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Algebra II - Linear Equations in Three Variables

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    Study Guide Algebra II – Linear Equa Ɵ ons in Three Variables 1. Quiz: Linear Equa Ɵ ons: Solu Ɵ ons Using Elimina Ɵ on with Three Variables 1. Question Solve the system of equations: Answer Choices  x = 1; y = 3; z = − 2  x = − 1; y = − 3; z = 2  x = 1; y = − 3; z = 2 Correct Answer x = 1; y = 3; z = − 2 Why This Is Correct Substitute the values into each equation to verify: 1. 2(1) + 3(3) + 4( − 2) = 2 + 9 − 8 = 3 ✔ 2. 3(1) − 4(3) + 2( − 2) = 3 − 12 − 4 = − 13 ✔

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    Study Guide 3. 1 + 5(3) − 3( − 2) = 1 + 15 + 6 = 22 ✔ Since the values satisfy all three equations, the correct solution is x = 1, y = 3, z = − 2 . 2. Question Solve the system: Answer Choices  x = 2; y = 4; z = 5  x = − 2; y = − 4; z = 5  x = − 2; y = 4; z = − 5 Correct Answer x = 2; y = 4; z = 5 Why This Is Correct Check the values: 1. 2(2) − 3(4) = 4 − 12 = − 8 3(5) − 7 = 15 − 7 = 8 Rearranging confirms the equality. 2. 4(5) − 2(2) + 3(4)

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    Study Guide = 20 − 4 + 12 = 28 3. 4(4) + 2(5) − 3(2) = 16 + 10 − 6 = 20 The equations hold for x = 2, y = 4, z = 5 , confirming the correct solution. 3. Question Solve the system: Answer Choices  x = 5; y = 7; z = − 6  x = − 5; y = 7; z = − 6  x = 5; y = − 7; z = 6 Correct Answer x = 5; y = 7; z = − 6 Why This Is Correct Substitute: 1. 3(7) + 2(5) − 5( − 6) = 21 + 10 + 30 = 61 ✔ 2. 4( − 6) + 2(7) + 5

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    Study Guide − 24 + 14 + 5 = − 5 ✔ 3. 3(5) + 2(7) − 3( − 6) = 15 + 14 + 18 = 47 ✔ All equations are satisfied, confirming the solution. 4. Question Solve the system: Answer Choices  x = 3; y = − 5; z = − 4  x = − 3; y = − 5; z = 4  x = − 3; y = 5; z = − 4 Correct Answer x = 3; y = − 5; z = − 4 Why This Is Correct Substitute the values: 1. 6( − 5) + 2(3) + 4( − 4) = − 30 + 6 − 16 = − 40 (matches the balanced equation form) 2. 5( − 4) + 7( − 5) − 4(3)

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    Study Guide = − 20 − 35 − 12 = − 67 3. 5( − 5) − 8( − 4) + 6(3) = − 25 + 32 + 18 = 25 These values satisfy the system, confirming the correct solution. 5. Question Solve the system: Answer Choices  x = − 4; y = 6; z = 12  x = − 4; y = − 6; z = 12  x = 4; y = − 6; z = − 12 Correct Answer x = − 4; y = 6; z = 12 Why This Is Correct Substitute values: 1. 4( − 4) + 3(6) + 3(12) = − 16 + 18 + 36 = 38 (matches equation balance) 2. 3( − 4) + 7(6)

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    Study Guide = − 12 + 42 = 30 2(12) − 78 = 24 − 78 = − 54 3. 5(12) + 4( − 4) + 4(6) = 60 − 16 + 24 = 68 Thus x = − 4, y = 6, z = 12 satisfies the system. 2. Linear Equa Ɵ ons: Solu Ɵ ons Using Elimina Ɵ on with Three Variables When solving a system of equations with three variables, it’s a little more challenging than with two variables, but you can still use the elimination method, which works like the one used for two variables. There’s also another method using a 3x3 matrix, but let’s focus on elimination for now. Steps for Elimination with Three Variables: 1. Write the equations in standard form : Make sure the equations don’t have any decimals or fractions. 2. Choose a variable to eliminate : Pick one of the variables (like x, y, or z) to eliminate, and choose two equations to use. 3. Eliminate the variable : Use the chosen two equations to eliminate the variable you picked in Step 2. 4. Solve for the other two variables : After eliminating one variable, you'll be left with two equations and two variables. Solve them to find values for the remaining variables. 5. Substitute the values back : Take the values you found in Step 4 and put them back into any of the original equations to find the last variable. 6. Check your work : Double - check by plugging all your answers into the original equations to make sure everything adds up.

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    Study Guide Example 1: Solving with Elimination Consider this system of equations: Step 1: Choose a variable to eliminate Let's choose x to eliminate. We’ll use equations (1) and (2): This gives us the new equation: Now, adding these two equations: The result is: Step 2: Choose two more equations Next, let's eliminate x again but use equations (2) and (3). After multiplying the appropriate terms, you'll get the new equation:

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    Study Guide Step 3: Solve the system of equations (4) and (5) Now solve the two equations: By solving, you find z = 3 . Step 4: Substitute back to find y and x Now substitute z = 3 into one of the earlier equations (like equation (4)) to find y = 2 . Then use the value of y and z in equation (2) to find x = - 1 . Step 5: Final check Substitute x = - 1 , y = 2 , and z = 3 into all original equations to ensure they are correct. The solution is: Example 2: Another Set of Equations Given this system of equations: Step 1: Eliminate y First, use equations (2) and (3) to eliminate y . After manipulating these, substitute the results to solve for x and z . Step 2: Final Substitution Once x = 4 and z = 3 are found, substitute these into the remaining equations to find y = - 2 .

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