1. Holt McDougal Algebra 1 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 1: Solving Inequalities with Variables on Both Sides Solve the inequality and graph the solutions. 5.2

2. Holt McDougal Algebra 1 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. 5.2

3. Holt McDougal Algebra 1 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 5.2

4. Holt McDougal Algebra 1 Solving Inequalities with Variables on Both Sides Check It Out! Example 3 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 5.2

5. Holt McDougal Algebra 1 Solving Inequalities with Variables on Both Sides Check It Out! Example 3 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. 5.2

6. Holt McDougal Algebra 1 Solving Inequalities with Variables on Both Sides Check It Out! Example 4 Solve the inequality and graph the solutions. 5(2 – r) ≥ 3(r – 2) 5(2) – 5(r) ≥ 3(r) + 3(–2) 10 – 5r ≥ 3r – 6 +6 16 − 5r ≥ 3r + 5r +5r 16 ≥ 8r Distribute 5 on the left side of the inequality and distribute 3 on the right side of the inequality. Since 6 is subtracted from 3r, add 6 to both sides to undo the subtraction. Since 5r is subtracted from 16 add 5r to both sides to undo the subtraction. 5.2

7. Holt McDougal Algebra 1 Solving Inequalities with Variables on Both Sides Check It Out! Example 4 Continued –6 –202 –4 4 16 ≥ 8r Since r is multiplied by 8, divide both sides by 8 to undo the multiplication. 2 ≥ r 5.2

8. Holt McDougal Algebra 1 Solving Inequalities with Variables on Both Sides Some inequalities are true no matter what value is substituted for the variable. For these inequalities, all real numbers are solutions. Some inequalities are false no matter what value is substituted for the variable. These inequalities have no solutions. If both sides of an inequality are fully simplified and the same variable term appears on both sides, then the inequality has all real numbers as solutions or it has no solutions. Look at the other terms in the inequality to decide which is the case. 5.2

9. Holt McDougal Algebra 1 Solving Inequalities with Variables on Both Sides Additional Example 5: All Real Numbers as Solutions or No Solutions Solve the inequality. 2x – 7 ≤ 5 + 2x The same variable term (2x) appears on both sides. Look at the other terms. For any number 2x, subtracting 7 will always result in a lower number than adding 5. All values of x make the inequality true. All real numbers are solutions. 5.2

10. Holt McDougal Algebra 1 Solving Inequalities with Variables on Both Sides 2(3y – 2) – 4 ≥ 3(2y + 7) Solve the inequality. Additional Example 6: All Real Numbers as Solutions or No Solutions Distribute 2 on the left side and 3 on the right side and combine like terms. 6y – 8 ≥ 6y + 21 The same variable term (6y) appears on both sides. Look at the other terms. For any number 6y, subtracting 8 will never result in a higher number than adding 21. No values of y make the inequality true. There are no solutions. 5.2

11. Holt McDougal Algebra 1 Solving Inequalities with Variables on Both Sides 5.2

12. Holt McDougal Algebra 1 Solving Inequalities with Variables on Both Sides 5.2

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15. Holt McDougal Algebra 1 Solving Inequalities with Variables on Both Sides 5.2