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Solving Inequalities with Variables on Both Sides. 2-5. Holt Algebra 1. Warm Up. Lesson Presentation. Lesson Quiz. Holt McDougal Algebra 1. Warm Up Solve each equation. 1. 2 x = 7 x + 15 2. . 3 y – 21 = 4 – 2 y. 3. 2(3z + 1) = – 2( z + 3).

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  1. Solving Inequalities with Variables on Both Sides 2-5 Holt Algebra 1 Warm Up Lesson Presentation Lesson Quiz Holt McDougal Algebra 1
  2. Warm Up Solve each equation. 1. 2x = 7x + 15 2. 3y – 21 = 4 – 2y 3. 2(3z + 1) = –2(z + 3) 4. 3(p –1) = 3p + 2 5. Solve and graph 5(2 –b) > 52.
  3. Objective Solve inequalities that contain variable terms on both sides.
  4. Steps to solve inequalities with variable terms on both sides of the inequality symbol. Use the properties of inequality to “collect” all the variable terms on one side. Then “collect” all the constant terms on the other side.
  5. y ≤ 4y + 18 –y –y 0 ≤ 3y + 18 –18 – 18 –18 ≤ 3y –8 –10 –6 –4 0 2 4 6 8 10 –2 Let’s try this together Solve the inequality and graph the solutions. y ≤ 4y + 18 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. –6 ≤ y (or y –6)
  6. Let’s try another one together Solve the inequality and graph the solutions. 4m – 3 < 2m + 6
  7. Try these yourself!!! Solve the inequality and graph the solutions. b.) 5t + 1 < –2t – 6 a.) 4x≥ 7x + 6
  8. Sample Word Problem 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.
  9. Home Cleaning Company siding charge Power Clean cost per window is less than # of windows. # of windows times $12 per window plus times – 12w –12w Example 2 Continued 312 + 12 • w < 36 • w 312 + 12w < 36w To collect the variable terms, subtract 12w from both sides. 312 < 24w 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.
  10. One more word problem 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?
  11. *Don’t forget you may need to simplify one or both sides of an inequality before solving it. Combine like terms Use Distributive Property.
  12. –12 –9 –6 –3 0 3 –2k –2k –3 –3 Let’s do this together Solve the inequality and graph the solutions. 2(k – 3) > 6 + 3k – 3 Distribute 2 on the left side of the inequality. 2(k – 3) > 3 + 3k 2k+ 2(–3)> 3 + 3k 2k –6 > 3 + 3k To collect the variable terms, subtract 2k from both sides. –6 > 3 + k Since 3 is added to k, subtract 3 from both sides to undo the addition. –9 > k
  13. Let’s do this together Solve the inequality and graph the solution. 0.9y ≥ 0.4y – 0.5
  14. Now you try!!! Solve the inequality and graph the solutions. d.) 0.5x– 0.3 + 1.9x < 0.3x + 6 c.) 5(2 – r) ≥ 3(r – 2)
  15. All Real Solutions Inequalities that are true no matter what value is substituted for the variable. No Solutions Inequalities that are false no matter what value is substituted for the variable. *To determine if the answer is all real solutions or no solutions check the final answer. If it is a true statement (all real solutions), if false statement (no solutions).
  16. 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.
  17. Additional Example 4B: All Real Numbers as Solutions or No Solutions Solve the inequality. 2(3y –2) – 4 ≥ 3(2y + 7) 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.
  18. Now you Try!!! Solve each inequality. e.) 4(y – 1) ≥ 4y + 2 f.) x – 2 < x + 1
  19. HOMEWORK: MIDTERM REVIEW WORKSHEETS
  20. HOMEWORK PG. 129-131 #20-35, 49, 56, 58-60
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