1. Holt Algebra 1 3-5 Solving Inequalities with Variables on Both Sides Solve inequalities that contain variable terms on both sides. Objective

2. Holt Algebra 1 3-5 Solving Inequalities with Variables on Both Sides 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. Use the properties of inequality to “ collect ” all the variable terms on one side and all the constant terms on the other side.

3. Holt Algebra 1 3-5 Solving Inequalities with Variables on Both Sides Example 1A: Solving Inequalities with Variables on Both Sides Solve the inequality and graph the solutions. y ≤ 4y + 18 –y 0 ≤ 3y + 18 –18 – 18 –18 ≤ 3y –6 ≤ y (or y  –6) To collect the variable terms on one side, subtract y from both sides. Since 18 is added to 3y, subtract 18 from both sides to undo the addition. Since y is multiplied by 3, divide both sides by 3 to undo the multiplication. –10 –8 –6–4 –2 0246810

4. Holt Algebra 1 3-5 Solving Inequalities with Variables on Both Sides 4m – 3 < 2m + 6 To collect the variable terms on one side, subtract 2m from both sides. –2m – 2m 2m – 3 < + 6 Since 3 is subtracted from 2m, add 3 to both sides to undo the subtraction + 3 2m < 9 Since m is multiplied by 2, divide both sides by 2 to undo the multiplication. 4 5 6 Example 1B: Solving Inequalities with Variables on Both Sides Solve the inequality and graph the solutions.

5. Holt Algebra 1 3-5 Solving Inequalities with Variables on Both Sides Example 2: Business Application 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? Let w be the number of windows.

6. Holt Algebra 1 3-5 Solving Inequalities with Variables on Both Sides Example 2 Continued 312 + 12 w < 36 w – 12w –12w 312 < 24w 13 < w The Home Cleaning Company is less expensive for houses with more than 13 windows. To collect the variable terms, subtract 12w from both sides. Since w is multiplied by 24, divide both sides by 24 to undo the multiplication. Home Cleaning Company siding charge plus $12 per window # of windows is less than Power Clean cost per window # of windows. times

7. Holt Algebra 1 3-5 Solving Inequalities with Variables on Both Sides Check It Out! Example 2 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? Let f represent the number of flyers printed. 24 + 0.10 f < 0.25 f plus $0.10 per flyer is less than # of flyers. A-Plus Advertising fee of $24 Print and More’s cost per flyer # of flyers times

8. Holt Algebra 1 3-5 Solving Inequalities with Variables on Both Sides Check It Out! Example 2 Continued 24 + 0.10f < 0.25f –0.10f 24 < 0.15f 160 < f To collect the variable terms, subtract 0.10f from both sides. Since f is multiplied by 0.15, divide both sides by 0.15 to undo the multiplication. More than 160 flyers must be delivered to make A-Plus Advertising the lower cost company.

9. Holt Algebra 1 3-5 Solving Inequalities with Variables on Both Sides You may need to simplify one or both sides of an inequality before solving it. Look for like terms to combine and places to use the Distributive Property.

10. Holt Algebra 1 3-5 Solving Inequalities with Variables on Both Sides Example 3A: Simplify Each Side Before Solving Solve the inequality and graph the solutions. 2(k – 3) > 6 + 3k – 3 2(k – 3) > 3 + 3k Distribute 2 on the left side of the inequality. 2k + 2(–3) > 3 + 3k 2k – 6 > 3 + 3k –2k – 2k –6 > 3 + k To collect the variable terms, subtract 2k from both sides. –3 –9 > k Since 3 is added to k, subtract 3 from both sides to undo the addition.

11. Holt Algebra 1 3-5 Solving Inequalities with Variables on Both Sides Example 3A Continued –9 > k –12–9–6–303

12. Holt Algebra 1 3-5 Solving Inequalities with Variables on Both Sides There are special cases of inequalities called identities and contradictions.

13. Holt Algebra 1 3-5 Solving Inequalities with Variables on Both Sides

14. Holt Algebra 1 3-5 Solving Inequalities with Variables on Both Sides Example 4A: Identities and Contradictions Solve the inequality. 2x – 7 ≤ 5 + 2x –2x –7 ≤ 5 Subtract 2x from both sides. 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.

15. Holt Algebra 1 3-5 Solving Inequalities with Variables on Both Sides 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 Distribute 2 on the left side and 3 on the right side. Example 4B: Identities and Contradictions Solve the inequality. –6y –8 ≥ 21 Subtract 6y from both sides. False statement.  No values of y make the inequality true. There are no solutions.

16. Holt Algebra 1 3-5 Solving Inequalities with Variables on Both Sides 4(y – 1) ≥ 4y + 2 4(y) + 4(–1) ≥ 4y + 2 4y – 4 ≥ 4y + 2 Distribute 4 on the left side. Check It Out! Example 4a Solve the inequality. –4y –4 ≥ 2 Subtract 4y from both sides. False statement.  No values of y make the inequality true. There are no solutions.

17. Holt Algebra 1 3-5 Solving Inequalities with Variables on Both Sides Solve the inequality. x – 2 < x + 1 –x –x –2 < 1 Subtract x from both sides. True statement. All values of x make the inequality true. All real numbers are solutions. Check It Out! Example 4b

18. Holt Algebra 1 3-5 Solving Inequalities with Variables on Both Sides Lesson Quiz: Part I Solve each inequality and graph the solutions. 1. t < 5t + 24t > –6 2. 5x – 9 ≤ 4.1x – 81x ≤ –80 b < 133. 4b + 4(1 – b) > b – 9

19. Holt Algebra 1 3-5 Solving Inequalities with Variables on Both Sides Lesson Quiz: Part II 4. 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.

20. Holt Algebra 1 3-5 Solving Inequalities with Variables on Both Sides Lesson Quiz: Part III Solve each inequality. 5. 2y – 2 ≥ 2(y + 7) contradiction, no solution 6. 2(–6r – 5) < –3(4r + 2) identity, all real numbers