What Is a 3-Variable System?
A system of linear equations with 3 variables is a set of three equations that each contain some combination of the unknowns x, y, and z.
A typical system looks like this: 2x + y − z = 3, x − 3y + 2z = 1, 4x + y + z = 9.
(x, y, z) that satisfies all three equations simultaneously. Geometrically, each equation represents a plane in three-dimensional space. Solving the system means finding the point (or points) where all three planes intersect.Geometric Cases
| Case | Geometry | Result |
|---|---|---|
| 1 | Three planes meet at a single point | The system has exactly one solution. |
| 2 | They intersect along a line | There are infinitely many solutions. |
| 3 | No single point lies on all three planes | The system has no solution. |
While two-variable systems are solved with two equations and two unknowns, adding a third variable requires a third equation to pin down a unique answer.
The core strategy for solving these systems is systematic elimination — reducing the three-variable problem to a two-variable problem, and then to a single equation in one variable.
Elimination Method for 3 Variables
The elimination method for a 3-variable system works by strategically combining pairs of equations to eliminate one variable at a time. The process has three main phases.
- Phase 1: Eliminate one variable.
Choose one variable to eliminate first — pick whichever has the simplest coefficients. Combine equation 1 with equation 2 to eliminate that variable, producing a new equation with only two unknowns. Then combine equation 1 with equation 3 (or equation 2 with equation 3) to eliminate the same variable, producing a second two-variable equation.
- Phase 2: Solve the 2-variable system.
You now have two equations in two unknowns. Use elimination or substitution to solve for one variable, then back-substitute to find the other.
- Phase 3: Back-substitute into an original equation.
Plug both known values into any of the original three equations to solve for the third variable.
The key rule is that you must eliminate the same variable in Phase 1 both times. If you eliminate z from equations 1 and 2, you must also eliminate z from another pair. Mixing different variables in this phase is the most common mistake students make.
Worked Example Step by Step
Let's solve the system: (1) x + y + z = 6, (2) 2x − y + z = 3, (3) x + 2y − z = 5.
Step 1 — Eliminate z from equations (1) and (3): Add them directly: (x + y + z) + (x + 2y − z) = 6 + 5, which gives 2x + 3y = 11. Call this equation (4).
Step 2 — Eliminate z from equations (2) and (3): Add them: (2x − y + z) + (x + 2y − z) = 3 + 5, which gives 3x + y = 8. Call this equation (5).
Step 3 — Solve the 2×2 system (4) and (5): Multiply equation (5) by 3: 9x + 3y = 24. Subtract equation (4): 9x + 3y − (2x + 3y) = 24 − 11, so 7x = 13 and x = 13/7. Substitute into (5): 3(13/7) + y = 8, so y = 8 − 39/7 = 17/7.
Step 4 — Find z: Substitute x and y into equation (1): 13/7 + 17/7 + z = 6, so z = 6 − 30/7 = 12/7.
(13/7, 17/7, 12/7)Always verify by plugging into all three original equations to confirm.
When There Is No Solution or Infinite Solutions
Not every 3-variable system has a unique solution. There are two special cases to watch for.
No Solution (Inconsistent System)
0 = 5. For example, if after eliminating variables you get 0x + 0y = 7, this is impossible and the system has no solution. Geometrically, at least two of the planes are parallel, or all three form a triangular "prism" with no common intersection.Infinitely Many Solutions (Dependent System)
0 = 0, meaning you have lost an equation. The solution is typically expressed in terms of a free variable (parameter). For example, you might write x = 2t − 1, y = t + 3, z = t, where t can be any real number. Each value of t gives a different valid solution — the three planes intersect along a line.Recognizing these special cases early saves time: if two of your original equations are multiples of each other, you already know the system is dependent.
Matrix Method (Gaussian Elimination)
For larger or more complex systems, Gaussian elimination using an augmented matrix is more efficient and systematic.
x + y + z = 6, 2x − y + z = 3, x + 2y − z = 5, the matrix is [1 1 1 | 6; 2 −1 1 | 3; 1 2 −1 | 5].Now use row operations to convert the matrix to row echelon form (upper triangular). The three allowed operations are: swap two rows, multiply a row by a nonzero constant, and add a multiple of one row to another.
- Step 1
Subtract 2×Row1 from Row2 to get
[0 −3 −1 | −9]. - Step 2
Subtract Row1 from Row3 to get
[0 1 −2 | −1]. - Step 3
Use the new Row3 to eliminate the y-entry in Row2. The matrix is now in triangular form and you can back-substitute from the bottom row up to find z, then y, then x.
Gaussian elimination is the backbone of linear algebra and scales to systems of any size. Solver AI uses this exact method when you scan a system of equations — it displays each row operation so you can follow the logic step by step.
Practice Problems
Solve: x + y + z = 10, 2x − y + 3z = 16, −x + 2y − z = −2.
Hint: Eliminate x from equations 1 and 3 by adding them.
(3, 2, 5)Solve: 3x − y + 2z = 12, x + 2y − z = 1, 2x − 3y + 3z = 11.
Hint: Eliminate z from equations 1 and 2 by multiplying equation 2 by 2 and adding.
(3, −1, 1)Determine if the system has a solution: x + y + z = 4, 2x + 2y + 2z = 8, x − y + z = 2.
Equation 2 is just 2 times equation 1, so the system is dependent with infinitely many solutions.Solve using an augmented matrix: x − 2y + z = 0, 3x + y − 2z = 7, 2x + 3y − 3z = 7.
(1, 1, 1)For instant step-by-step solutions, scan any of these problems with Solver AI to see every elimination step worked out in detail.