Extra Example 1 Solving an Inequality with a Variable on One Side

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1 LARSON ALGEBRA 2 CHAPTER 1, LESSON 6, EXTRA EXAMPLES Extra Example 1 Solving an Inequality with a Variable on One Side Solve 11y 9 > y 9 > 13 Write original inequality. 11y > 22 Add 9 to each side. y > 2 Divide each side by 11. The solutions are all real numbers greater than 2, as shown in the graph below CHECK As a check, try several numbers that are greater than 2 in the original inequality. Also, try checking some numbers that are less than or equal to 2 to see that they are not solutions of the original inequality. Extra Example 2 Solving an Inequality with a Variable on Both Sides Solve 7x x º 12. 7x x º 12 Write original inequality. 7x 10x º 21 Subtract 9 from each side. º3x º21 Subtract 10x from each side. x 7 Divide each side by º3 and reverse the inequality. The solutions are all real numbers less than or equal to 7. Check several numbers less than or equal to 7 in the original inequality. Algebra 2, Chapter 1, Lesson 6, Extra Examples 1

2 Extra Example 3 Using a Simple Inequality The percent of households h with cable television is modeled by h = 2.3y + 44 where y is the number of years since Describe the years when the percent is less than h < 53.2 The percent of households is less than y + 44 < 53.2 Substitute for h. 2.3y < 9.2 Subtract 44 from each side. y < 4 Divide each side by 2.3. Because y, which is the number of years since 1988, is less than 4, the percentage of households with cable television is less than 53.2 in years prior to Extra Example 4 Solving an And Compound Inequality Solve º9 < t + 4 < 10. To solve, you must isolate the variable between the two inequality signs. º9 < t + 4 < 10 Write original inequality. º13 < t < 6 Subtract 4 from each expression. The solution is all real numbers greater than º13 and less than 6. Check several numbers between º13 and 6 in the original inequality. The graph is shown below Algebra 2, Chapter 1, Lesson 6, Extra Examples 2

3 Extra Example 5 Solving an Or Compound Inequality Solve 6x + 9 < 3 or 3x º 8 > 13. A solution of this compound inequality is a solution of either of its simple parts, so you should solve each part separately. Solution of first inequality: 6x + 9 < 3 Write first inequality. 6x < º6 Subtract 9 from each side. x < º1 Divide each side by 6. Solution of second inequality: 3x º 8 > 13 Write second inequality. 3x > 21 Add 8 to each side. x > 7 Divide each side by 3. The solutions are all real numbers less than º1 or greater than 7. Check several numbers in the solution to see that they satisfy one of the simple parts of the original inequality. The graph is shown below Algebra 2, Chapter 1, Lesson 6, Extra Examples 3

4 Extra Example 6 Using an And Compound Inequality Milk will keep until its expiration date and will not freeze when stored between a minimum temperature of º1 C and a maximum temperature of 5 C. The temperature C (in degrees Celsius) satisfies the inequality º1 < C < 5. Write the inequality in degrees Fahrenheit. Let F represent the temperature in degrees Fahrenheit and use the formula C = 5 (F º 32). 9 º1 < C < 5 Write original inequality. º1 < 5 9 (F º 32) < 5 Substitute 5 (F º 32) for C. 9 º1.8 < F º 32 < 9 Multiply each expression by 1.8, which equals 9 5, the reciprocal of < F < 41 Add 32 to each expression. The milk will keep until its expiration date as long as the temperature stays between 30.2 F and 41 F. Algebra 2, Chapter 1, Lesson 6, Extra Examples 4

5 Extra Example 7 Solving an Or Compound Inequality The feeding instructions on your dog s food recommend to 33 cups of food daily 4 a. Write the conditions that represent underfeeding or overfeeding your dog as a compound inequality. b. Rewrite the conditions in ounces of dog food. a. Let c represent the number of cups per day. The conditions you need to describe are given by c < or c > b. Let z be the amount of dog food in ounces per day. The relationship between cups and and ounces is c = 0.125z. You can rewrite the conditions in ounces per day by substituting 0.125z for c in each inequality and solving for z. c < or c > z < or 0.125z > z < 20 or z > 30 If you give your dog less than 20 ounces or more than 30 ounces of food, you are underfeeding or overfeeding your dog. Algebra 2, Chapter 1, Lesson 6, Extra Examples 5