Linear Systems

Grade 12 ยท algebra ยท 87 practice problems ยท read aloud

๐Ÿ”Š Listen to this explanation

Linear Systems in Algebra

๐Ÿ” What Are Linear Systems & Why Are They Useful?

A linear system is a set of two or more linear equations involving the same variables. Solving the system means finding the values for the variables that make all equations true simultaneously. This is crucial for modeling real-world scenarios with multiple constraints, like optimizing business profits, calculating chemical mixtures, or analyzing electrical circuits.

๐Ÿ“ How to Solve: A Step-by-Step Guide

For Grade 12, we focus on systems with two or three variables, primarily using Elimination and Substitution.

  1. Choose a Method: Elimination is often best when coefficients are easily aligned. Substitution is ideal when one variable is already isolated.
  2. Solve for One Variable: Manipulate the equations to find the value of one variable.
  3. Back-Substitute: Substitute the value you found back into one of the original equations to solve for the remaining variable(s).
  4. Check Your Solution: Verify your solution satisfies all original equations.

๐Ÿงฎ Visual Examples

Example 1: Elimination Method

Solve: 2x + 3y = 7 and x - y = 1

  1. Multiply the second equation by 3: 3(x - y = 1) โ†’ 3x - 3y = 3
  2. Add to the first equation: (2x + 3y) + (3x - 3y) = 7 + 3 โ†’ 5x = 10
  3. Solve for x: x = 2
  4. Substitute x=2 into x - y = 1: 2 - y = 1 โ†’ y = 1
  5. Solution: (2, 1)

Example 2: Substitution Method (3 Variables)

Solve: x + y + z = 6, 2y + 5z = -4, and 2x + 5y - z = 27

  1. From the first equation, solve for x: x = 6 - y - z
  2. Substitute this into the third equation: 2(6 - y - z) + 5y - z = 27 โ†’ 12 + 3y - 3z = 27
  3. Simplify: 3y - 3z = 15 โ†’ y - z = 5
  4. Now you have a system with the second equation: 2y + 5z = -4 and y - z = 5. Solve this smaller system (e.g., y = 5 + z, substitute...) to find y=3, z=-2.
  5. Substitute back to find x=5. Solution: (5, 3, -2)

โš ๏ธ Common Mistakes to Avoid

  • Sign Errors: The most common mistake! Be meticulous with negative signs when adding or subtracting equations in the elimination method.
  • Inconsistent Systems: Not all systems have a single solution. If all variables cancel and you get a false statement (e.g., 0=5), the system has no solution (the lines are parallel).
  • Forgetting to Check: Always plug your final answers back into the original equations to catch arithmetic errors.

๐Ÿ’ก Tips & Tricks

  • Method Choice: Use elimination for "messy" coefficients and substitution for "simple" coefficients (like 1 or -1).
  • Matrix Shortcut: For larger systems (3+ variables), learning to use matrices and Gaussian elimination is a powerful Grade 12 skill.
  • Visualize: Sketching the lines (for 2 variables) or planes (for 3 variables) can help you understand the number of solutions (one, infinite, or none).

๐ŸŽฏ Practice Suggestions

Mastery comes from consistent practice.

  • Start with 2-variable systems using both methods until you're fluent.
  • Progress to 3-variable systems, which require more organized work.
  • Create your own problems with known solutions and solve them.
  • Look for word problems that translate into linear systems to see the real-world application.

Practice problems

6 of the 87, worked through step by step โ€” try them before opening the answer.

1 2x + 3y - z = 9, x - 2y + 2z = 1, 3x + y - 3z = 4

Hint: Try using elimination by combining equations to eliminate one variable at a time. Start by eliminating the same variable from two different pairs of equations.

Show the answer

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

  1. Write the system: 2x + 3y - z = 9 (1) x - 2y + 2z = 1 (2) 3x + y - 3z = 4 (3)
  2. Multiply equation (2) by 2: 2x - 4y + 4z = 2 (4)
  3. Subtract equation (4) from equation (1): (2x + 3y - z) - (2x - 4y + 4z) = 9 - 2 7y - 5z = 7 (5)
  4. Multiply equation (2) by 3: 3x - 6y + 6z = 3 (6)
  5. Subtract equation (6) from equation (3): (3x + y - 3z) - (3x - 6y + 6z) = 4 - 3 7y - 9z = 1 (7)
  6. Subtract equation (7) from equation (5): (7y - 5z) - (7y - 9z) = 7 - 1 4z = -8 z = -2
  7. Substitute z = -2 into equation (5): 7y - 5(-2) = 7 7y + 10 = 7 7y = -3 y = 1
  8. Substitute y = 1 and z = -2 into equation (2): x - 2(1) + 2(-2) = 1 x - 2 - 4 = 1 x - 6 = 1 x = 2
  9. Verify with equation (1): 2(2) + 3(1) - (-2) = 4 + 3 + 2 = 9 โœ“ The solution is x = 2, y = 1, z = -2

2 2x + 3y - z = 9, x - 2y + 2z = 3, 3x + y - 3z = 0

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

Show the answer

Answer: x = 2, y = 3, z = 4

  1. Write the system: 2x + 3y - z = 9 (1) x - 2y + 2z = 3 (2) 3x + y - 3z = 0 (3)
  2. Eliminate z from equations (1) and (2): Multiply (1) by 2: 4x + 6y - 2z = 18 Add to (2): (4x + 6y - 2z) + (x - 2y + 2z) = 18 + 3 5x + 4y = 21 (4)
  3. Eliminate z from equations (1) and (3): Multiply (1) by 3: 6x + 9y - 3z = 27 Subtract (3): (6x + 9y - 3z) - (3x + y - 3z) = 27 - 0 3x + 8y = 27 (5)
  4. Solve system (4) and (5): Multiply (4) by 2: 10x + 8y = 42 Subtract (5): (10x + 8y) - (3x + 8y) = 42 - 27 7x = 15 x = 15/7
  5. Substitute x = 15/7 into (4): 5(15/7) + 4y = 21 75/7 + 4y = 21 4y = 21 - 75/7 = 147/7 - 75/7 = 72/7 y = 18/7
  6. Substitute x = 15/7 and y = 18/7 into (1): 2(15/7) + 3(18/7) - z = 9 30/7 + 54/7 - z = 9 84/7 - z = 9 12 - z = 9 z = 3 The solution is x = 15/7, y = 18/7, z = 3.

3 2x + 3y - z = 11, x - 2y + 3z = 6, 3x + y - 2z = 5

Hint: Use elimination method to reduce the system to two equations with two variables, then solve for the remaining variable

Show the answer

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

  1. Multiply the second equation by 2: 2(x - 2y + 3z) = 2(6) โ†’ 2x - 4y + 6z = 12
  2. Subtract the first equation from this result: (2x - 4y + 6z) - (2x + 3y - z) = 12 - 11 โ†’ -7y + 7z = 1
  3. Multiply the second equation by 3: 3(x - 2y + 3z) = 3(6) โ†’ 3x - 6y + 9z = 18
  4. Subtract the third equation from this result: (3x - 6y + 9z) - (3x + y - 2z) = 18 - 5 โ†’ -7y + 11z = 13
  5. Now we have: -7y + 7z = 1 and -7y + 11z = 13
  6. Subtract the first from the second: (-7y + 11z) - (-7y + 7z) = 13 - 1 โ†’ 4z = 12 โ†’ z = 3
  7. Substitute z = 3 into -7y + 7z = 1: -7y + 7(3) = 1 โ†’ -7y + 21 = 1 โ†’ -7y = -20 โ†’ y = 20/7
  8. Substitute y = 20/7 and z = 3 into the first equation: 2x + 3(20/7) - 3 = 11 โ†’ 2x + 60/7 - 3 = 11 โ†’ 2x + 60/7 - 21/7 = 11 โ†’ 2x + 39/7 = 11 โ†’ 2x = 11 - 39/7 โ†’ 2x = 77/7 - 39/7 โ†’ 2x = 38/7 โ†’ x = 19/7 Final answer: x = 19/7, y = 20/7, z = 3

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

Hint: Use elimination method to reduce the system to two equations with two variables, then solve for the remaining variable

Show the answer

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

  1. Multiply the first equation by 2: 4x + 2y - 2z = 6
  2. Add this to the second equation: (4x + 2y - 2z) + (x - 3y + 2z) = 6 + (-4) โ†’ 5x - y = 2
  3. Multiply the first equation by 4: 8x + 4y - 4z = 12
  4. Add this to the third equation: (8x + 4y - 4z) + (3x + 2y - 4z) = 12 + 1 โ†’ 11x + 6y - 8z = 13
  5. From Step 1, we have z = 2x + y - 3
  6. Substitute into
  7. 11x + 6y - 8(2x + y - 3) = 13 โ†’ 11x + 6y - 16x - 8y + 24 = 13 โ†’ -5x - 2y = -11
  8. Multiply Step 2 by 2: 10x - 2y = 4
  9. Add to
  10. (-5x - 2y) + (10x - 2y) = -11 + 4 โ†’ 5x - 4y = -7
  11. Solve the system: 5x - y = 2 and 5x - 4y = -7
  12. Subtract the second from the first: (5x - y) - (5x - 4y) = 2 - (-7) โ†’ 3y = 9 โ†’ y = -1
  13. Substitute y = -1 into 5x - (-1) = 2 โ†’ 5x + 1 = 2 โ†’ 5x = 1 โ†’ x = 2
  14. Substitute x = 2, y = -1 into z = 2(2) + (-1) - 3 = 4 - 1 - 3 = 0 Final answer: x = 2, y = -1, z = 0

5 2x + y - z = 6, x - 3y + 2z = -4, 3x + 2y - 4z = 5

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

Show the answer

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

  1. Write the system: 2x + y - z = 6 (1) x - 3y + 2z = -4 (2) 3x + 2y - 4z = 5 (3)
  2. Multiply equation (1) by 2: 4x + 2y - 2z = 12 Add this to equation (2): (4x + 2y - 2z) + (x - 3y + 2z) = 12 + (-4) 5x - y = 8 (4)
  3. Multiply equation (1) by 4: 8x + 4y - 4z = 24 Subtract equation (3): (8x + 4y - 4z) - (3x + 2y - 4z) = 24 - 5 5x + 2y = 19 (5)
  4. Subtract equation (4) from equation (5): (5x + 2y) - (5x - y) = 19 - 8 3y = 11
  5. Solve for y: y = 11/3
  6. Substitute y = 11/3 into equation (4): 5x - 11/3 = 8 5x = 8 + 11/3 = 24/3 + 11/3 = 35/3 x = 7/3
  7. Substitute x = 7/3 and y = 11/3 into equation (1): 2(7/3) + 11/3 - z = 6 14/3 + 11/3 - z = 6 25/3 - z = 6 z = 25/3 - 18/3 = 7/3 Final answer: x = 7/3, y = 11/3, z = 7/3

6 3x + 2y - z = 11, 2x - y + 3z = 7, x + 4y - 2z = 5

Hint: Use elimination method by strategically combining equations to eliminate one variable at a time. Start by choosing which variable to eliminate first.

Show the answer

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

  1. Label the equations: (1) 3x + 2y - z = 11 (2) 2x - y + 3z = 7 (3) x + 4y - 2z = 5
  2. Eliminate z from equations (1) and (2): Multiply (1) by 3: 9x + 6y - 3z = 33 Add to (2): 2x - y + 3z = 7 Result: 11x + 5y = 40 (Equation A)
  3. Eliminate z from equations (1) and (3): Multiply (1) by 2: 6x + 4y - 2z = 22 Subtract (3): x + 4y - 2z = 5 Result: 5x = 17 (Equation B)
  4. Solve Equation B: x = 17/5 = 3.4
  5. Substitute x = 3.4 into Equation A: 11(3.4) + 5y = 40 37.4 + 5y = 40 5y = 2.6 y = 0.52
  6. Substitute x = 3.4 and y = 0.52 into equation (1): 3(3.4) + 2(0.52) - z = 11 10.2 + 1.04 - z = 11 11.24 - z = 11 z = 0.24
  7. Verify with equation (2): 2(3.4) - 0.52 + 3(0.24) = 6.8 - 0.52 + 0.72 = 7 โœ“ Final answer: x = 3.4, y = 0.52, z = 0.24
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