1 / 29

Warm Up Solve each equation. 1. 2 x = 7 x + 15 2.

–4. –3. –2. –1. –5. –6. 0. Warm Up Solve each equation. 1. 2 x = 7 x + 15 2. x = –3. 3 y – 21 = 4 – 2 y. y = 5. z = –1. 3. 2(3z + 1) = – 2( z + 3). 4. 3( p – 1) = 3 p + 2. no solution. b < –3. 5. Solve and graph 5(2 – b ) > 5 2. Objective.

sealey
Download Presentation

Warm Up Solve each equation. 1. 2 x = 7 x + 15 2.

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. –4 –3 –2 –1 –5 –6 0 Warm Up Solve each equation. 1. 2x = 7x + 15 2. x = –3 3y – 21 = 4 – 2y y = 5 z = –1 3. 2(3z + 1) = –2(z + 3) 4. 3(p –1) = 3p + 2 no solution b < –3 5. Solve and graph 5(2 –b) > 52.

  2. Objective Solve inequalities that contain variable terms on both sides.

  3. Some inequalities have variable terms on both sides of the inequality symbol. You can solve these inequalities like you solved equations with variables on both sides. Use the properties of inequality to “collect” all the variable terms on one side and all the constant terms on the other side.

  4. Directions: Solve the inequality and graph the solutions.

  5. y ≤ 4y + 18 –y –y 0 ≤ 3y + 18 –18 – 18 –18 ≤ 3y –8 –10 –6 –4 0 2 4 6 8 10 –2 Example 1 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. –2m –2m 2m – 3 < + 6 + 3 + 3 2m < 9 4 5 6 Example 2 4m – 3 < 2m + 6 To collect the variable terms on one side, subtract 2m from both sides. 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.

  7. 4x ≥ 7x + 6 –7x –7x x ≤ –2 –8 –10 –6 –4 0 2 4 6 8 10 –2 Example 3 4x ≥ 7x + 6 To collect the variable terms on one side, subtract 7x from both sides. –3x ≥ 6 Since x is multiplied by –3, divide both sides by –3 to undo the multiplication. Change ≥ to ≤.

  8. 5t + 1 < –2t – 6 +2t +2t 7t + 1 < –6 – 1 < –1 7t < –7 7t < –7 7 7 t < –1 –4 –1 5 –3 –2 0 1 2 3 4 –5 Example 4 5t + 1 < –2t – 6 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.

  9. Example 5: 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.

  10. 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 5 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.

  11. $0.10 per flyer Print and More’s cost per flyer A-Plus Advertising fee of $24 # of flyers. # of flyers is less than times times plus Example 6 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

  12. –0.10f –0.10f Example 6 Continued 24 + 0.10f < 0.25f To collect the variable terms, subtract 0.10f from both sides. 24 < 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.

  13. You may need to simplify one or both sides of an inequality before solving it. Look for like terms to combine and places to use the Distributive Property.

  14. –2k –2k –3 –3 Example 7 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

  15. –12 –9 –6 –3 0 3 Example 7 Continued –9 > k

  16. –0.4y –0.4y 0.5y ≥–0.5 0.5 0.5 –4 –1 5 –3 –2 0 1 2 3 4 –5 Example 8 0.9y ≥ 0.4y – 0.5 0.9y ≥ 0.4y – 0.5 To collect the variable terms, subtract 0.4y from both sides. 0.5y ≥ – 0.5 Since y is multiplied by 0.5, divide both sides by 0.5 to undo the multiplication. y ≥ –1

  17. +6 +6 + 5r +5r Example 9 5(2 – r) ≥ 3(r – 2) Distribute 5 on the left side of the inequality and distribute 3 on the right side of the inequality. 5(2 – r) ≥ 3(r – 2) 5(2) –5(r) ≥ 3(r) + 3(–2) Since 6 is subtracted from 3r, add 6 to both sides to undo the subtraction. 10 – 5r ≥ 3r – 6 16 − 5r ≥ 3r Since 5r is subtracted from 16 add 5r to both sides to undo the subtraction. 16 ≥ 8r

  18. –6 –4 –2 0 2 4 Example 9 Continued 16 ≥ 8r Since r is multiplied by 8, divide both sides by 8 to undo the multiplication. 2 ≥ r

  19. + 0.3 + 0.3 –0.3x –0.3x Example 10 0.5x – 0.3 + 1.9x < 0.3x + 6 2.4x –0.3 < 0.3x + 6 Simplify. Since 0.3 is subtracted from 2.4x, add 0.3 to both sides. 2.4x –0.3 < 0.3x + 6 2.4x < 0.3x + 6.3 Since 0.3x is added to 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

  20. –4 –1 5 –3 –2 0 1 2 3 4 –5 Example 10 Continued x < 3

  21. There are special cases of inequalities called identities and contradictions.

  22. –2x –2x Example 11 2x – 7 ≤ 5 + 2x 2x – 7 ≤ 5 + 2x Subtract 2x from both sides.  –7 ≤ 5 True statement. The inequality 2x − 7 ≤ 5 + 2x is an identity. All values of x make the inequality true. Therefore, all real numbers are solutions.

  23. –6y –6y Example 12 2(3y – 2) – 4 ≥ 3(2y + 7) Distribute 2 on the left side and 3 on the right side. 2(3y – 2) – 4 ≥ 3(2y + 7) 2(3y) – 2(2) – 4 ≥ 3(2y) + 3(7) 6y – 4 – 4 ≥ 6y + 21 6y – 8 ≥ 6y + 21 Subtract 6y from both sides.  –8 ≥ 21 False statement. No values of y make the inequality true. There are no solutions.

  24. –4y –4y Example 13 Solve the inequality. 4(y – 1) ≥ 4y + 2 4(y – 1) ≥ 4y + 2 Distribute 4 on the left side. 4(y) + 4(–1) ≥ 4y + 2 4y – 4 ≥ 4y + 2 Subtract 4y from both sides.  –4 ≥ 2 False statement. No values of y make the inequality true. There are no solutions.

  25. –x –x Example 14 Solve the inequality. x – 2 < x + 1 x – 2 < x + 1 Subtract x from both sides.  True statement. –2 < 1 All values of x make the inequality true. All real numbers are solutions.

  26. Lesson Summary: Part I Solve each inequality and graph the solutions. 1. t < 5t + 24 t > –6 2. 5x – 9 ≤ 4.1x –81 x ≤–80 3. 4b + 4(1 – b) > b – 9 b < 13

  27. Lesson Summary: Part II 4. Rick bought a photo printer and supplies for $186.90, which will allow him to print photos for $0.29 each. A photo store charges $0.55 to print each photo. How many photos must Rick print before his total cost is less than getting prints made at the photo store? Rick must print more than 718 photos.

  28. Lesson Summary: Part III Solve each inequality. 5. 2y – 2 ≥ 2(y + 7) contradiction, no solution 6. 2(–6r – 5) < –3(4r + 2) identity, all real numbers

More Related