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Solving with Three Variables

Solving with Three Variables. Section 3-6 pages 148-155. Solving Using Elimination. Solve the system. 2x + y - z = 5 3x - y + 2z = -1 x - y - z = 0. Solving the system. Step 1: Pick two of the equations and eliminate the variable of your choice: Let ’ s pick  &  2x + y - z = 5

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Solving with Three Variables

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  1. Solving with Three Variables Section 3-6 pages 148-155

  2. Solving Using Elimination Solve the system. • 2x + y - z = 5 3x - y + 2z = -1 x - y - z = 0

  3. Solving the system Step 1: Pick two of the equations and eliminate the variable of your choice: Let’s pick  &  • 2x + y - z = 5 • 3x - y + 2z = -1 Let’s eliminate z: Multiply first line by 2 4x + 2y - 2z = 10 3x - y + 2z = -1 7x + 1y = 9

  4. Step 2: Pick two other equations and eliminate the same variable you eliminated in step 1. Let’s pick  &  2x + y - z = 5 x - y - z = 0 Multiply  by -1. -2x - y + z = -5 x - y - z = 0 -1x - 2y = -5

  5. Step 3: Take the two equations that you just eliminated z from: 7x + 1y = 9 -1x - 2y = -5 Pick one of the variables to eliminate. Let’s pick x. 7x + 1y = 9 -7x - 14y = -35 -13y = -26 Solve for y: y = 2

  6. Step 4: Plug the y value into one of the equations from step 3 and solve for x. 1x + 1(2) = 9 x = 7 Step 5: Plug the x and y into one of the original equations and solve for z. 2(7) + 2 - z = 5 16 - z = 5 -z = -11 z = 11

  7. Final Step Plug the answers into a (x, y, z) (7, 2, 11)

  8. Try Some x – 3y + 2z = 11 -x + 4y +3z = 5 2x – 2y – 4z = 2

  9. x – 3y + z = 6 2x – 5y – z = -2 -x + y + 2z = 7

  10. 3x + 2y – z = 12 -4x + y – 2z = 4 x – 3y + z + -4

  11. x + y + 2z = 3 2x + y + 3z = 7 -x – 2y + z = 10

  12. Using to solve word problems A stadium has 49,000 seats. Seats sell for $25 in Section A, $20 in Section B, and $15 in Section C. The number of seats in Section A equals the total number of seats in Sections B and C. Suppose the stadium takes in $1,052,000 from each sold out event. How many seats does each section hold?

  13. Assign your variables: x = Section A, y = Section B, z = Section C Set up your equations: x + y + z = 49,000 25x + 20y + 15z = 1,052,000 x = y + z Make sure all variables are on one side: x + y + z = 49,000 25x + 20y + 15z = 1,052,000 x - y - z = 0

  14. Pick two equations to eliminate one of the variables: -25(x + y + z = 49,000) 25x + 20y + 15z = 1,052,000 -25x -25y - 25z = -1,225,000 -5y - 10z = -173,000 Pick two different equations and eliminate same variable: x + y + z = 49,000 -1(x - y - z = 0) -x + y + z = 0 2y + 2z = 49,000

  15. Add your two new equations together: -5y - 10z = -173,000 5(2y + 2z = 49,000) 10y + 10z = 245,000 5y = 72,000 y = 14,400 Plug into one of the equations above and solve for z. 10(14,400) + 10z = 245,000 144,000 + 10z = 245,000 10z = 101,000 z = 10,100

  16. Plug the y and z into one of the original equations to find x: x + 14,400 + 10,100 = 49,000 x + 24,500 = 49,000 x = 24,500 Section A has 24,500 seats Section B has 14,400 seats Section C has 10,100 seats

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