Solving Linear Inequalities in One Variable
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Learn how to solve linear inequalities in one variable, graph solutions, and apply properties of inequalities. Examples and interval notation included for easy understanding.
Solving Linear Inequalities in One Variable
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Linear Inequalities in One Variable A linear inequality in one variableis 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 .
Solutions to Linear Inequalities Represents the set {xx 7}. Represents the set {xx > – 4}. 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. • 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..
Solutions to Linear Inequalities -5 -5 -5 -4 -4 -4 -3 -3 -3 -2 -2 -2 -1 -1 -1 0 0 0 1 1 1 2 2 2 3 3 3 4 4 4 5 5 5 x < 3 (–∞, 3) Interval notation: NOT included –2 < x < 0 (–2, 0) Interval notation: Included –1.5 x 3 [–1.5, 3) Interval notation:
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 • If a, b, and c are real numbers, and c is positive, then a < b and ac < bc are equivalent inequalities. • If a, b, and c are real numbers, and c is negative, then a < b and ac > bc are equivalent inequalities.
Solving Linear Inequalities Solving Linear Inequalities in One Variable • 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.
Solving Linear Inequalities -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 Example:Solve the inequality. Graph the solution and give your answer in interval notation. 2(x – 3) < 4x + 10 2x – 6 < 4x + 10 Distribute. – 6 < 2x + 10Subtract 2x from both sides. – 16 < 2x Subtract 10 from both sides. – 8 < x or x > – 8Divide both sides by 2. (–8, ∞)
Solving Linear Inequalities -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 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 – 7Subtract 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.) (–∞∞)
Solving Linear Inequalities Example: Solve the inequality. Give your answer in interval notation. a.) 9 < z + 5 < 13 b.) –7 < 2p – 3 ≤ 5 a.) 9 < z + 5 < 13 4 < z < 8Subtract 5 from all three parts. (4, 8) b.) –7 < 2p – 3 ≤ 5 –4 < 2p ≤ 8 Add 3 to all three parts. –2 < p ≤ 4 Divide all three parts by 2. (–2, 4]
Solution: Solving Linear Inequalities Example: Solve the inequality. Give your answer in graph form. 3x + 9 5(x – 1) 3x + 9 5x – 5 Use distributive property on right side. 3x – 3x + 9 5x – 3x – 5 Subtract 3x from both sides. 9 2x – 5 Simplify both sides. 9 + 5 2x – 5 + 5 Add 5 to both sides. 14 2x Simplify both sides. 7 x Divide both sides by 2.
Compound Inequalities Acompound 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).
Solving Compound Inequalities Example: Solve the inequality. Give your answer 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
is at least decreased Six times a number 10 by 2 6x – ≥ 10 2 Finding an Unknown Number Example continued: 2.) TRANSLATE 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.