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3-4 Solving Systems of Linear Equations in 3 Variables

3-4 Solving Systems of Linear Equations in 3 Variables. Algebra 2 Textbook. Terms to remember. Visualizing what this means. Using elimination to solve. STEP 1. Rewrite the system as a linear system in two variables. 4 x + 2 y + 3 z = 1. Add 2 times Equation 3.

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3-4 Solving Systems of Linear Equations in 3 Variables

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  1. 3-4 Solving Systems of Linear Equations in 3 Variables Algebra 2 Textbook

  2. Terms to remember

  3. Visualizing what this means

  4. Using elimination to solve

  5. 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 A

  6. 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 A – 16x – 7z = –11 and new Equation B. EXAMPLE 1 Use the elimination method 2x – 3y + 5z = – 14 – 16x – 7z = – 11 New Equation B 16x+ 11z = –1 4z = –12 z = – 3 Solve for z. x = 2 Substitute into new Equation Aor Bto find x.

  7. 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.

  8. 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.

  9. – 4x – 4y – 4z = – 12 Add –4 times Equation 1 to Equation 2. 4x + 4y + 4z = 7 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. New Equation A 0 = – 5 Because you obtain a false equation, you can conclude that the original system has no solution.

  10. 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 A

  11. 3x + 3y + z = 12 Solve the new linear system for both of its variables. STEP 2 – 4x – 4y = – 16 Add –2 times new Equation A 4x + 4y = 16 to new Equation B. 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 B Because you obtain the identity 0 = 0, the system has infinitely many solutions.

  12. 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.

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

  14. 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

  15. 3.20 4.80 3.68 Step 1 Write the Equations Eq. 1 p + c = 100 Eq. 2 3.2p + 4.8c = 368 Step 2 Use substitution to solve the system Solve Eq. 1 for pp = 100-c Substitute 100-c in for p in Eq. 2 3.2(100-c) + 4.8c = 368 Use distribute property to simplify 320 – 3.20c + 4.8c = 368 Combine like terms 320 + 1.6c = 368 Subtract 320 from each side 1.6c = 48 Solve for c then p by substitution c = 30 p = 100 – 30 = 70 Step 3 Write the answers down with correct labels 30 pounds of cashews & 70 pounds of peanuts

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