Systems of Equations
Systems in Three Variables
Extending elimination to three unknowns, the order in which variables are removed, and back-substitution.
What this page covers
- Eliminate one variable from a three-equation system
- Reduce to two equations in two unknowns and finish
- Back-substitute to recover all three values
- Check a three-variable solution properly
The strategy
Three unknowns are handled by reducing to two, then to one. Each stage uses the elimination already established for two variables.
Step 2 must use two different pairs. Eliminating from equations (1) and (2), then again from (1) and (2), gives the same equation twice and no progress. A common choice is (1) with (2), then (1) with (3).
Worked example: the source's system
Solve the three-equation system
Eliminate x, since it has coefficient 1 in the first two equations.
Now eliminate y from these two:
Back-substitute into the reduced equation:
And into an original equation:
In the third equation, which was not used for back-substitution: -2(7) + 2(2) + 5(-4) = -14 + 4 - 20 = -30 ✓.
Triangular form
The reduction produces a system in triangular form: three unknowns in the first equation, two in the second, one in the third.
Triangular form is trivial to solve from the bottom up, and it is exactly what row-echelon form is when the system is written as a matrix. The matrix version simply drops the variable names and works with the coefficients alone.
- Solve the last equation, which has one unknown.
- Substitute into the second-last, which then has one unknown.
- Continue upward until every variable is known.
- Check in an equation not used in the back-substitution.
A second worked example
From the typed supplementary notes
Solve x + 4y - z = 4, x + 3y + z = 8, 2x + 6y + z = 13.
2(-1) + 6(2) + 3 = -2 + 12 + 3 = 13 ✓.
Choosing well
| Choice | Guidance |
|---|---|
| Which variable to eliminate | The one with a coefficient of 1 or -1 somewhere |
| Which pairs to use | The two whose coefficients match most easily |
| Add or subtract | Add when the signs oppose; subtract when they agree |
| When to multiply both equations | Only when no single multiplier will match the coefficients |
| Where to back-substitute | The simplest equation containing the needed variable |
| Where to check | An equation not used in the back-substitution |
These are not rules but habits that keep the numbers small. A poor choice at the first step can turn a clean system into a page of fractions.
Common mistakes
| Mistake | Correct | Why |
|---|---|---|
| Using the same pair twice | Use two different pairs | The same equation gives no new information |
| Eliminating different variables in the two pairings | Eliminate the same variable both times | Otherwise the reduced system still has three unknowns |
| Multiplying one side only | Multiply the whole equation | Otherwise the equation changes |
| Back-substituting into an equation already used | It works, but the check must use a different one | Otherwise the check is circular |
| Losing a sign when subtracting | Write the negated equation on its own line | Every term changes sign |
| Stopping at two variables | Recover all three | A solution is an ordered triple |
Frequently asked questions
Which variable should I eliminate first?
Whichever has the simplest coefficients — ideally one appearing with coefficient 1 or -1 somewhere. It keeps the arithmetic manageable.
How many pairings do I need?
Two. Eliminate the chosen variable from one pair of equations and from a different pair, giving two equations in the remaining two unknowns.
Why must the two pairings be different?
Because using the same pair twice produces the same equation and no new information. Each original equation should be used at least once.
What does triangular form mean?
The reduced system has three variables in the first equation, two in the second and one in the third. Back-substitution then unwinds it from the bottom.
Source. Handwritten teaching notes, Week 10 pages 11-12 and Week 11 pages 1-3, with the typed supplementary notes on Gaussian elimination.
This page is an original exposition prepared for the KEVOS® knowledge library. It restates, reorganises and verifies the mathematics of the supplied teaching notes; it is not a reproduction of them. Numerical values taken from the notes are identified as source examples and are not presented as engineering standards.