1 / 19

3.9 Solving Systems of Equations in Three Variables

3.9 Solving Systems of Equations in Three Variables. Algebra II Mrs. Spitz Fall 2006. Objective. Solve a system of equations in three variables. Assignment. Pp. 147-148 #11-23 all. Application.

robert-page
Download Presentation

3.9 Solving Systems of Equations in Three Variables

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. 3.9 Solving Systems of Equations in Three Variables Algebra II Mrs. Spitz Fall 2006

  2. Objective • Solve a system of equations in three variables.

  3. Assignment • Pp. 147-148 #11-23 all

  4. Application • Courtney has a total of 256 points on three Algebra tests. His score on the first test exceeds his score on the second by 6 points. His total score before taking the third test was 164 points. What were Courtney’s test scores on the three tests?

  5. Explore • Problems like this one can be solved using a system of equations in three variables. Solving these systems is very similar to solving systems of equations in two variables. Try solving the problem • Let f = Courtney’s score on the first test • Let s = Courtney’s score on the second test • Let t = Courtney’s score on the third test.

  6. Plan • Write the system of equations from the information given. f + s + t = 256 f – s = 6 f + s = 164 The total of the scores is 256. The difference between the 1st and 2nd is 6 points. The total before taking the third test is the sum of the first and second tests..

  7. Solve • Now solve. First use elimination on the last two equations to solve for f. f – s = 6 f + s = 164 2f = 170 f = 85 The first test score is 85.

  8. Solve • Then substitute 85 for f in one of the original equations to solve for s. f + s = 164 85 + s = 164 s = 79 The second test score is 79.

  9. Solve • Next substitute 85 for f and 79 for s in f + s + t = 256. f + s + t = 256 85 + 79 + t = 256 164 + t = 256 t = 92 The third test score is 92. Courtney’s test scores were 85, 79, and 92.

  10. Examine • Now check your results against the original problem. • Is the total number of points on the three tests 256 points? 85 + 79 + 92 = 256 ✔ • Is one test score 6 more than another test score? 79 + 6 = 85 ✔ • Do two of the tests total 164 points? 85 + 79 =164 ✔ • Our answers are correct.

  11. Solutions? • You know that a system of two linear equations doesn’t necessarily have a solution that is a unique ordered pair. Similarly, a system of three linear equations in three variables doesn’t always have a solution that is a unique ordered triple.

  12. Graphs • The graph of each equation in a system of three linear equations in three variables is a plane. Depending on the constraints involved, one of the following possibilities occurs.

  13. The three planes intersect at one point. So the system has a unique solution. 2. The three planes intersect in a line. There are an infinite number of solutions to the system. Graphs

  14. Graphs 3. Each of the diagrams below shows three planes that have no points in common. These systems of equations have no solutions.

  15. Substitute 4 for z and 1 for y in the first equation, x + 2y + z = 9 to find x. x + 2y + z = 9 x + 2(1) + 4 = 9 x + 6 = 9 x = 3 Solution is (3, 1, 4) Check: 1st 3 + 2(1) +4 = 9 ✔ 2nd 3(1) -4 = 1 ✔ 3rd 3(4) = 12 ✔ Solve the third equation, 3z = 12 3z = 12 z = 4 Substitute 4 for z in the second equation 3y – z = -1 to find y. 3y – (4) = -1 3y = 3 y = 1 Ex. 1: Solve this system of equations

  16. Set the next two equations together and multiply the first times 2. 2(x + 3y – 2z = 11) 2x + 6y – 4z = 22 3x - 2y + 4z = 1 5x + 4y = 23 Next take the two equations that only have x and y in them and put them together. Multiply the first times -1 to change the signs. Set the first two equations together and multiply the first times 2. 2(2x – y + z = 3) 4x – 2y +2z = 6 x + 3y -2z = 11 5x + y = 17 Ex. 2: Solve this system of equations

  17. Now you have y = 2. Substitute y into one of the equations that only has an x and y in it. 5x + y = 17 5x + 2 = 17 5x = 15 x = 3 Now you have x and y. Substitute values back into one of the equations that you started with. 2x – y + z = 3 2(3) - 2 + z = 3 6 – 2 + z = 3 4 + z = 3 z = -1 Next take the two equations that only have x and y in them and put them together. Multiply the first times -1 to change the signs. -1(5x + y = 17) -5x - y = -17 5x + 4y = 23 3y = 6 y = 2 Ex. 2: Solve this system of equations

  18. Ex. 2: Check your work!!! Solution is (3, 2, -1) Check: 1st 2x – y + z = 2(3) – 2 – 1 = 3 ✔ 2nd x + 3y – 2z = 11 3 + 3(2) -2(-1) = 11 ✔ 3rd 3x – 2y + 4z 3(3) – 2(2) + 4(-1) = 1 ✔

  19. Now you have y = 2. Substitute y into one of the equations that only has an x and y in it. 5x + y = 17 5x + 2 = 17 5x = 15 x = 3 Now you have x and y. Substitute values back into one of the equations that you started with. 2x – y + z = 3 2(3) - 2 + z = 3 6 – 2 + z = 3 4 + z = 3 z = -1 Next take the two equations that only have x and y in them and put them together. Multiply the first times -1 to change the signs. -1(5x + y = 17) -5x - y = -17 5x + 4y = 23 3y = 6 y = 2 Ex. 2: Solve this system of equations

More Related