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Graphing Systems of Inequalities

Learn how to graph systems of inequalities and identify common regions. Solve multi-step problems using graphs. Examples provided.

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Graphing Systems of Inequalities

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  1. EXAMPLE 1 Graph a system of two inequalities Graph the system of inequalities. y > –2x – 5 Inequality 1 y <x + 3 Inequality 2

  2. STEP 1 Graph each inequality in the system. Use red for y > – 2x – 5and bluefory≤ x + 3. STEP 2 Identify the region that is common to both graphs. It is the region that is shaded purple. EXAMPLE 1 Graph a system of two inequalities SOLUTION

  3. ` 2 y >– x + 4 3 EXAMPLE 2 Graph a system with no solution Graph the system of inequalities. 2x + 3y < 6 Inequality 1 Inequality 2

  4. STEP 1 Graph each inequality in the system. Use red for2x + 3y <6and bluefory > – x + 4. 2 3 STEP 2 Identify the region that is common to both graphs. There is no region shaded both red and blue. So, the system has no solution. EXAMPLE 2 Graph a system with no solution SOLUTION

  5. y > x + 4 EXAMPLE 3 Graph a system with an absolute value inequality Graph the system of inequalities. y < 3 Inequality 1 Inequality 2

  6. STEP 1 Graph each inequality in the system. Use red for y ≤3and bluefor y > x + 4 . STEP 2 Identify the region that is common to both graphs. It is the region that is shaded purple. EXAMPLE 3 Graph a system with an absolute value inequality SOLUTION

  7. EXAMPLE 4 Solve a multi-step problem SHOPPING A discount shoe store is having a sale, as described in the advertisement shown. • Use the information in the ad to write a system of inequalities for the regular footwear prices and possible sale prices. • Graph the system of inequalities. • Use the graph to estimate the range of possible sale prices for footwear that is regularly priced at $70.

  8. STEP 1 Write a system of inequalities. Let xbe the regular footwear price and let ybe the sale price. From the information in the ad, you can write the following four inequalities. EXAMPLE 4 Solve a multi-step problem SOLUTION x ≥ 20 Regular price must be at least $20. x ≤ 80 Regular price can be at most $80. y ≥ 0.4x Sale price is at least (100 – 60)% = 40% of regular price. y ≤ 0.9x Sale price is at most (100 – 10)% = 90% of regular price.

  9. STEP 2 Graph each inequality in the system.Then identify the region that is common to all the graphs. It is the region that is shaded. Identify the range of possible sale prices for $70 footwear. From the graph you can see that when x = 70, the value of yis between these values: STEP 3 EXAMPLE 4 Solve a multi-step problem 0.4(70) = 28 and 0.9(70) = 63 So, the value ofysatisfies28 ≤ y ≤ 63.

  10. ANSWER Therefore, footwear regularly priced at $70 sells for between $28 and $63, inclusive, during the sale. EXAMPLE 4 Solve a multi-step problem

  11. STEP 1 Rewrite the system as a linear system in two variables. 4x + 2y + 3z = 1 Add 2 times Equation 3 12x – 2y + 8z = – 2 to Equation 1. EXAMPLE 1 Use the elimination method Solve the system. 4x + 2y + 3z = 1 Equation 1 2x – 3y + 5z = – 14 Equation 2 6x – y + 4z = – 1 Equation 3 SOLUTION 16x + 11z = – 1 New Equation 1

  12. Add – 3 times Equation 3 to Equation 2. – 18x + 3y – 12z = 3 STEP 2 Solve the new linear system for both of its variables. Add new Equation 1 – 16x – 7z = –11 and new Equation 2. EXAMPLE 1 Use the elimination method 2x – 3y + 5z = – 14 – 16x – 7z = – 11 New Equation 2 16x+ 11z = –1 4z = –12 z = – 3 Solve for z. x = 2 Substitute into new Equation 1 or 2 to find x.

  13. Substitute x = 2 and z = – 3 into an original equation and solve for y. STEP 3 EXAMPLE 1 Use the elimination method 6x–y + 4z = – 1 Write original Equation 3. 6(2) –y + 4(– 3) = – 1 Substitute 2 for xand –3 for z. y = 1 Solve for y.

  14. – 4x – 4y – 4z = – 12 Add –4 times Equation 1 4x + 4y + 4z = 7 to Equation 2. EXAMPLE 2 Solve a three-variable system with no solution Solve the system. x + y + z = 3 Equation 1 4x + 4y + 4z = 7 Equation 2 3x – y + 2z = 5 Equation 3 SOLUTION When you multiply Equation 1 by – 4 and add the result to Equation 2, you obtain a false equation. 0 = – 5 New Equation 1

  15. EXAMPLE 2 Solve a three-variable system with no solution Because you obtain a false equation, you can conclude that the original system has no solution.

  16. STEP 1 Rewrite the system as a linear system in two variables. x + y + z = 4 Add Equation 1 x + y – z = 4 to Equation 2. EXAMPLE 3 Solve a three-variable system with many solutions x + y + z = 4 Solve the system. Equation 1 x + y – z = 4 Equation 2 3x + 3y + z = 12 Equation 3 SOLUTION 2x + 2y = 8 New Equation 1

  17. 3x + 3y + z = 12 Solve the new linear system for both of its variables. STEP 2 – 4x – 4y = – 16 Add –2 times new Equation 1 4x + 4y = 16 to new Equation 2. EXAMPLE 3 Solve a three-variable system with many solutions x + y – z = 4 Add Equation 2 to Equation 3. 4x + 4y = 16 New Equation 2 Because you obtain the identity 0 = 0, the system has infinitely many solutions.

  18. EXAMPLE 3 Solve a three-variable system with many solutions STEP 3 Describe the solutions of the system. One way to do this is to divide new Equation 1 by 2 to get x + y = 4, or y = – x + 4. Substituting this into original Equation 1 produces z = 0. So, any ordered triple of the form (x, –x + 4, 0) is a solution of the system.

  19. Marketing The marketing department of a company has a budget of $30,000 for advertising. A television ad costs $1000, a radio ad costs $200, and a newspaper ad costs $500. The department wants to run 60 ads per month and have as many radio ads as television and newspaper ads combined. How many of each type of ad should the department run each month? EXAMPLE 4 Solve a system using substitution

  20. Write verbal models for the situation. STEP 1 EXAMPLE 4 Solve a system using substitution SOLUTION

  21. STEP 2 Write a system of equations. Let xbe the number of TV ads, ybe the number of radio ads, and zbe the number of newspaper ads. Rewrite the system in Step 2 as a linear system in twovariables by substituting x+ zfor yin Equations 1 and 2. STEP 3 EXAMPLE 4 Solve a system using substitution Equation1 x + y + z = 60 1000x + 200y + 500z = 30,000 Equation 2 y = x + z Equation 3

  22. EXAMPLE 4 Solve a system using substitution x + y + z = 60 Write Equation 1. x + (x + z) + z = 60 Substitute x + zfor y. 2x + 2z = 60 New Equation 1 1000x + 200y + 500z = 30,000 Write Equation 2. 1000x + 200(x + z) + 500z = 30,000 Substitute x + zfor y. 1200x + 700z = 30,000 New Equation 2

  23. STEP 4 Solve the linear system in two variables from Step 3. – 1200x – 1200z = – 36,000 Add 2600 times new Equation 1 1200x +700z = 30,000 to new Equation2. EXAMPLE 4 Solve a system using substitution – 500z = – 6000 z = 12 Solve for z. x = 18 Substitute into new Equation 1 or 2 to find x. y = 30 Substitute into an original equation to find y.

  24. ANSWER The solution is x = 18, y = 30, and z = 12, or (18, 30,12). So, the department should run 18 TV ads, 30 radio ads, and 12 newspaper ads each month. EXAMPLE 4 Solve a system using substitution

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