1 / 16

Solving Systems of Equations with Three Variables Using Elimination Method

This guide walks through the step-by-step process of solving a system of equations with three variables using the elimination method. The example provided illustrates how to eliminate variables systematically to find the values of x, y, and z. The method ensures clarity by demonstrating how to manipulate the equations to isolate and solve for each variable effectively. This is further complemented by a real-world application involving a stadium's seating arrangement, enriching the learning experience through context.

cassie
Télécharger la présentation

Solving Systems of Equations with Three Variables Using Elimination Method

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. 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

More Related