Systems with Three Variables

Solve by elimination x – 3y +3z = -4 2x + 3y – z = 15 4x – 3y –z = 19 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 #2 x + 2y + z =10 2x – y + 3z = -5 2x – 3y – 5z = 27

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