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Inequalities with 2-5 Solving Variables on Both Sides Objective Solve inequalities that contain variable terms on both sides. Holt McDougal Algebra 1 Inequalities with 2-5 Solving Variables on Both Sides Example 1A: Solving Inequalities with Variables on Both Sides Solve the inequality and graph the solutions. y ≤ 4y + 18 To collect the variable terms on one y ≤ 4y + 18 side, subtract y from both sides. –y –y 0 ≤ 3y + 18 –18 – 18 Since 18 is added to 3y, subtract 18 from both sides to undo the addition. –18 ≤ 3y Since y is multiplied by 3, divide both sides by 3 to undo the multiplication. –6 ≤ y (or y –6) Holt McDougal Algebra 1 –10 –8 –6 –4 –2 0 2 4 6 8 10 Inequalities with 2-5 Solving Variables on Both Sides Example 1B: Solving Inequalities with Variables on Both Sides Solve the inequality and graph the solutions. 4m – 3 < 2m + 6 To collect the variable terms on one –2m – 2m side, subtract 2m from both sides. 2m – 3 < +6 +3 +3 2m < 9 Since 3 is subtracted from 2m, add 3 to both sides to undo the subtraction Since m is multiplied by 2, divide both sides by 2 to undo the multiplication. 4 Holt McDougal Algebra 1 5 6 Inequalities with 2-5 Solving Variables on Both Sides Check It Out! Example 1a Solve the inequality and graph the solutions. 4x ≥ 7x + 6 4x ≥ 7x + 6 –7x –7x To collect the variable terms on one side, subtract 7x from both sides. –3x ≥ 6 x ≤ –2 Since x is multiplied by –3, divide both sides by –3 to undo the multiplication. Change ≥ to ≤. –10 –8 –6 –4 –2 Holt McDougal Algebra 1 0 2 4 6 8 10 Inequalities with 2-5 Solving Variables on Both Sides Check It Out! Example 1b Solve the inequality and graph the solutions. 5t + 1 < –2t – 6 5t + 1 < –2t – 6 +2t +2t 7t + 1 < –6 – 1 < –1 7t < –7 7t < –7 7 7 t < –1 –5 –4 –3 –2 –1 Holt McDougal Algebra 1 0 1 2 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. 3 4 5 Inequalities with 2-5 Solving 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. Holt McDougal Algebra 1 Inequalities with 2-5 Solving Variables on Both Sides Example 2 Continued Home Cleaning Company siding charge 312 plus + $12 per window 12 times # of windows is less than Power Clean cost per window • w < 36 312 + 12w < 36w – 12w –12w 312 < 24w times # of windows. • w 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. 13 < w The Home Cleaning Company is less expensive for houses with more than 13 windows. Holt McDougal Algebra 1 Inequalities with 2-5 Solving 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. A-Plus Advertising plus fee of $24 24 + $0.10 per flyer times 0.10 • Holt McDougal Algebra 1 Print and # of flyers is less than More’s cost f < 0.25 times # of flyers. per flyer • f Inequalities with 2-5 Solving Variables on Both Sides Check It Out! Example 2 Continued 24 + 0.10f < 0.25f –0.10f –0.10f 24 To collect the variable terms, subtract 0.10f from both sides. < 0.15f Since f is multiplied by 0.15, divide both sides by 0.15 to undo the multiplication. 160 < f More than 160 flyers must be delivered to make A-Plus Advertising the lower cost company. Holt McDougal Algebra 1 Inequalities with 2-5 Solving Variables on Both Sides Example 3A: Simplify Each Side Before Solving Solve the inequality and graph the solutions. 2(k – 3) > 6 + 3k – 3 Distribute 2 on the left side of 2(k – 3) > 3 + 3k the inequality. 2k + 2(–3) > 3 + 3k 2k – 6 > 3 + 3k –2k – 2k –6 > 3 + k –3 –3 –9 > k Holt McDougal Algebra 1 To collect the variable terms, subtract 2k from both sides. Since 3 is added to k, subtract 3 from both sides to undo the addition. Inequalities with 2-5 Solving Variables on Both Sides Example 3A Continued –9 > k –12 –9 Holt McDougal Algebra 1 –6 –3 0 3 Inequalities with 2-5 Solving Variables on Both Sides Example 3B: Simplify Each Side Before Solving Solve the inequality and graph the solution. 0.9y ≥ 0.4y – 0.5 0.9y ≥ 0.4y – 0.5 –0.4y –0.4y To collect the variable terms, subtract 0.4y from both sides. 0.5y ≥ – 0.5 0.5y ≥ –0.5 0.5 0.5 y ≥ –1 –5 –4 –3 –2 –1 Holt McDougal Algebra 1 0 1 2 3 Since y is multiplied by 0.5, divide both sides by 0.5 to undo the multiplication. 4 5 Inequalities with 2-5 Solving Variables on Both Sides Check It Out! Example 3a 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 5(2 – r) ≥ 3(r – 2) the right side of the inequality. 5(2) – 5(r) ≥ 3(r) + 3(–2) Since 6 is subtracted from 3r, 10 – 5r ≥ 3r – 6 add 6 to both sides to undo +6 +6 the subtraction. 16 − 5r ≥ 3r Since 5r is subtracted from 16 + 5r +5r add 5r to both sides to undo the subtraction. 16 ≥ 8r Holt McDougal Algebra 1 Inequalities with 2-5 Solving Variables on Both Sides Check It Out! Example 3a Continued 16 ≥ 8r Since r is multiplied by 8, divide both sides by 8 to undo the multiplication. 2≥r –6 –4 Holt McDougal Algebra 1 –2 0 2 4 Inequalities with 2-5 Solving 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 Simplify. 2.4x – 0.3 < 0.3x + 6 Since 0.3 is subtracted 2.4x – 0.3 < 0.3x + 6 from 2.4x, add 0.3 to + 0.3 + 0.3 both sides. 2.4x < 0.3x + 6.3 Since 0.3x is added to –0.3x –0.3x 6.3, subtract 0.3x from both sides. 2.1x < 6.3 Since x is multiplied by 2.1, divide both sides by 2.1. x<3 Holt McDougal Algebra 1 Inequalities with 2-5 Solving Variables on Both Sides Check It Out! Example 3b Continued x<3 –5 –4 –3 –2 –1 Holt McDougal Algebra 1 0 1 2 3 4 5 Inequalities with 2-5 Solving 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. Holt McDougal Algebra 1 Inequalities with 2-5 Solving Variables on Both Sides Additional Example 4B: All Real Numbers as Solutions or No Solutions Solve the inequality. 2(3y – 2) – 4 ≥ 3(2y + 7) 6y – 8 ≥ 6y + 21 Distribute 2 on the left side and 3 on the right side and combine like terms. 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. Holt McDougal Algebra 1 Inequalities with 2-5 Solving Variables on Both Sides Check It Out! Example 4a Solve the inequality. 4(y – 1) ≥ 4y + 2 4y – 4 ≥ 4y + 2 Distribute 4 on the left side. 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. Holt McDougal Algebra 1 Inequalities with 2-5 Solving Variables on Both Sides Check It Out! Example 4b Solve the inequality. x–2<x+1 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. Holt McDougal Algebra 1