 # 3-6 Solving Systems of Linear Equations in Three Variables Objective: CA 2.0: Students solve systems of linear equations and inequalities in three variables.

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3-6 Solving Systems of Linear Equations in Three Variables Objective: CA 2.0: Students solve systems of linear equations and inequalities in three variables by substitution, with graphs.

The linear combination method (3- variable systems): Step 1: Use the linear combination method to rewrite the linear system in three variables as a linear system in two variables. Step 2: Solve the new linear system for both of its variables.

Step 3 Substitute the value found in step 2 into one of the original equations and solves for the remaining variable. If you obtain a false equation, such as: 0 = 1, then the system has no solutions. 0 = 0, then the system has infinitely many solutions.

Example 1: Solve the system using the Linear Combination Method.

Step 1: Eliminate one of the variables in two of the original equations. 3x + 2y + 4z = 11 Equation 1 2(2x – y + 3z = 4) Equation 2 4x – 2y + 6z = 8 7x + 0y + 10z = 19 7x + 10z = 19 new Equation 1

Step 1: Eliminate one of the variables in two of the original equations. -3(2x – y + 3z = 4) Equation 2 5x – 3y + 5z = -1Equation 3 -6x + 3y – 9z = -12 -x – 4z = -18 new Equation 2

Step 2: Solve the new system of equations 7x + 10z = 19 -x – 4z = -13 7x + 10z = 19 7(-x – 4z) = -13(7) + -7x - 28z = -91 -7x – 28x = -91 -18z = -72 z = 4

Substitute z = 4 into new equation 1 or 2 and solve for x -x – 4(4) = -13 -x -16 = -13 -x = 3 x = -3

Step 3: Substitute x = -3 and z = 4 into any of the original equations and solve for y. 2x – y + 3z = 4 2(-3) – y + 3(4) = 4 -6 – y + 12 = 4 -y + 6 = 4 -y = -2 y = 2

The solution is x = -3, y = 2, z = 4 or (-3, 2, 4)

Example 2: Solve the system of Equations x + y + z = 2 3x + 3y + 3z = 14 x – 2y + z = 4

Step 1: Eliminate one of the variables in two of the original equations. -3(x + y + z = 2) -3x – 3y -3z = -6 Add -3 times Equation 1 to Equation 2 -3x- 3y - 3z = -6 new Eq.1 3x + 3y + 3z = 14 Eq. 2 0 = 8 Since 0 ≠ 8 the result is a false equation. This system has NO SOLUTIONS.

Example 3: Solve the system of equations. x + y + z = 2 x + y – z = 2 2x + 2y + z = 4

Step 1: Use the linear combination method to rewrite the linear system in three variables as a linear system in two variables. x + y + z = 2 + x + y – z = 2 2x + 2y = 4 New Equation 1

x + y – z = 2 + 2x + 2y + z = 4 3x + 3y = 6 Equation 2 Equation 3 New Equation 2

Step 2: Solve the new linear system for both of its variables. 3(2x + 2y) = 4(3) -2(3x + 3y) = 6(-2) 6x + 6y = 12 -6x – 6y = -12 0 = 0 0 = 0 is an identity therefore there are infinitely many solutions to this system of equations. 3 (New equation 1) -2 (New equation 2)

Home work page 181 12 –30 every third one, 35 – 39 odd, 45 – 71 every other odd.

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