3.6 Systems with Three Variables

There are three strategies for solving systems with three equations: Solve by elimination Solve an equivalent system Solving by substitution

Solve by elimination x – 3y +3z = -4 2x + 3y – z = 15 4x – 3y –z = 19 Why pick elimination? Note that the y values in this set are inverses. They will go away once you add the equations together. x – 3y +3z = -4 2x + 3y – z = 15 4x – 3y –z = 19 Step 1: Pair the equation to eliminate y, because the y terms are already additive inverses x – 3y +3z = -4 2x + 3y – z = 15 3x +2z = 11 2x + 3y – z = 15 4x – 3y –z = 19 6x -2z =34

3x +2z = 11 6x – 2z = 34 9x = 45 x=5 3 (5) + 2z = 11 15 + 2z = 11 Step 2: Take the two new equations and use elimination to solve for the remaining two variables (in this case, x and z) Plug in x into either equation to solve for z 3x +2z = 11 6x – 2z = 34 9x = 45 x=5 3 (5) + 2z = 11 15 + 2z = 11 2z = -4 z = -2

x=5, z =-2 x – 3y +3z = -4 5 – 3y + 3(-2)=-4 5 – 3y -6 = -4 -3y = -3 Step 3: Substitute the values for x and z back into one of the original equations (whichever seems easier) and solve for y x=5, z =-2 x – 3y +3z = -4 5 – 3y + 3(-2)=-4 5 – 3y -6 = -4 -3y = -3 y=1 x – 3y +3z = -4 2x + 3y – z = 15 4x – 3y –z = 19 (5, 1, -2) is the solution

3x3 Systems #1 2x + y – z = 5 3x - y + 2z = -1 x - y - z = 0

3x3 Systems #1 2x + y – z = 5 3x - y + 2z = -1 x - y - z = 0

Solving an Equivalent System What is an equivalent system? Very similar to the direct elimination method, but in this case you may need to multiply one or more of the equations by a constant in order for a variable to drop out 2x + y – z = 5 x + 4y + 2z = 16 15x + 6y - 2z = 12 Step 1: Pair the equation to eliminate z, because the z terms are already additive inverses in equations #2 2x + y - z = 5 x + 4y +2z = 16 x + 4y + 2z = 16 15x + 6y - 2z = 12 Note: the Zs are inverses in system #2 but not in #1. Multiply the first equation by 2 to get the Zs so that they will drop out

(2x + y - z = 5)2 x + 4y +2z = 16 x + 4y + 2z = 16 4x + 2y -2 z = 10 Step 2: Manipulate the first equation by multiplying by 2. Now the z variable will drop out of the equations when you combine them. (2x + y - z = 5)2 x + 4y +2z = 16 x + 4y + 2z = 16 15x + 6y - 2z = 12 16x + 10y = 28 4x + 2y -2 z = 10 x + 4y +2z = 16 5x + 6y =26 These equations are all ready set up to allow us to eliminate z. Simply add together to get your two variable equation.

5x + 6y =16 (5x + 6y =16) 5 16x + 10y = 28 (16x + 10y = 28) -3 Step 3: Take the two new equations and use elimination to solve for the remaining two variables (in this case, x and y) Again, we do not have a variable that simply drops out for us. Get a common multiple that you can eliminate. In this case the easiest is to get the Ys to cancel 5x + 6y =16 16x + 10y = 28 (5x + 6y =16) 5 (16x + 10y = 28) -3 25x + 30y = 130 -48x - 30y = -84 -23x = 46 x = -2

5(-2) + 6y =26 5x + 6y =26 -10 + 6y = 26 16x + 10y = 28 6y = 36 Y = 6 Plug x= -2 into either equation to solve for y. 5(-2) + 6y =26 -10 + 6y = 26 6y = 36 Y = 6 5x + 6y =26 16x + 10y = 28

2x + y – z = 5 2 (-2) + (6) – z = 5 -4 +6 –z =5 2 – z = 5 -z = 3 Step 4: Substitute the values for x and y back into one of the original equations (whichever seems easier) and solve for y 2x + y – z = 5 2 (-2) + (6) – z = 5 -4 +6 –z =5 2 – z = 5 -z = 3 z = -3 2x + y – z = 5 x + 4y + 2z = 16 15x + 6y - 2z = 12 (-2, 6, -3) is the solution

3x3 Systems #2 x + 2y + z =10 2x – y + 3z = -5 2x – 3y – 5z = 27 Note: this ALMOST Looks like we could solve The system by elimination. What do we have to do to make one if the variables Inverse of one another? -2(x + 2y + z =10) 2x – y + 3z = -5 2x – 3y – 5z = 27

Now try on your own -2x - 4y -2z =-20 2x – y + 3z = -5

Now try on your own -2x - 4y -2z =-20 2x – y + 3z = -5

Solve by Substitution X – 2y + z = -4 -4x + y -2z = 1 2x + 2y – z = 10 Why pick substitution? Note in each equation a different variable is by itself. X – 2y + z = -4 -4x + y -2z = 1 2x + 2y – z = 10 Step 1: Choose one equation to solve for one of its variables. x - 2y + z = -4 x – 2y = -z -4 x = 2y – z - 4 X = 2y – z - 4

-4x + y -2z = 1 -4(2y – z – 4) + y -2z = 1 -8y + 4z + 16 + y -2z = 1 Step 2: Substitute the expression for x into each of the other two equations. -4x + y -2z = 1 -4(2y – z – 4) + y -2z = 1 -8y + 4z + 16 + y -2z = 1 -7y + 2z + 16 = 1 -7y+ 2z = -15 2x + 2y –z =10 2(2y – z – 4) + 2y –z =10 4y – 2z – 8 + 2y –z =10 6y – 3z – 8 =10 6y -3z = 18

-7y + 2z = -15 6y – 3z = 18 (-7y + 2z = -15)3 (6y – 3z = 18)2 Step 3: Take the remaining equations and solve the system to solve for z and y. We need to eliminate a variable. Let’s get rid of z since on of the terms is already negative -7y + 2z = -15 6y – 3z = 18 (-7y + 2z = -15)3 (6y – 3z = 18)2 -21y + 6z = -45 12y – 6z = 36 -9y = -9 y =1 -7y + 2z = -15 -7(1) + 2z = -15 2z = -8 z = -4 Plug y into one of the equations to solve for z

x – 2y + z = -4 x- 2y + z = -4 -4x + y -2z = 1 x -2(1) + (-4) = -4 Step 4: Substitute the values for y and z into one of the original equations x – 2y + z = -4 -4x + y -2z = 1 2x + 2y – z = 10 x- 2y + z = -4 x -2(1) + (-4) = -4 x – 2 – 4 = -4 x – 6 = -4 x = 2 (2, 1, -4) is the solution to the system of equations.

3x3 Systems #3 x – 3y + z =6 2x – 5y – z = -2 -x + y + 2z = 7

3x3 Systems #3 x – 3y + z = 6 2x – 5y – z = -2 -x + y + 2z = 7

3x3 Systems #4 6a + 12b – 8c = 24 9a + 18b – 12c = 30 Hint: can you factor?

3x3 Systems #4 6a + 12b – 8c = 24 9a + 18b – 12c = 30 The a, b, and c terms of the first two equations are the same. They exist as parallel planes, so there will be no solution to this set of equations!

3x3 Systems #5 4x – 6y + 4z = 12 6x – 9y + 6z = 18 5x – 8y + 10z =20

3x3 Systems #5 4x – 6y + 4z = 12 6x – 9y + 6z = 18 5x – 8y + 10z =20 The first two equations will be planes that lie on top of each other. Equations 1 and 3 (or 2 and 3) are planes that will intersect. How do planes intersect? A Line We are looking for one point where all three planes will intersect. In this case we say that there is no unique solution.