Solving Systems of Three Linear Equations in Three Variables

Solving Systems of Three Linear Equations in Three Variables The Elimination Method

Solutions of a system with 3 equations • The solution to a system of three linear equations in three variables is an ordered triple. • (x, y, z) • The solution must be a solution of all 3 equations.

Is (–3, 2, 4) a solution of this system? • 3x + 2y + 4z = 11 • 2x – y + 3z = 4 • 5x – 3y + 5z = –1 3(–3) + 2(2) + 4(4) = 11 2(–3) – 2 + 3(4) = 4 5(–3) – 3(2) + 5(4) = –1 P P P Yes, it is a solution to the system because it is a solution to all 3 equations.

This lesson will focus on the Elimination Method.

Use elimination to solve the following system of equations. x – 3y + 6z = 21 3x + 2y – 5z = –30 2x – 5y + 2z = –6

Step 1 Rewrite the system as two smaller systems, each containing two of the three equations.

x – 3y + 6z = 21 3x + 2y – 5z = –30 2x – 5y + 2z = –6 x – 3y + 6z = 21 x – 3y + 6z = 21 3x + 2y – 5z = –30 2x – 5y + 2z = –6

Step 2 Eliminate THE SAME variable in each of the two smaller systems. Any variable will work, but sometimes one may be a bit easier to eliminate. I choose x for this system.

(x – 3y + 6z = 21) 3x + 2y – 5z = –30 –3x + 9y – 18z = –63 3x + 2y – 5z = –30 11y – 23z = –93 (x – 3y + 6z = 21) 2x – 5y + 2z = –6 –2x + 6y – 12z = –42 2x – 5y + 2z = –6 y – 10z = –48 (–3) (–2)

Step 3 Write the resulting equations in two variables together as a system of equations. Solve the system for the two remaining variables.

11y – 23z = –93 y – 10z = –48 11y – 23z = –93 –11y + 110z = 528 87z = 435 z = 5 y – 10(5) = –48 y – 50 = –48 y = 2 (–11)

Step 4 Substitute the value of the variables from the system of two equations in one of the ORIGINAL equations with three variables.

x – 3y + 6z = 21 3x + 2y – 5z = –30 2x – 5y + 2z = –6 I choose the first equation. x – 3(2) + 6(5) = 21 x – 6 + 30 = 21 x + 24 = 21 x = –3

Step 5 CHECK the solution in ALL 3 of the original equations. Write the solution as an ordered triple.

P –3 – 3(2) + 6(5) = 21 3(–3) + 2(2) – 5(5) = –30 2(–3) – 5(2) + 2(5) = –6 x – 3y + 6z = 21 3x + 2y – 5z = –30 2x – 5y + 2z = –6 P P The solution is (–3, 2, 5).

It is very helpful to neatly organize your work on your paper in the following manner. (x, y, z)

x+3y-z=-11 2x+y+z=1 z’s are easy to cancel! 3x+4y=-10 2. 2x+y+z=1 5x-2y+3z=21 Must cancel z’s again! -6x-3y-3z=-3 5x-2y+3z=21 -x-5y=18 2(2)+(-4)+z=1 4-4+z=1 3. 3x+4y=-10 -x-5y=18 Solve for x & y. 3x+4y=-10 -3x-15y+54 -11y=44 y=- 4 3x+4(-4)=-10 x=2 (2, - 4, 1) Solve the system. x+3y-z=-11 2x+y+z=1 5x-2y+3z=21 z=1

2x+2y+z=5 4x+4y+2z=6 Cancel z’s again. -4x-4y-2z=-10 4x+4y+2z=6 0=- 4 Doesn’t make sense! No solution -x+2y+z=3 2x+2y+z=5 z’s are easy to cancel! -x+2y+z=3 -2x-2y-z=-5 -3x=-2 x=2/3 Solve the system. -x+2y+z=3 2x+2y+z=5 4x+4y+2z=6

3. x+y=3 2x+2y=6 Cancel the x’s. -2x-2y=-6 2x+2y=6 0=0 This is true. ¸ many solutions -2x+4y+z=1 3x-3y-z=2 z’s are easy to cancel! x+y=3 3x-3y-z=2 5x-y-z=8 Cancel z’s again. -3x+3y+z=-2 5x-y-z=8 2x+2y=6 Solve the system. -2x+4y+z=1 3x-3y-z=2 5x-y-z=8

Try this one. x – 6y – 2z = –8 –x + 5y + 3z = 2 3x – 2y – 4z = 18 (4, 3, –3)

Here’s another one to try. –5x + 3y + z = –15 10x + 2y + 8z = 18 15x + 5y + 7z = 9 (1, –4, 2)

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?

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.

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..

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.

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.

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.

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.

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.

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.

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

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

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

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

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

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 ✔

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

Solving Systems of Three Linear Equations in Three Variables