9.2Solving Three-Variable Linear Systems
Solve 3×3 systems using Gaussian elimination and back-substitution. Write augmented matrices and use row operations to reach row-echelon form.
Three-variable systems model 3D problems in physics, economics, and engineering. Gaussian elimination is the algorithm behind every linear algebra solver and is the precursor to matrix methods.
Essential Question
How does back-substitution after elimination allow you to solve a 3×3 linear system, and what does it mean geometrically when a 3-variable system has no solution or infinitely many solutions?
Lesson Overview
Three planes intersect at one point → unique solution.
A 3×3 linear system consists of three equations in three unknowns (x, y, z).
Geometric interpretation: each equation represents a plane in 3D space.
- Unique solution: three planes intersect at exactly one point.
- No solution: planes are parallel or form a "triangular prism" arrangement with no common point.
- Infinitely many solutions: planes intersect along a line, or all three are the same plane.
Strategy — Elimination with back-substitution:
- Choose a variable to eliminate; use pairs of equations to eliminate it.
- Reduce to a 2×2 system in two variables.
- Solve the 2×2 system.
- Back-substitute to find the third variable.
Row operations (same as matrix row operations):
- Swap two equations.
- Multiply an equation by a nonzero constant.
- Add a multiple of one equation to another.
Applications: 3-commodity mixture problems, curve fitting through 3 points, network flow.
Worked Examples
Solve: x + y + z = 6, 2x − y + z = 3, x + 2y − z = 2.
Label: (1) x+y+z=6 (2) 2x−y+z=3 (3) x+2y−z=2
Eliminate y from (1)+(3): (x+y+z)+(x+2y−z) = 6+2 → 2x+3y = 8 (A)
Eliminate y from (2)+(3): (2x−y+z)+(x+2y−z) = 3+2 → 3x+y = 5 (C)
From (C): y = 5−3x. Substitute into (A): 2x+3(5−3x)=8 → 2x+15−9x=8 → −7x=−7 → x=1
y = 5−3(1) = 2
From (1): 1+2+z=6 → z=3
Solve: 2x + y − z = 8, −3x − y + 2z = −11, −2x + y + 2z = −3.
Label: (1) 2x+y−z=8 (2) −3x−y+2z=−11 (3) −2x+y+2z=−3
(1)+(2): −x+z = −3 (A)
(1)+(3): 2y+z = 5 (B)
(2)+(3): −5x+4z = −14 (C)
From (A): z = x−3. Substitute into (C): −5x+4(x−3)=−14 → −x−12=−14 → x=2
z = 2−3 = −1
From (B): 2y+(−1)=5 → 2y=6 → y=3
Identify the system type: x + 2y − z = 4, 2x + 4y − 2z = 8, −x − 2y + z = −4.
Equation (2) = 2 × Equation (1): 2(x+2y−z)=2(4)=8 ✓
Equation (3) = −1 × Equation (1): −(x+2y−z)=−4 ✓
All three equations represent the same plane.
Identify the system type: x + y + z = 3, 2x + 2y + 2z = 7, x − y + z = 1.
If (2) were consistent with (1), it would equal 2×3 = 6, but (2) gives 7.
2×(Eq 1) gives 2x+2y+2z=6, yet Eq 2 says 2x+2y+2z=7. Contradiction.
Curve fitting: Find a, b, c so that y = ax² + bx + c passes through (1, 2), (2, 5), (3, 10).
Substitute each point:
(1,2): a + b + c = 2 (1)
(2,5): 4a + 2b + c = 5 (2)
(3,10): 9a + 3b + c = 10 (3)
(2)−(1): 3a + b = 3 (A)
(3)−(2): 5a + b = 5 (B)
(B)−(A): 2a = 2 → a = 1
From (A): b = 3−3(1) = 0
From (1): c = 2−1−0 = 1
Guided Practice
Solve: x + y + z = 0, 2x − y + 3z = −1, x + 2y − z = 4.
Hint: Eliminate z by adding equations in pairs.
Solve: 2x + y + z = 7, x − y + 2z = 4, 3x + 2y − z = 10.
Hint: Start by eliminating y from two pairs of equations.
Determine the type: x + y − z = 2, 2x + 2y − 2z = 4, 3x + 3y − 3z = 6.
Hint: Compare the equations — are they multiples of each other?
Determine the type: x + y + z = 5, 2x + 2y + 2z = 9, x − y + z = 1.
Hint: Check if equation 2 is consistent with equation 1.
A parabola y = ax² + bx + c passes through (−1, 6), (0, 1), and (2, 3). Find a, b, c.
Hint: Substitute each point to get three equations in a, b, c.
Key Vocabulary
3×3 linear system
Three linear equations in three unknowns x, y, z.
Example: x+y+z=6, 2x−y+z=3, x+2y−z=2
Back-substitution
After solving for one variable, substituting its value into other equations to find the remaining variables.
Row operations
Swapping equations, multiplying an equation by a nonzero constant, or adding a multiple of one equation to another — all preserve the solution set.
Triangular form
A system where the first equation has 3 variables, the second has 2, and the third has 1 — making back-substitution straightforward.
Dependent system (3D)
The three planes intersect along a line or are all the same plane — infinitely many solutions.
Inconsistent system (3D)
The planes have no common intersection point — no solution exists.
Quick Check Quiz
Interactive Practice — 5 Questions
The solution to a 3×3 system represents geometrically:
In a 3×3 system, if elimination gives 0 = 0, the system is:
Which is a valid row operation?
Solve: x + y = 3, y + z = 5, x + z = 4.
Back-substitution is used after:
Common Mistakes
Eliminating different variables in each step without a plan.
Choose ONE variable to eliminate first. Use it to reduce all three equations to a 2×2 system, then solve that.
Forgetting to use all three equations — only using two pairs.
To eliminate one variable from a 3×3 system, you need to use the variable in at least two different pairs of equations.
Sign errors when subtracting equations.
When subtracting, distribute the negative sign to EVERY term. Consider adding the negative of an equation instead.
Not checking the solution in all three original equations.
A 3-variable solution must satisfy all three equations. Always verify by substituting back.
Math Tips
Organize your work: label each equation (1), (2), (3) and each derived equation (A), (B). This prevents confusion.
The goal is to reduce the 3×3 system to a 2×2 system, then to a 1×1 equation. Work systematically.
For curve fitting through 3 points, substitute each point into y = ax² + bx + c to get three equations in a, b, c.
If two equations are multiples of each other, the system is either dependent (same equation) or inconsistent (contradictory constants).
Always verify your solution in ALL THREE original equations — not just the last one you used.