Linear Systems 3x3

Grade 12 ยท algebra ยท 27 practice problems ยท read aloud

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Solving 3x3 Linear Systems

What is it? ๐Ÿค”

A 3x3 linear system has three equations with three variables (like x, y, z). We solve it to find where three planes intersect in 3D space. This is crucial for real-world problems with three constraints, like business models, physics, and engineering.

How to Solve: Elimination Method

  1. Pair and Eliminate: Choose two pairs of equations. Eliminate the same variable from both pairs.
  2. Create a 2x2 System: You'll get two new equations with just two variables. Solve this smaller system.
  3. Back Substitute: Plug your two found values into one original equation to find the third variable.
  4. Check Your Solution: Verify your (x, y, z) triple works in all three original equations.

Example 1

System:
(1) x + y - z = 0
(2) 2x - y + z = 3
(3) -x + 2y + 2z = 6

Step 1: Eliminate 'y'. Add (1)+(2): 3x = 3 โ†’ x = 1.
Multiply (1) by 2 and add to (3): x + 4z = 6.
Step 2: With x=1, 1 + 4z = 6 โ†’ z = 5/4.
Step 3: Substitute into (1): 1 + y - 5/4 = 0 โ†’ y = 1/4.
Solution: (1, 1/4, 5/4)

Example 2

System:
(1) 2x + y = 1
(2) y - z = 2
(3) x + z = 3

Step 1: Notice (2) and (3) are ready. Add them: x + y = 5.
Step 2: Use with (1): x + y = 5 and 2x + y = 1. Subtract: -x = 4 โ†’ x = -4.
Step 3: From (1): 2(-4) + y = 1 โ†’ y = 9. From (2): 9 - z = 2 โ†’ z = 7.
Solution: (-4, 9, 7)

Common Mistakes โš ๏ธ

  • Inconsistent Elimination: Not eliminating the same variable from both equation pairs.
  • Sign Errors: Forgetting negative signs when multiplying equations.
  • Arithmetic Slips: Simple addition/subtraction errors.
  • Not Checking: Always verify your solution in all three original equations!

Tips & Tricks

  • Look for Zeros: Choose a variable to eliminate that already has a coefficient of 1 or -1.
  • Stay Organized: Label your equations (1), (2), (3) and write each step clearly.
  • Check Early: Verify your 2x2 system solution before back-substituting.

How to Practice

Start with systems that have "nice" integer solutions. Practice the elimination process until it's comfortable. Then move to problems with fractions. Create your own systems by choosing an (x, y, z) solution and making up three equations. Try online generators for unlimited practice!

Practice problems

6 of the 27, worked through step by step โ€” try them before opening the answer.

1 x + 3y - z = 6, 2x - y + 4z = 1, 3x + 2y - z = 11

Hint: Try using elimination by first eliminating one variable from pairs of equations. For example, you might multiply the first equation by a number to help eliminate x when combined with the second equation.

Show the answer

Answer: x = 3, y = 1, z = 0

  1. Let's use elimination. Start with the equations: (1) x + 3y - z = 6 (2) 2x - y + 4z = 1 (3) 3x + 2y - z = 11
  2. Multiply equation (1) by 2: 2x + 6y - 2z = 12 Subtract equation (2): (2x + 6y - 2z) - (2x - y + 4z) = 12 - 1 This gives: 7y - 6z = 11 (Equation A)
  3. Multiply equation (1) by 3: 3x + 9y - 3z = 18 Subtract equation (3): (3x + 9y - 3z) - (3x + 2y - z) = 18 - 11 This gives: 7y - 2z = 7 (Equation B)
  4. Subtract Equation B from Equation A: (7y - 6z) - (7y - 2z) = 11 - 7 This gives: -4z = 4, so z = -1
  5. Substitute z = -1 into Equation B: 7y - 2(-1) = 7 7y + 2 = 7 7y = 5 y = 5/7
  6. Substitute y = 5/7 and z = -1 into equation (1): x + 3(5/7) - (-1) = 6 x + 15/7 + 1 = 6 x + 15/7 + 7/7 = 6 x + 22/7 = 42/7 x = 42/7 - 22/7 x = 20/7 The solution is x = 20/7, y = 5/7, z = -1.

2 3x + y - 5z = 7, x - 3y + z = -5, 5x - y + 3z = 11

Hint: Try using elimination by combining equations to eliminate one variable at a time. Start by looking for coefficients that can cancel each other out.

Show the answer

Answer: x = 1, y = 3, z = -1

  1. Write the system: (1) 3x + y - 5z = 7 (2) x - 3y + z = -5 (3) 5x - y + 3z = 11
  2. Multiply equation (2) by 3: 3x - 9y + 3z = -15 Subtract from equation (1): (3x + y - 5z) - (3x - 9y + 3z) = 7 - (-15) 10y - 8z = 22 Divide by 2: 5y - 4z = 11 (equation 4)
  3. Multiply equation (2) by 5: 5x - 15y + 5z = -25 Subtract from equation (3): (5x - y + 3z) - (5x - 15y + 5z) = 11 - (-25) 14y - 2z = 36 Divide by 2: 7y - z = 18 (equation 5)
  4. Multiply equation (5) by 4: 28y - 4z = 72 Subtract equation (4): (28y - 4z) - (5y - 4z) = 72 - 11 23y = 61 y = 61/23 = 3
  5. Substitute y = 3 into equation (5): 7(3) - z = 18 21 - z = 18 z = 21 - 18 = -1
  6. Substitute y = 3, z = -1 into equation (2): x - 3(3) + (-1) = -5 x - 9 - 1 = -5 x - 10 = -5 x = 5 Final answer: x = 1, y = 3, z = -1

3 Solve: 3x + y - 2z = 6, x - 4y + z = -1, 2x + y + z = 11

Hint: Try using elimination by first eliminating one variable from pairs of equations. For example, you might eliminate z from the first and second equations, then from the first and third equations.

Show the answer

Answer: x = 4, y = 1, z = 2

  1. Write the system: 3x + y - 2z = 6 (1) x - 4y + z = -1 (2) 2x + y + z = 11 (3)
  2. Eliminate z from equations (1) and (2): Multiply (2) by 2: 2x - 8y + 2z = -2 Add to (1): (3x + y - 2z) + (2x - 8y + 2z) = 6 + (-2) 5x - 7y = 4 (4)
  3. Eliminate z from equations (1) and (3): Multiply (3) by 2: 4x + 2y + 2z = 22 Add to (1): (3x + y - 2z) + (4x + 2y + 2z) = 6 + 22 7x + 3y = 28 (5)
  4. Solve the system of (4) and (5): Multiply (4) by 3: 15x - 21y = 12 Multiply (5) by 7: 49x + 21y = 196 Add: (15x - 21y) + (49x + 21y) = 12 + 196 64x = 208 x = 208/64 = 13/4
  5. Substitute x = 13/4 into (5): 7(13/4) + 3y = 28 91/4 + 3y = 28 3y = 28 - 91/4 = 112/4 - 91/4 = 21/4 y = (21/4)/3 = 21/12 = 7/4
  6. Substitute x = 13/4 and y = 7/4 into (3): 2(13/4) + (7/4) + z = 11 26/4 + 7/4 + z = 11 33/4 + z = 11 z = 11 - 33/4 = 44/4 - 33/4 = 11/4
  7. Check the solution: (1): 3(13/4) + (7/4) - 2(11/4) = 39/4 + 7/4 - 22/4 = 24/4 = 6 โœ“ (2): (13/4) - 4(7/4) + (11/4) = 13/4 - 28/4 + 11/4 = -4/4 = -1 โœ“ (3): 2(13/4) + (7/4) + (11/4) = 26/4 + 7/4 + 11/4 = 44/4 = 11 โœ“ The solution is x = 13/4, y = 7/4, z = 11/4.

4 Solve: 3x + y - 2z = 6, x - 4y + z = 1, 2x + y - 3z = -1

Hint: Try using elimination by first eliminating y between the first and third equations, then between the first and second equations.

Show the answer

Answer: x = 2, y = 1, z = 1/2

When solving systems of three equations, elimination involves strategically combining equations to eliminate one variable at a time, reducing the system to two equations with two variables, then solving that smaller system.

5 Solve: 3x + 5y - z = 7, x - 3y + 5z = -1, 5x + y - 3z = 11

Hint: Try using elimination by first eliminating one variable from pairs of equations. For example, multiply equations to get matching coefficients for one variable, then subtract.

Show the answer

Answer: x = 2, y = 1, z = 2

  1. Write the system: (1) 3x + 5y - z = 7 (2) x - 3y + 5z = -1 (3) 5x + y - 3z = 11
  2. Eliminate z from equations (1) and (2): Multiply (1) by 5: 15x + 25y - 5z = 35 Add to (2): (15x + 25y - 5z) + (x - 3y + 5z) = 35 + (-1) 16x + 22y = 34 Divide by 2: 8x + 11y = 17 (Equation A)
  3. Eliminate z from equations (1) and (3): Multiply (1) by 3: 9x + 15y - 3z = 21 Subtract (3): (9x + 15y - 3z) - (5x + y - 3z) = 21 - 11 4x + 14y = 10 Divide by 2: 2x + 7y = 5 (Equation B)
  4. Solve the system of equations A and B: A: 8x + 11y = 17 B: 2x + 7y = 5 Multiply B by 4: 8x + 28y = 20 Subtract A: (8x + 28y) - (8x + 11y) = 20 - 17 17y = 3 y = 3/17
  5. Substitute y = 3/17 into B: 2x + 7(3/17) = 5 2x + 21/17 = 5 2x = 5 - 21/17 = 85/17 - 21/17 = 64/17 x = 32/17
  6. Substitute x = 32/17 and y = 3/17 into equation (1): 3(32/17) + 5(3/17) - z = 7 96/17 + 15/17 - z = 7 111/17 - z = 119/17 -z = 119/17 - 111/17 = 8/17 z = -8/17 The solution is x = 32/17, y = 3/17, z = -8/17.

6 Solve: 2x + 3y - z = 7, x - 2y + 4z = -2, 3x + y - 2z = 12

Hint: Try using elimination by first eliminating one variable from pairs of equations, then solving the resulting two-variable system.

Show the answer

Answer: x = 3, y = 2, z = 1

  1. Write the system: 2x + 3y - z = 7 (1) x - 2y + 4z = -2 (2) 3x + y - 2z = 12 (3)
  2. Eliminate z from equations (1) and (2): Multiply (1) by 4: 8x + 12y - 4z = 28 Add to (2): (8x + 12y - 4z) + (x - 2y + 4z) = 28 + (-2) 9x + 10y = 26 (4)
  3. Eliminate z from equations (1) and (3): Multiply (1) by 2: 4x + 6y - 2z = 14 Subtract (3): (4x + 6y - 2z) - (3x + y - 2z) = 14 - 12 x + 5y = 2 (5)
  4. Solve the system of (4) and (5): From (5): x = 2 - 5y Substitute into (4): 9(2 - 5y) + 10y = 26 18 - 45y + 10y = 26 -35y = 8 y = -8/35 = -2/7
  5. Find x: x = 2 - 5(-2/7) = 2 + 10/7 = 14/7 + 10/7 = 24/7
  6. Find z using equation (1): 2(24/7) + 3(-2/7) - z = 7 48/7 - 6/7 - z = 7 42/7 - z = 7 6 - z = 7 z = -1
  7. Verify with equation (3): 3(24/7) + (-2/7) - 2(-1) = 72/7 - 2/7 + 2 = 70/7 + 2 = 10 + 2 = 12 โœ“ The solution is x = 24/7, y = -2/7, z = -1.
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