols such as parentheses by using the distributive property. Step 3: Simplify each side of inequality by combining like terms. Step 4: Write the inequality with variable terms on one side and numbers on the other side by using the addition property of inequality. Step 5: Get the variable alone by using the multiplication property of inequality.

Example Solve: 3x + 8 ≥ 5. Graph the solution set.

Example Solve: 3x + 9 ≥ 5(x – 1). Graph the solution set. 3x + 9 ≥ 5x – 5 Apply the distributive property. 3x – 3x + 9 ≥ 5x – 3x – 5 Subtract 3x from both sides. 9 ≥ 2x – 5 Simplify. 9 + 5 ≥ 2x – 5 + 5 Add 5 to both sides. 14 ≥ 2x Simplify. 7 ≥ x Divide both sides by 2. The graph of solution set is{x|x ≤ 7}.

Example Solve: 7(x – 2) + x > –4(5 – x) – 12. Graph the solution set. 7(x – 2) + x > –4(5 – x) – 12 7x – 14 + x > –20 + 4x – 12 Apply the distributive property. 8x – 14 > 4x – 32 Combine like terms. 8x – 4x – 14 > 4x – 4x – 32 Subtract 4x from both sides. 4x – 14 > –32 Simplify. 4x – 14 + 14 > –32 + 14 Add 14 to both sides. 4x > –18 Simplify. Divide both sides by 4. The graph of solution set is {x|x > –9/2}.

Compound Inequalities A compound inequality contains two inequality symbols. 0  4(5 – x) < 8 This means 0  4(5 – x) and 4(5 – x) < 8. To solve the compound inequality, perform operations simultaneously to all three parts of the inequality (left, middle and right).

Example Graph:

Example Solve the inequality. Graph the solution set and write it in interval notation. 9 < z + 5 < 13 9 < z + 5 < 13 9 – 5 < z + 5 – 5 < 13 – 5 Subtract 5 from all three parts. 4 < z < 8 (4, 8)

Example Solve the inequality. Graph the solution set and write it in interval notation. –7 < 2p – 3 ≤ 5 –7 < 2p – 3 ≤ 5 –4 < 2p ≤ 8 Add 3 to all three parts. –2 < p ≤ 4 Divide all three parts by 2. (–2, 4]

Example Solve the inequality. Graph the solution set and write it in interval notation. 0  20 – 4x < 8 0  20 – 4x < 8 Use the distributive property. 0 – 20  20 – 20 – 4x < 8 – 20 Subtract 20 from each part. – 20  – 4x < – 12 Simplify each part. 5  x > 3 Divide each part by –4. Remember that the sign changes direction when you divide by a negative number. The solution is (3,5].

Inequality Applications Example: Six times a number, decreased by 2, is at least 10. Find the number. 1.) UNDERSTAND Let x = the unknown number. “Six times a number” translates to 6x, “decreased by 2” translates to 6x – 2, “is at least 10” translates ≥ 10. Continued

Finding an Unknown Number Example continued: 2.) TRANSLATE Six times a number 6x decreased – by 2 2 is at least ≥ 10 Continued

Finding an Unknown Number Example continued: 3.) SOLVE 6x – 2 ≥ 10 6x ≥ 12 Add 2 to both sides. x ≥ 2 Divide both sides by 6. 4.) INTERPRET Check: Replace “number” in the original statement of the problem with a number that is 2 or greater. Six times 2, decreased by 2, is at least 10 6(2) – 2 ≥ 10 10 ≥ 10 State: The number is 2.

Example You are having a catered event. You can spend at most $1200. The set up fee is $250 plus $15 per person, find the greatest number of people that can be invited and still stay within the budget. Let x represent the number of people Set up fee + cost per person × number of people ≤ 1200 250 + 15x ≤ 1200 continued

continued You are having a catered event. You can spend at most $1200. The set up fee is $250 plus $15 per person, find the greatest number of people that can be invited and still stay within the budget. The number of people who can be invited must be 63 or less to stay within the budget.