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

Terms to remember

Visualizing what this means

Using elimination to solve

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

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

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

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

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

Solve a three-variable system with many solutions EXAMPLE 3 Solve a three-variable system with 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 linear system in two variables. x + y + z = 4 x + y – z = 4 Add Equation 1 to Equation 2. 2x + 2y = 8 New Equation A

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

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.

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

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

Step 1 Write the Equations 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 p p = 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