Solving 3 Variable Systems of Equations by Elimination

A 3 variable system of equations is a set of three linear equations that all share the same three unknowns, usually written as \(x\), \(y\), and \(z\). Solving the system means finding the one combination of values for \(x\), \(y\), and \(z\) that makes all three equations true at the same time.

This builds directly on the two-variable case. If you have not worked with a system of linear equations before, it helps to get comfortable there first, since a 3 variable system is really just one extra equation and one extra unknown layered on top of the same idea.

The elimination method for two variables cancels a variable by adding or subtracting equations. With three variables, you use the exact same idea, just twice. First you eliminate one variable from two different pairs of equations, which leaves you with a smaller 2 variable system. Then you eliminate a second variable from that smaller system to solve for the last two unknowns. Finally you substitute back into an original equation to find the variable you eliminated first.

Given a 3 variable system such as:

\(x + y + z = 6\)

\(2x - y + z = 3\)

\(x + 2y - z = 5\)

Step 1: Pick a variable to eliminate first

Look for a variable with matching or opposite coefficients across two of the equations. Here, \(z\) has coefficients of \(1\), \(1\), and \(-1\), which makes it a convenient first target.

Step 2: Eliminate that variable from one pair

Add the first and second equations. Since \(z\) and \(z\) do not cancel, first subtract instead: \((x + y + z) - (2x - y + z) = 6 - 3\), which gives \(-x + 2y = 3\). Call this Equation A.

Step 3: Eliminate the same variable from a second pair

Add the first and third equations: \((x + y + z) + (x + 2y - z) = 6 + 5\), which gives \(2x + 3y = 11\). Call this Equation B.

Step 4: Solve the new 2 variable system

Equations A and B now form a 2 variable system in \(x\) and \(y\):

\(-x + 2y = 3\)

\(2x + 3y = 11\)

Multiply Equation A by \(2\) so the \(x\) terms cancel: \(-2x + 4y = 6\). Add this to Equation B: \(7y = 17\), so \(y = \frac{17}{7}\). Substituting back into Equation A gives \(x = 2y - 3 = \frac{34}{7} - 3 = \frac{13}{7}\).

Step 5: Back-substitute to find the third variable

Use the original first equation \(x + y + z = 6\) to solve for \(z\): \(z = 6 - x - y = 6 - \frac{13}{7} - \frac{17}{7} = \frac{12}{7}\).

Step 6: Check the solution

Substitute \(x = \frac{13}{7}\), \(y = \frac{17}{7}\), \(z = \frac{12}{7}\) into all three original equations to confirm each one balances. This check is the best way to catch an arithmetic slip before calling the answer final.

Elimination is easiest when a variable already has equal or opposite coefficients in two equations, since then you can add or subtract without multiplying anything first. When no variable lines up that neatly, multiply one or both equations by a constant so that a variable's coefficients become opposites, just as you would in a two-variable elimination problem, before adding the equations together.

Elimination tends to be faster when the equations are already lined up in \(x\), \(y\), \(z\) form with reasonably simple coefficients. If one equation is already solved for a single variable, or isolating a variable is quick, the alternative solving 3 variable systems of equations by substitution approach can be just as efficient. Both methods always produce the same solution, so use whichever keeps the arithmetic simplest for the system in front of you.

Forgetting to eliminate the same variable in both equation pairs is the most common error, since mixing which variable you cancel leaves you with a 2 variable system that still has three unknowns. Another frequent slip is dropping a sign when subtracting equations. Writing out each step of the subtraction fully, rather than doing it in your head, helps avoid this.

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