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Chapter3.6
Chapter3.6
Chapter3.6
Chapter3.6
Chapter3.6
Chapter3.6
Chapter3.6
Chapter3.6
Chapter3.6
Chapter3.6
Chapter3.6
Chapter3.6
Chapter3.6
Chapter3.6
Chapter3.6
Chapter3.6
Chapter3.6
Chapter3.6
Chapter3.6
Chapter3.6
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Chapter3.6

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  • 1. Warm Up California Standards Lesson Presentation Preview
  • 2. Warm Up Compare. Use < or >. 1. 5 7 2. –3 –4 3. 2.5 –2.7 4. –8 –7 Solve. 5. 4 + y = 16 6. m – 7 = 14 7. –3 = 8 + w 8. 7 = t + 10 < > > < 12 21 – 11 – 3
  • 3. AF4.0 Students solve simple linear equations and inequalities over the rational numbers. California Standards
  • 4. When you add or subtract the same number on both sides of an inequality, the resulting statement will still be true. – 2 < 5 +7 +7 5 < 12 You can find solution sets of inequalities the same way you find solutions of equations, by isolating the variable.
  • 5. Solve and graph the inequality. Additional Example 1A: Solving Inequalities by Adding or Subtracting x + 3 > –5 x + 3 > –5 – 3 –3 x > –8 Since 3 is added x, subtract 3 from both sides. – 9  8 –7  6  5  4  3  2  1 0 1 2 3 4 5
  • 6. Additional Example 1A Continued Check x + 3 > –5 – 4 + 3 > –5 ?  – 1 > –5 ? Substitute –4 for x. According to the graph –4 should be a solution and –9 should not be a solution. So –4 is a solution. x + 3 > –5 – 9 + 3 > –5 ?  – 6 > –5 ? Substitute –9 for x. So –9 is not a solution. 
  • 7. m – 4 ≥ –2 m – 4 ≥ –2 + 4 + 4 m ≥ 2 Since 4 is subtracted from m, add 4 to both sides. Solve and graph the inequality. Additional Example 1B: Solving Inequalities by Adding or Subtracting – 1 0 1 2 3 4 5 6 7 8 9 10 11 12
  • 8. r + 3 ≤ – 3 r + 3 ≤ –3 – 3 –3 r ≤ –6 Since 3 is added to r, subtract 3 from both sides. Additional Example 1C: Solving Inequalities by Adding or Subtracting Solve and graph the inequality. – 9  8 –7  6  5  4  3  2  1 0 1 2 3 4 5
  • 9. 5 > n + 1 Since 1¼ is added to n, subtract 1¼ from both sides. Additional Example 1D: Solving Inequalities by Adding or Subtracting Solve and graph the inequality. 3 4 1 4 5 > n + 1 3 4 1 4 – 1 – 1 1 4 1 4 4 > n 1 2 – 7  6 –5  4  3  2  1 0 1 2 3 4 5 6 7
  • 10. Solve and graph the inequality. Check It Out! Example 1A x + 4 > –2 x + 4 > –2 – 4 –4 x > –6 Since 4 is added x, subtract 4 from both sides. – 9  8 –7  6  5  4  3  2  1 0 1 2 3 4 5
  • 11. Check It Out! Example 1A Continued Check x + 4 > –2 2 + 4 > –2 ?   6 > –2 ? Substitute 2 for x. According to the graph 2 should be a solution and –8 should not be a solution. So 2 is a solution. x + 4 > –2 – 8 + 4 > –2 ?  – 4 > –2 ? Substitute –8 for x. So –8 is not a solution. 
  • 12. w – 8 ≥ –3 w – 8 ≥ –3 + 8 + 8 w ≥ 5 Since 8 is subtracted from w, add 8 to both sides. Solve and graph the inequality. Check It Out! Example 1B – 1 0 1 2 3 4 5 6 7 8 9 10 11 12
  • 13. c + 6 ≤ – 1 c + 6 ≤ –1 – 6 – 6 c ≤ –7 Since 6 is added to c, subtract 6 from both sides. Check It Out! Example 1C Solve and graph the inequality. – 9  8 –7  6  5  4  3  2  1 0 1 2 3 4 5
  • 14. 3 > n + 1 Since 1 is added to n, subtract 1 from both sides. Check It Out! Example 1D Solve and graph the inequality. 2 3 1 3 3 > n + 1 2 3 1 3 – 1 – 1 1 3 1 3 2 > n 1 3 1 3 1 3 – 7  6 –5  4  3  2  1 0 1 2 3 4 5 6 7
  • 15. While training for a race, Ann’s goal is to run at least 3.5 miles each day. She has already run 1.8 miles today. Write and solve an inequality to find out how many more miles she must run today. Additional Example 2: Sports Application Let m = the number of additional miles. 1.8 + m ≥ 3.5 – 1.8 –1.8 m ≥ 1.7 Ann should run at least 1.7 more miles. Since 1.8 is added to m, subtract 1.8 from both sides. 1.8 miles plus additional miles is at least 3.5 miles. 1.8 + m ≥ 3.5
  • 16. 2 is greater than 1.7. Substitute 2 for m. Check Additional Example 2 Continued x 1 is less than 1.7. Substitute 1 for m. 1.8 + m ≥ 3.5 1.8 + 2 ≥ 3.5 ? 3.8 ≥ 3.5 ? 1.8 + m ≥ 3.5 1.8 + 1 ≥ 3.5 ? 2.8 ≥ 3.5 ?
  • 17. Tim’s company produces recycled paper. They produce 60.5 lb of paper each day. They have already produced at least 20.2 lb today. Write and solve an inequality to find out how many more pounds Tim’s company must produce. Check It Out! Example 2 Let p = the number of additional pounds of paper. 20.2 + p ≥ 60.5 – 20.2 –  20.2 p ≥ 40.3 Tim’s company should produce at least 40.3 lb more of paper. Since 20.2 is added to p, subtract 20.2 from both sides. 20.2 lbs plus additional pounds is at least 60.5 lb. 20.2 + p ≥ 60.5
  • 18. 41 is greater than 40.3. Substitute 41 for p. Check Check It Out! Example 2 Continued x 40 is less than 40.3. Substitute 40 for p. 20.2 + p ≥ 60.5 20.2 + 41 ≥ 60.5 ? 61.2 ≥ 60.5 ? 20.2 + p ≥ 60.5 20.2 + 40 ≥ 60.5 ? 60.2 ≥ 60.5 ?
  • 19. Lesson Quiz: Part I Solve and graph each inequality. 1. g – 7 < –3 2. 5 + s ≥ 4 3. –5.1 ≤ x – 5.1 4. 3 + y > 4 1 5 g < 4  1 0 1 2 3 4 5  4  3  2  1 0 1 2 • s ≥ –1  4  3  2  1 0 1 2 • x ≥ 0 y < 1/5 2/5 3/5 4/5 1 1 1/5 4 5
  • 20. Lesson Quiz: Part II 5. Tasha is folding letters for a fundraiser. She knows there are at least 300 letters, and she has already folded 125 of them. Write and solve an inequality to show how many more letters she must fold. 125 + x ≥ 300; x ≥ 175

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