Linear Systems 3x3

Grade 10 · algebra · 24 practice problems · read aloud

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Solving 3x3 Linear Systems

What is it? 🤔

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 → x = 1.
Multiply (1) by 2 and add to (3): x + 4z = 6.
Step 2: With x=1, 1 + 4z = 6 → z = 5/4.
Step 3: Substitute into (1): 1 + y - 5/4 = 0 → 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 → x = -4.
Step 3: From (1): 2(-4) + y = 1 → y = 9. From (2): 9 - z = 2 → z = 7.
Solution: (-4, 9, 7)

Common Mistakes ⚠️

  • 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 24, worked through step by step — try them before opening the answer.

1 x + y + z = 8, 2x - y + z = 7, x - 2y + 2z = 4

Hint: Try using elimination by adding or subtracting equations to eliminate one variable at a time. Start by looking for equations where adding them cancels out one variable.

Show the answer

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

  1. Label the equations: (1) x + y + z = 8 (2) 2x - y + z = 7 (3) x - 2y + 2z = 4
  2. Add equations (1) and (2) to eliminate y: (1) + (2): (x + y + z) + (2x - y + z) = 8 + 7 3x + 2z = 15 (call this equation 4)
  3. Multiply equation (1) by 2 and add to equation (3) to eliminate y: 2*(1): 2x + 2y + 2z = 16 Add to (3): (2x + 2y + 2z) + (x - 2y + 2z) = 16 + 4 3x + 4z = 20 (call this equation 5)
  4. Subtract equation (4) from equation (5) to eliminate x: (5) - (4): (3x + 4z) - (3x + 2z) = 20 - 15 2z = 5 z = 2.5
  5. Substitute z = 2.5 into equation (4): 3x + 2(2.5) = 15 3x + 5 = 15 3x = 10 x = 10/3
  6. Substitute x = 10/3 and z = 2.5 into equation (1): 10/3 + y + 2.5 = 8 y + 10/3 + 5/2 = 8 y + 20/6 + 15/6 = 8 y + 35/6 = 8 y = 8 - 35/6 y = 48/6 - 35/6 y = 13/6 The solution is x = 10/3, y = 13/6, z = 5/2.

2 2x + 3y - z = 7, x - y + 2z = 1, 3x + y + z = 9

Hint: Try using elimination by adding or subtracting equations to eliminate one variable at a time. Start by looking for equations where adding them cancels out one variable.

Show the answer

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

  1. Label the equations: (1) 2x + 3y - z = 7 (2) x - y + 2z = 1 (3) 3x + y + z = 9
  2. Add equations (1) and (3) to eliminate z: (1) + (3): (2x + 3y - z) + (3x + y + z) = 7 + 9 5x + 4y = 16 (equation 4)
  3. Multiply equation (2) by 1 and add to equation (1) to eliminate z: (1) + (2): (2x + 3y - z) + (x - y + 2z) = 7 + 1 3x + 2y + z = 8 (equation 5)
  4. Now use equations (3) and (5) to eliminate z: (3) - (5): (3x + y + z) - (3x + 2y + z) = 9 - 8 -y = 1 y = -1
  5. Substitute y = -1 into equation (4): 5x + 4(-1) = 16 5x - 4 = 16 5x = 20 x = 4
  6. Substitute x = 4 and y = -1 into equation (3): 3(4) + (-1) + z = 9 12 - 1 + z = 9 11 + z = 9 z = -2
  7. Verify with equation (1): 2(4) + 3(-1) - (-2) = 8 - 3 + 2 = 7 ✓ Final answer: x = 4, y = -1, z = -2

3 2x + y - z = 5, x - 2y + z = -5, 3x + y + 2z = 10

Hint: Try using elimination by adding or subtracting equations to eliminate one variable at a time. Start by looking for equations that can be combined to cancel out one variable.

Show the answer

Answer: x = 2, y = 1, z = 0

  1. Label the equations: (1) 2x + y - z = 5 (2) x - 2y + z = -5 (3) 3x + y + 2z = 10
  2. Add equations (1) and (2) to eliminate z: (1) + (2): (2x + x) + (y - 2y) + (-z + z) = 5 + (-5) 3x - y = 0 So y = 3x
  3. Substitute y = 3x into equation (1): 2x + (3x) - z = 5 5x - z = 5 So z = 5x - 5
  4. Substitute y = 3x and z = 5x - 5 into equation (3): 3x + (3x) + 2(5x - 5) = 10 3x + 3x + 10x - 10 = 10 16x - 10 = 10 16x = 20 x = 20/16 = 5/4
  5. Find y and z: y = 3x = 3(5/4) = 15/4 z = 5x - 5 = 5(5/4) - 5 = 25/4 - 20/4 = 5/4
  6. Verify with equation (3): 3(5/4) + 15/4 + 2(5/4) = 15/4 + 15/4 + 10/4 = 40/4 = 10 ✓ The solution is x = 5/4, y = 15/4, z = 5/4.

4 2x + y - z = 5, x - 3y + 2z = -5, 3x + 2y + z = 10

Hint: Try eliminating one variable first by adding or subtracting equations. For example, if you had equations a + b - c = 4 and 2a - b + c = 1, adding them would eliminate both b and c.

Show the answer

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

  1. Label the equations: (1) 2x + y - z = 5 (2) x - 3y + 2z = -5 (3) 3x + 2y + z = 10
  2. Add equations (1) and (3) to eliminate z: (1) + (3): (2x + y - z) + (3x + 2y + z) = 5 + 10 5x + 3y = 15 (equation 4)
  3. Multiply equation (1) by 2 and add to equation (2): 2*(1): 4x + 2y - 2z = 10 Add to (2): (4x + 2y - 2z) + (x - 3y + 2z) = 10 + (-5) 5x - y = 5 (equation 5)
  4. Solve the system of equations (4) and (5): (4) 5x + 3y = 15 (5) 5x - y = 5 Subtract (5) from (4): (5x + 3y) - (5x - y) = 15 - 5 4y = 10 y = 2.5
  5. Substitute y = 2.5 into equation (5): 5x - 2.5 = 5 5x = 7.5 x = 1.5
  6. Substitute x = 1.5 and y = 2.5 into equation (1): 2(1.5) + 2.5 - z = 5 3 + 2.5 - z = 5 5.5 - z = 5 z = 0.5
  7. Verify with equation (3): 3(1.5) + 2(2.5) + 0.5 = 4.5 + 5 + 0.5 = 10 ✓ The solution is x = 1.5, y = 2.5, z = 0.5

5 2x + y - z = 5, x - 3y + 2z = -10, 3x + 2y - z = 10

Hint: Try using elimination by combining equations to eliminate one variable at a time. Start by eliminating z between the first and third equations.

Show the answer

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

  1. Write the system: (1) 2x + y - z = 5 (2) x - 3y + 2z = -10 (3) 3x + 2y - z = 10
  2. Eliminate z using equations (1) and (3): (1) 2x + y - z = 5 (3) 3x + 2y - z = 10 Subtract (1) from (3): (3x - 2x) + (2y - y) + (-z + z) = 10 - 5 x + y = 5 (Equation A)
  3. Eliminate z using equations (1) and (2): Multiply (1) by 2: 4x + 2y - 2z = 10 Add to (2): (4x + x) + (2y - 3y) + (-2z + 2z) = 10 + (-10) 5x - y = 0 (Equation B)
  4. Solve the system of equations A and B: (A) x + y = 5 (B) 5x - y = 0 Add (A) and (B): 6x = 5 x = 5/6
  5. Substitute x = 5/6 into (A): 5/6 + y = 5 y = 5 - 5/6 = 25/6
  6. Substitute x = 5/6 and y = 25/6 into (1): 2(5/6) + 25/6 - z = 5 10/6 + 25/6 - z = 5 35/6 - z = 5 -z = 5 - 35/6 = -5/6 z = 5/6
  7. Verify with original equations: (1) 2(5/6) + 25/6 - 5/6 = 10/6 + 25/6 - 5/6 = 30/6 = 5 ✓ (2) 5/6 - 3(25/6) + 2(5/6) = 5/6 - 75/6 + 10/6 = -60/6 = -10 ✓ (3) 3(5/6) + 2(25/6) - 5/6 = 15/6 + 50/6 - 5/6 = 60/6 = 10 ✓ The solution is x = 5/6, y = 25/6, z = 5/6.

6 2x + y - z = 5, x - 3y + 2z = -10, 3x + 2y + z = 15

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

Show the answer

Answer: x = 5, y = 0, z = 5

  1. Write the system: (1) 2x + y - z = 5 (2) x - 3y + 2z = -10 (3) 3x + 2y + z = 15
  2. Multiply equation (1) by 2: 4x + 2y - 2z = 10 Add this to equation (2): (4x + 2y - 2z) + (x - 3y + 2z) = 10 + (-10) 5x - y = 0, so y = 5x
  3. Substitute y = 5x into equation (1): 2x + (5x) - z = 5 7x - z = 5 z = 7x - 5
  4. Substitute y = 5x and z = 7x - 5 into equation (3): 3x + 2(5x) + (7x - 5) = 15 3x + 10x + 7x - 5 = 15 20x - 5 = 15 20x = 20 x = 1
  5. Find y and z: y = 5x = 5(1) = 5 z = 7x - 5 = 7(1) - 5 = 2
  6. Check in all equations: (1) 2(1) + 5 - 2 = 2 + 5 - 2 = 5 ✓ (2) 1 - 3(5) + 2(2) = 1 - 15 + 4 = -10 ✓ (3) 3(1) + 2(5) + 2 = 3 + 10 + 2 = 15 ✓ The solution is x = 1, y = 5, z = 2.
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