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EXAMPLE 1

STEP 1. Rewrite the system as a linear system in two variables. 4 x + 2 y + 3 z = 1. Add 2 times Equation 3. 12 x – 2 y + 8 z = –2. to Equation 1. EXAMPLE 1. Use the elimination method. Solve the system. 4 x + 2 y + 3 z = 1. Equation 1. 2 x – 3 y + 5 z = –14.

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EXAMPLE 1

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  1. STEP 1 Rewrite the system as a linear system in two variables. 4x + 2y + 3z = 1 Add 2 times Equation 3 12x – 2y + 8z = –2 to Equation 1. EXAMPLE 1 Use the elimination method Solve the system. 4x + 2y + 3z = 1 Equation 1 2x – 3y + 5z = –14 Equation 2 6x – y + 4z = –1 Equation 3 SOLUTION 16x + 11z = –1 New Equation 1

  2. Add – 3 times Equation 3 to Equation 2. –18x + 3y –12z = 3 STEP 2 Solve the new linear system for both of its variables. Add new Equation 1 –16x – 7z = –11 and new Equation 2. EXAMPLE 1 Use the elimination method 2x – 3y + 5z = –14 –16x – 7z = –11 New Equation 2 16x+ 11z = –1 4z = –12 z = –3 Solve for z. x = 2 Substitute into new Equation 1 or 2 to find x.

  3. Substitute x = 2 and z = – 3 into an original equation and solve for y. STEP 3 EXAMPLE 1 Use the elimination method 6x–y + 4z = –1 Write original Equation 3. 6(2) –y + 4(–3) = –1 Substitute 2 for xand –3 for z. y = 1 Solve for y.

  4. –4x – 4y – 4z = –12 Add –4 times Equation 1 4x + 4y + 4z = 7 to Equation 2. EXAMPLE 2 Solve a three-variable system with no solution Solve the system. x + y + z = 3 Equation 1 4x + 4y + 4z = 7 Equation 2 3x – y + 2z = 5 Equation 3 SOLUTION When you multiply Equation 1 by – 4 and add the result to Equation 2, you obtain a false equation. 0 = –5 New Equation 1

  5. EXAMPLE 2 Solve a three-variable system with no solution Because you obtain a false equation, you can conclude that the original system has no solution.

  6. STEP 1 Rewrite the system as a linear system in two variables. x + y + z = 4 Add Equation 1 x + y – z = 4 to Equation 2. EXAMPLE 3 Solve a three-variable system with many solutions x + y + z = 4 Solve the system. Equation 1 x + y – z = 4 Equation 2 3x + 3y + z = 12 Equation 3 SOLUTION 2x + 2y = 8 New Equation 1

  7. 3x + 3y + z = 12 Solve the new linear system for both of its variables. STEP 2 –4x – 4y = –16 Add –2 times new Equation 1 4x + 4y = 16 to new Equation 2. EXAMPLE 3 Solve a three-variable system with many solutions x + y – z = 4 Add Equation 2 to Equation 3. 4x + 4y = 16 New Equation 2 Because you obtain the identity 0 = 0, the system has infinitely many solutions.

  8. EXAMPLE 3 Solve a three-variable system with many solutions STEP 3 Describe the solutions of the system. One way to do this is to divide new Equation 1 by 2 to get x + y = 4, or y = –x + 4. Substituting this into original Equation 1 produces z = 0. So, any ordered triple of the form (x, –x + 4, 0) is a solution of the system.

  9. 1. 3x + y – 2z = 10 6x – 2y + z = –2 x + 4y + 3z = 7 (1, 3, –2) ANSWER for Examples 1, 2 and 3 GUIDED PRACTICE Solve the system.

  10. 2. x + y – z = 2 2x + 2y – 2z = 6 5x + y – 3z = 8 ANSWER no solution for Examples 1, 2 and 3 GUIDED PRACTICE

  11. 3. x + y + z = 3 x + y – z = 3 2x + 2y + z = 6 Infinitely many solutions ANSWER for Examples 1, 2 and 3 GUIDED PRACTICE

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