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Solving Systems of Linear Equations in Three Variables AII, 2.0: Students solve systems of linear equations and inequalities (in two or three variables) by substitution, with graphs, or with matrices. LA, 6.0: Students demonstrate an understanding that linear systems are inconsistent (have no solutions), have exactly one solution, or have infinitely many solutions Before we begin QUIZ ON SOLVING A SYSTEM OF TWO EQUATIONS (30 MINUTES) Solving Systems of Linear Equations in Three Variables Objectives 1. Solve systems of linear equations in three variables a) Use the linear combination method b) Solve a system with NO solution c) Solve a system with Infinitely many solutions 2. Solve systems of three variables by substitution Key Words • Ordered triple • Linear combination • System of equations The Linear Combination Method (3-Variable System) 1) a) WRITE THIS DOWN!!! The steps to solving a system of linear equations in three variables. Example: REWRITE the linear system in three variables as a linear system in two variables using the linear combination method. b) 2) SOLVE the new linear system for both of its variables. a) 2 − + 6 = −4 6 + 4 − 5 = −7 −4 − 2 + 5 = 9 b) c) d) 3) −3,4,1 4) Combine two equations and eliminate one variable Combine a different pair of equations to eliminate the same variable Combine the equations that resulted from steps 1 and 2 to eliminate one of the two variables that are left Solve for the remaining variable Substitute that value into one of the equations with two variables Solve for the second variable SUBSTITUTE the values found in Step 2 into on of the original equations and solve for the remaining variable. CHECK the solution in each of the original equations Example 1 Use the Linear Combination Method Solve the system. 3x + 2y + 4z = 11 Equation 1 2x – y + 3z = 4 Equation 2 5x – 3y + 5z = – 1 Equation 3 SOLUTION STEP 1 Rewrite the system as a system in two variables. First, add 2 times Equation 2 to Equation 1 to eliminate y. 3x + 2y + 4z = 11 3x + 2y + 4z = 11 2x – y + 3z = 4 4x – 2y + 6z = 8 7x + 10z = 19 New Equation 1 Example 1 Use the Linear Combination Method Now add –3 times Equation 2 to Equation 3 to eliminate y. 5x – 3y + 5z = – 1 5x – 3y + 5z = – 1 2x – y + 3z = 4 – 6x + 3y – 9z = – 12 – x – 4z = – 13 New Equation 2 STEP 2 Solve the new system of linear equations in two variables. First, add 7 times new Equation 2 to new Equation 1 to eliminate x. 7x + 10z = 19 – x – 4z = – 13 7x + 10z = 19 – 7x – 28z = – 91 – 18z = – 72 Example 1 Use the Linear Combination Method z = 4 Solve for z. Substitute 4 for z in new Equation 1 or 2 and solve for x to =3. – get x STEP 3 Substitute –3 for x and 4 for z in one 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 Equation 2 Substitute –3 for x and 4 for z. Multiply. Combine like terms. Solve for y. Example 1 Use the Linear Combination Method STEP 4 Check by substituting 3–for x, 2 for y, and 4 for z in each of the original equations. ANSWER The solution is x triple ( –3, 2, 4). – = 3, y =2, and z =4, or the ordered Example 2 Solve a System with No Solution x + y + z = 2 Solve the system. Equation 1 3x + 3y + 3z = 14 Equation 2 x – 2y + z = 4 Equation 3 SOLUTION Multiply Equation 1 by 3– and add the result to Equation 2. – 3x – 3y – 3z = – 6 3x + 3y + 3z = 14 0 = 8 Add –3 times Equation 1 to Equation 2. False statement Example 2 Solve a System with No Solution ANSWER Because solving the system resulted in the false statement = 0 8, the original system of equations has no solution. Example 3 Solve a System with Infinitely Many Solutions Solve the system. x + y + z = 4 Equation 1 x + y – z = 4 Equation 2 3x + 3y + z = 12 Equation 3 SOLUTION STEP 1 Rewrite the system as a system in two variables. x + y + z = 4 Add Equation 1 x + y – z = 4 to Equation 2. 2x + 2y = 8 New Equation 1 Example 3 Solve a System with Infinitely Many Solutions x + y – z = 4 3x + 3y + z = 12 4x + 4y = 16 Add Equation 2 to Equation 3. New Equation 2 STEP 2 Solve the new system of linear equations in two variables. – 4x – 4y = – 16 4x + 4y = 16 0 = 0 Add –2 times new Equation 1 to new Equation 2. Example 3 Solve a System with Infinitely Many Solutions ANSWER Because solving the system resulted in the true statement = 0 0, the original system of equations has infinitely many solutions. The three planes intersect in a line. Checkpoint Solve Systems Tell how many solutions the linear system has. If the system has one solution, solve the system. Then check your solution. 1. x + y + 3z = 1 x + y – z = 1 ANSWER 1; (1, 0, 0) ANSWER no solution – x – 3y + 4z = – 1 2. x – y + z = 4 3x + y – 2z = – 2 2x – 2y + 2z = 5 Checkpoint Solve Systems Tell how many solutions the linear system has. If the system has one solution, solve the system. Then check your solution. 3. x + y + 2z = 10 – x + 2y + z = 5 – x + 4y + 3z = 15 ANSWER infinitely many solutions The Substitution Method (3-Variable System) WRITE THIS DOWN!!! The steps to solving a system of linear equations in three variables. Example: −− =3 − + 2 + 5 = −1 + + 4 = 4 2, −2,1 1) SOLVE for one of the variables in any of the equations 2) SUBSTITUTE the variable found in Step 1 into another equation. 3) SUBSTITUTE the variable found in Step 1 into the last equation. 4) SOLVE the new linear system of two equations in two variables for both of its variables. NOTE: use preferred method. 5) SUBSTITUTE the values found in Step 4 into original equation to find the last variable. 6) CHECK the solution in each of the original equations Conclusion Summary • How do you solve a system of linear equations in three variables? – Rewrite the system as a linear system in two variables by eliminating one variable. Solve that system for each variable. Substitute those two values into an equation in the original system to get the values of the third variable. Assignment • To be completed in class – Pg156 #(14, 18, 22, 26, 31, 32) • Extra Practice Assignment to be completed by the end of the week. – Systems of two equations