Solving Inequalities with Multiple Operations

Warm Up Solve each equation. 1. 2x – 5 = –17 2. –6 14 Solve each inequality and graph the solutions. t > –4 3. 5 < t + 9 4. a ≤ –8

Objective Solve inequalities that contain more than one operation.

Inequalities that contain more than one operation require more than one step to solve. Use inverse operations to undo the operations in the inequality one at a time.

Directions: Solve the inequality and graph the solutions.

45 + 2b > 61 –45 –45 b > 8 0 2 4 6 8 10 14 20 12 18 16 Example 1 45 + 2b > 61 Since 45 is added to 2b, subtract 45 from both sides to undo the addition. 2b > 16 Since b is multiplied by 2, divide both sides by 2 to undo the multiplication.

8 – 3y ≥ 29 –8 –8 –7 –8 –10 –6 –4 0 2 4 6 8 10 –2 Example 2 8 – 3y ≥ 29 Since 8 is added to –3y, subtract 8 from both sides to undo the addition. –3y ≥21 Since y is multiplied by –3, divide both sides by –3 to undo the multiplication. Change ≥ to ≤. y ≤ –7

–12 ≥ 3x + 6 – 6 – 6 –8 –10 –6 –4 0 2 4 6 8 10 –2 Example 3 –12 ≥ 3x + 6 Since 6 is added to 3x, subtract 6 from both sides to undo the addition. –18 ≥ 3x Since x is multiplied by 3, divide both sides by 3 to undo the multiplication. –6 ≥ x

x + 5 < –6 –5 –5 –11 –20 –16 –12 –8 –4 0 Example 4 Since x is divided by –2, multiply both sides by –2 to undo the division. Change > to <. Since 5 is added to x, subtract 5 from both sides to undo the addition. x < –11

–1 –1 –10 –20 –16 –12 –8 –4 0 Example 5 Since 1 – 2n is divided by 3, multiply both sides by 3 to undo the division. 1 – 2n ≥21 Since 1 is added to −2n, subtract 1 from both sides to undo the addition. –2n ≥ 20 Since n is multiplied by −2, divide both sides by −2 to undo the multiplication. Change ≥ to ≤. n ≤ –10

To solve more complicated inequalities, you may first need to simplify the expressions on one or both sides by using the order of operations, combining like terms, or using the Distributive Property.

10 –8 –10 –6 –4 0 2 4 6 8 –2 –3 Example 6 2 – (–10) > –4t 12 > –4t Combine like terms. Since t is multiplied by –4, divide both sides by –4 to undo the multiplication. Change > to <. –3 < t (or t > –3)

–8 + 4x ≤ 8 +8 +8 10 –8 –10 –6 –4 0 2 4 6 8 –2 Example 7 –4(2 – x) ≤ 8 −4(2 – x) ≤ 8 Distribute –4 on the left side. −4(2) − 4(−x) ≤ 8 Since –8 is added to 4x, add 8 to both sides. 4x ≤16 Since x is multiplied by 4, divide both sides by 4 to undo the multiplication. x ≤ 4

–3 –3 Example 8 Multiply both sides by 6, the LCD of the fractions. Distribute 6 on the left side. 4f + 3 > 2 Since 3 is added to 4f, subtract 3 from both sides to undo the addition. 4f > –1

0 Example 8 Continued 4f > –1 Since f is multiplied by 4, divide both sides by 4 to undo the multiplication.

– 5 > – 5 m > 10 0 2 4 6 8 10 14 20 12 18 16 Example 9 2m + 5 > 52 Simplify 52. 2m + 5 > 25 Since 5 is added to 2m, subtract 5 from both sides to undo the addition. 2m > 20 Since m is multiplied by 2, divide both sides by 2 to undo the multiplication.

– 11 –11 10 –8 –10 –6 –4 0 2 4 6 8 –2 Example 10 3 + 2(x + 4) > 3 Distribute 2 on the left side. 3 + 2(x + 4) > 3 3 + 2x + 8 > 3 Combine like terms. 2x + 11 > 3 Since 11 is added to 2x, subtract 11 from both sides to undo the addition. 2x > –8 Since x is multiplied by 2, divide both sides by 2 to undo the multiplication. x > –4

Example 11 Multiply both sides by 8, the LCD of the fractions. Distribute 8 on the right side. 5 < 3x –2 Since 2 is subtracted from 3x, add 2 to both sides to undo the subtraction. +2 + 2 7 < 3x

2 4 6 8 0 10 Example 11 Continued Solve the inequality and graph the solutions. 7 < 3x Since x is multiplied by 3, divide both sides by 3 to undo the multiplication.

daily cost at We Got Wheels Cost at Rent-A-Ride must be less than $0.20 per mile # of miles. plus times 55 < 38 m + 0.20  Example 12: Application To rent a certain vehicle, Rent-A-Ride charges $55.00 per day with unlimited miles. The cost of renting a similar vehicle at We Got Wheels is $38.00 per day plus $0.20 per mile. For what number of miles in the cost at Rent-A-Ride less than the cost at We Got Wheels? Let m represent the number of miles. The cost for Rent-A-Ride should be less than that of We Got Wheels.

55 < 38 + 0.20m –38 –38 Example 12 Continued 55 < 38 + 0.20m Since 38 is added to 0.20m, subtract 8 from both sides to undo the addition. 17 < 0.20m Since m is multiplied by 0.20, divide both sides by 0.20 to undo the multiplication. 85 < m Rent-A-Ride costs less when the number of miles is more than 85.

Check a number greater than 85. Check the endpoint, 85. 55 = 38 + 0.20m 55 < 38 + 0.20m 55 38 + 0.20(85) 55 < 38 + 0.20(90) 55 38 + 17 55 < 38 + 18  55 55  55 < 56 Example 12 Continued Check

is greater than or equal to First test score second test score divided by number of scores total score plus 90 ≥  (95 x) 2 + Example 13 The average of Jim’s two test scores must be at least 90 to make an A in the class. Jim got a 95 on his first test. What grades can Jim get on his second test to make an A in the class? Let x represent the test score needed. The average score is the sum of each score divided by 2.

–95 –95 Example 13 Continued Since 95 + x is divided by 2, multiply both sides by 2 to undo the division. 95 + x ≥ 180 Since 95 is added to x, subtract 95 from both sides to undo the addition. x ≥ 85 The score on the second test must be 85 or higher.

90 90   90 90 90.5 ≥ 90 Example 13 Continued Check Check a number greater than 85. Check the end point, 85.

Lesson Summary: Part I Solve each inequality and graph the solutions. 1. 13 – 2x ≥ 21 x ≤–4 2. –11 + 2 < 3p p > –3 t > 7 3. 23 < –2(3 –t) 4.

Lesson Summary: Part II 5. A video store has two movie rental plans. Plan A includes a $25 membership fee plus $1.25 for each movie rental. Plan B costs $40 for unlimited movie rentals. For what number of movie rentals is plan B less than plan A? more than 12 movies

Solving Inequalities with Multiple Operations