Inequalities with Variables on Both Sides

1. Inequalities with Variables on Both Sides

2. Some inequalities have variable terms on both sides of the inequality symbol. You can solve these inequalities like you solved equations with variables on both sides.

3. y ≤ 4y + 18 –4y –4y -3y ≤ 18 –8 –10 –6 –4 0 2 4 6 8 10 –2 Example 1 Solve the inequality and graph the solutions. y ≤ 4y + 18 Subtract 4y from both sides. ___ ___ -3 -3 Since y is multiplied by -3, divide both sides by -3 to undo the multiplication. y –6 Since you divided by a negative, you must flip the inequality sign.

4. 4x ≥ 7x + 6 –7x –7x x ≤ –2 –8 –10 –6 –4 0 2 4 6 8 10 –2 Example 2 Solve the inequality and graph the solutions. 4x ≥ 7x + 6 Subtract 7x from both sides. –3x ≥ 6 Since x is multiplied by –3, divide both sides by –3 to undo the multiplication. Change ≥ to ≤.

5. Example 3 • The Home Cleaning Company charges $312 to power-wash the siding of a house plus $12 for each window. Power Clean charges $36 per window, and the price includes power-washing the siding. How many windows must a house have to make the total cost from The Home Cleaning Company less expensive than Power Clean? • 312 + 12w < 36w • Subtract 12w from both sides • 312 < 24w • Divide both sides by 24 • 312/24 < 24w/24 • 13 < w or w > 13 • There will have to be 14 windows in order for The Home Cleaning Company to be less expensive than Power Clean.

6. Example 4 • A-Plus Advertising charges a fee of $24 plus $0.10 per flyer to print and deliver flyers. Print and More charges $0.25 per flyer. For how many flyers is the cost at A-Plus Advertising less than the cost of Print and More? • 24 + 0.10f < 0.25f • Subtract 0.10f from both sides. • 24 + 0.10f – 0.10f < 0.25f – 0.10f • 24 < 0.15f • Divide both sides by 0.15 • 24/0.15 < 0.15f/0.15 • 160 < f or f > 160 • 161 flyers or more will make A-Plus Advertising less than Print and More.

7. –12 –9 –6 –3 0 3 –3k –3k +6 +6 Example 5 Solve the inequality and graph the solutions. 2(k – 3) > 6 + 3k – 3 Distribute 2 on the left side of the inequality. And, combine the 6 and -3 on the right side. 2(k – 3) > 3 + 3k 2k+ 2(–3)> 3 + 3k 2k –6 > 3 + 3k Subtract 3k from both sides. -k – 6 > 3 Since 6 is subtracted from k, add 6 to both sides to undo the subtraction. -k > 9 Divide both sides by -1 (flip the inequality). k < -9

8. -3r -3r –6 –4 –2 0 2 4 -10 -10 Example 6 Solve the inequality and graph the solutions. 5(2 – r) ≥ 3(r – 2) Distribute 5 on the left side of the inequality and distribute 3 on the right side of the inequality. 5(2 – r) ≥ 3(r – 2) 5(2) –5(r) ≥ 3(r) + 3(–2) Subtract 3r from both sides. 10 – 5r ≥ 3r – 6 10 − 8r ≥ -6 Subtract 10 from both sides -8r ≥ -16 ___ ___ -8 -8 Divide both sides by -8 (flip the inequality). r < 2

9. There are special cases of inequalities called identities and contradictions.

11. –2x –2x Example 7 Solve the inequality. 2x – 7 ≤ 5 + 2x 2x – 7 ≤ 5 + 2x Subtract 2x from both sides.  –7 ≤ 5 True statement. The inequality 2x − 7 ≤ 5 + 2x is an identity. All values of x make the inequality true. Therefore, all real numbers are solutions.

12. –6y –6y Example 8 Solve the inequality. 2(3y – 2) – 4 ≥ 3(2y + 7) Distribute 2 on the left side and 3 on the right side. 2(3y – 2) – 4 ≥ 3(2y + 7) 2(3y) – 2(2) – 4 ≥ 3(2y) + 3(7) 6y – 4 – 4 ≥ 6y + 21 6y – 8 ≥ 6y + 21 Subtract 6y from both sides.  –8 ≥ 21 False statement. No values of y make the inequality true. There are no solutions.

13. Try these… Lesson Quiz: Part I Solve each inequality and graph the solutions. 1. t < 5t + 24 t > –6 2. 5x – 9 ≤ 4.1x –81 x ≤–80 3. 4b + 4(1 – b) > b – 9 b < 13 4. 2y – 2 ≥ 2(y + 7) no solution

14. Try these… 5. Rick bought a photo printer and supplies for $186.90, which will allow him to print photos for $0.29 each. A photo store charges $0.55 to print each photo. How many photos must Rick print before his total cost is less than getting prints made at the photo store? Rick must print more than 718 photos.