What if all variables cancel out and leave $0 = 0$?

The system has infinitely many solutions. This means the planes intersect in a line or are the exact same plane (a dependent system).

What if all variables cancel out and leave $0 = k$ (where $k eq 0$)?

The system has no solution. This means at least two of the planes are parallel and never intersect (an inconsistent system).

Can I use substitution instead of elimination?

Yes. However, elimination is generally less prone to algebraic errors and fraction complications for 3x3 systems, unless one equation already isolates a variable (e.g., $x = 2y - z$).

How to solve a 3x3 system of linear equations

Solving a 3x3 system of linear equations involves using the elimination method to systematically eliminate one variable, reducing the problem to a 2x2 system. This method applies whenever you have three linear equations with three distinct variables.

The setup

Write the three equations in standard form $Ax + By + Cz = D$ and align the variables vertically. Label the equations (1), (2), and (3) to track your work.

The steps

Choose two equations and use addition or subtraction to eliminate one variable. Label the resulting equation (4).
Choose a different pair of equations from the original three and eliminate the same variable. Label the resulting equation (5).
Solve the new 2x2 system composed of equations (4) and (5) for the two remaining variables using standard 2x2 elimination or substitution.
Substitute these two known values into any of the original 3x3 equations to solve for the third variable.

Checking the result

Substitute your final $(x, y, z)$ coordinate into all three original equations. The left side must equal the right side for each equation. If any equation fails, the solution is incorrect.

Common errors

The most frequent error is eliminating one variable in the first pair of equations, but a different variable in the second pair. This yields two equations with three variables, which cannot be solved. Arithmetic errors, especially dropping negative signs when multiplying equations by a scalar, are also common.

Worked example

Solve the system of equations: (1) $x + y + z = 6$ (2) $2x - y + z = 3$ (3) $x + 2y - z = 2$

Eliminate $z$ using equations (1) and (3): (1) $x + y + z = 6$ (3) $x + 2y - z = 2$ Add (1) and (3): (4) $2x + 3y = 8$

Eliminate $z$ using equations (2) and (3): (2) $2x - y + z = 3$ (3) $x + 2y - z = 2$ Add (2) and (3): (5) $3x + y = 5$

Solve the 2x2 system of (4) and (5): (4) $2x + 3y = 8$ (5) $3x + y = 5$

Multiply (5) by $-3$ : $-9x - 3y = -15$

Add this to (4): $(2x + 3y) + (-9x - 3y) = 8 - 15$ $-7x = -7$ $x = 1$

Substitute $x = 1$ into (5): $3(1) + y = 5$ $3 + y = 5$ $y = 2$

Substitute $x = 1$ and $y = 2$ into (1): $1 + 2 + z = 6$ $3 + z = 6$ $z = 3$

The solution is $(1, 2, 3)$ .

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References: OpenStax College Algebra, Chapter 7: Systems of Equations and Inequalities · Khan Academy, Unit: Systems of equations