# 2 8 Solving Linear Inequalities Linear Inequalities in

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§ 2. 8 Solving Linear Inequalities

Linear Inequalities in One Variable A linear inequality in one variable is an inequality that can be written in the form ax + b < c where a, b, and c are real numbers and a is not 0. This definition and all other definitions, properties and steps in this section also hold true for the inequality symbols >, , or . Martin-Gay, Beginning and Intermediate Algebra, 4 ed 2

Solutions to Linear Inequalities Graphing Solutions to Linear Inequalities • Use a number line. • Use a closed circle at the endpoint of an interval if you want to include the point. • Use an open circle at the endpoint if you DO NOT want to include the point. Represents the set {x x 7}. Represents the set {x x > – 4}. Interval notation is used to write solution sets of inequalities. • Use a parenthesis if you want to include the number. • Use a bracket if you DO NOT want to include the number. Martin-Gay, Beginning and Intermediate Algebra, 4 ed 3

Solutions to Linear Inequalities x<3 -5 -4 -3 -2 -1 0 1 2 3 4 Interval notation: (–∞, 3) 5 NOT included – 2 < x < 0 -5 -4 -3 -2 -1 0 1 2 3 4 5 Interval notation: (– 2, 0) Included – 1. 5 x 3 -5 -4 -3 -2 -1 0 1 2 Interval notation: [– 1. 5, 3) Martin-Gay, Beginning and Intermediate Algebra, 4 ed 4

Properties of Inequality 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 are equivalent inequalities. Martin-Gay, Beginning and Intermediate Algebra, 4 ed 5

Solving Linear Inequalities in One Variable 1) 2) 3) 4) 5) Clear the inequality of fractions by multiplying both sides by the LCD of all fractions of the inequality. Remove grouping symbols by using the distributive property. Simplify each side of equation by combining like terms. Write the inequality with variable terms on one side and numbers on the other side by using the addition property of equality. Get the variable alone by using the multiplication property of equality. Martin-Gay, Beginning and Intermediate Algebra, 4 ed 6

Solving Linear Inequalities Example: Solve the inequality. Graph the solution and give your answer in interval notation. 2(x – 3) < 4 x + 10 2 x – 6 < 4 x + 10 Distribute. – 6 < 2 x + 10 Subtract 2 x from both sides. – 16 < 2 x Subtract 10 from both sides. – 8 < x or x > – 8 Divide both sides by 2. -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 (– 8, ∞) Martin-Gay, Beginning and Intermediate Algebra, 4 ed 7

Solving Linear Inequalities Example: Solve the inequality. Graph the solution and give your answer in interval notation. x+5 x– 2 x x– 7 Subtract 5 from both sides. 0 – 7 Subtract x from both sides. (Always true!) Since 0 is always greater than – 7, the solution is all real numbers. (Any value we put in for x in the original statement will give us a true inequality. ) -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 (–∞, ∞) Martin-Gay, Beginning and Intermediate Algebra, 4 ed 8

Solving Linear Inequalities Example: Solve the inequality. Give your answer in graph form. 3 x + 9 5(x – 1) 3 x + 9 5 x – 5 3 x – 3 x + 9 5 x – 3 x – 5 Use distributive property on right side. Subtract 3 x from both sides. 9 2 x – 5 Simplify both sides. 9 + 5 2 x – 5 + 5 Add 5 to both sides. 14 2 x Simplify both sides. 7 x Divide both sides by 2. Solution: Martin-Gay, Beginning and Intermediate Algebra, 4 ed 9

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). Martin-Gay, Beginning and Intermediate Algebra, 4 ed 10

Solving Compound Inequalities Example: Solve the inequality. Give your answer in interval notation. 0 4(5 – x) < 8 0 20 – 4 x < 8 Use the distributive property. 0 – 20 – 4 x < 8 – 20 Subtract 20 from each part. – 20 – 4 x < – 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]. Martin-Gay, Beginning and Intermediate Algebra, 4 ed 11

Solving Compound Inequalities Example: Solve the inequality. Give your answer in interval notation. a. ) 9 < z + 5 < 13 b. ) – 7 < 2 p – 3 ≤ 5 a. ) 9 < z + 5 < 13 4<z<8 (4, 8) b. ) – 7 < 2 p – 3 ≤ 5 – 4 < 2 p ≤ 8 – 2 < p ≤ 4 Subtract 5 from all three parts. Add 3 to all three parts. Divide all three parts by 2. (– 2, 4] Martin-Gay, Beginning and Intermediate Algebra, 4 ed 12

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 6 x, “decreased by 2” translates to 6 x – 2, “is at least 10” translates ≥ 10. Continued Martin-Gay, Beginning and Intermediate Algebra, 4 ed 13

Finding an Unknown Number Example continued: 2. ) TRANSLATE Six times a number decreased by 2 is at least 10 6 x – 2 ≥ 10 Continued Martin-Gay, Beginning and Intermediate Algebra, 4 ed 14

Finding an Unknown Number Example continued: 3. ) SOLVE 6 x – 2 ≥ 10 6 x ≥ 12 x≥ 2 Add 2 to both sides. 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 or greater. Martin-Gay, Beginning and Intermediate Algebra, 4 ed 15

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