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Holt McDougal Algebra 1 2-5 Solving Inequalities with Variables on Both Sides Solve inequalities that contain variable terms on both sides. Objective.

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Presentation on theme: "Holt McDougal Algebra 1 2-5 Solving Inequalities with Variables on Both Sides Solve inequalities that contain variable terms on both sides. Objective."— Presentation transcript:

1 Holt McDougal Algebra 1 2-5 Solving Inequalities with Variables on Both Sides Solve inequalities that contain variable terms on both sides. Objective

2 Holt McDougal Algebra 1 2-5 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

3 Holt McDougal Algebra 1 2-5 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.

4 Holt McDougal Algebra 1 2-5 Solving Inequalities with Variables on Both Sides Solve the inequality and graph the solutions. Check It Out! Example 1a 4x ≥ 7x + 6 –7x –3x ≥ 6 x ≤ –2 To collect the variable terms on one side, subtract 7x from both sides. Since x is multiplied by –3, divide both sides by –3 to undo the multiplication. Change ≥ to ≤. –10 –8 –6–4 –2 0246810

5 Holt McDougal Algebra 1 2-5 Solving Inequalities with Variables on Both Sides Solve the inequality and graph the solutions. Check It Out! Example 1b 5t + 1 < –2t – 6 +2t 7t + 1 < –6 – 1 < –1 7t < –7 7 t < –1 –5 –4 –3–2 –1 01234 5 To collect the variable terms on one side, add 2t to both sides. Since 1 is added to 7t, subtract 1 from both sides to undo the addition. Since t is multiplied by 7, divide both sides by 7 to undo the multiplication.

6 Holt McDougal Algebra 1 2-5 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.

7 Holt McDougal Algebra 1 2-5 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

8 Holt McDougal Algebra 1 2-5 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

9 Holt McDougal Algebra 1 2-5 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.

10 Holt McDougal Algebra 1 2-5 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 McDougal Algebra 1 2-5 Solving Inequalities with Variables on Both Sides Example 3A Continued –9 > k –12–9–6–303

12 Holt McDougal Algebra 1 2-5 Solving Inequalities with Variables on Both Sides Example 3B: Simplify Each Side Before Solving Solve the inequality and graph the solution. 0.9y ≥ 0.4y – 0.5 –0.4y 0.5y ≥ – 0.5 0.5 y ≥ –1 To collect the variable terms, subtract 0.4y from both sides. Since y is multiplied by 0.5, divide both sides by 0.5 to undo the multiplication. –5 –4 –3–2 –1 01234 5

13 Holt McDougal Algebra 1 2-5 Solving Inequalities with Variables on Both Sides Check It Out! Example 3a 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.

14 Holt McDougal Algebra 1 2-5 Solving Inequalities with Variables on Both Sides Check It Out! Example 3a Continued –6 –202 –4 4 16 ≥ 8r Since r is multiplied by 8, divide both sides by 8 to undo the multiplication. 2 ≥ r

15 Holt McDougal Algebra 1 2-5 Solving Inequalities with Variables on Both Sides Check It Out! Example 3b Solve the inequality and graph the solutions. 0.5x – 0.3 + 1.9x < 0.3x + 6 2.4x – 0.3 < 0.3x + 6 + 0.3 2.4x < 0.3x + 6.3 –0.3x 2.1x < 6.3 Since 0.3 is subtracted from 2.4x, add 0.3 to both sides. Since 0.3x is added to 6.3, subtract 0.3x from both sides. x < 3 Since x is multiplied by 2.1, divide both sides by 2.1. Simplify. 2.4x – 0.3 < 0.3x + 6

16 Holt McDougal Algebra 1 2-5 Solving Inequalities with Variables on Both Sides Check It Out! Example 3b Continued x < 3 –5 –4 –3–2 –1 01234 5

17 Holt McDougal Algebra 1 2-5 Solving Inequalities with Variables on Both Sides Additional Example 4A: 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.

18 Holt McDougal Algebra 1 2-5 Solving Inequalities with Variables on Both Sides 2(3y – 2) – 4 ≥ 3(2y + 7) Solve the inequality. Additional Example 4B: 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.

19 Holt McDougal Algebra 1 2-5 Solving Inequalities with Variables on Both Sides 4(y – 1) ≥ 4y + 2 4y – 4 ≥ 4y + 2 Distribute 4 on the left side. Check It Out! Example 4a Solve the inequality. The same variable term (4y) appears on both sides. Look at the other terms. For any number 4y, subtracting 4 will never result in a higher number than adding 2. No values of y make the inequality true. There are no solutions.

20 Holt McDougal Algebra 1 2-5 Solving Inequalities with Variables on Both Sides Solve the inequality. x – 2 < x + 1 Check It Out! Example 4b The same variable term (x) appears on both sides. Look at the other terms. For any number x, subtracting 2 will always result in a lesser number than adding 1. All values of x make the inequality true. All real numbers are solutions.


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