Systems with More Variables

Intuition

You can already crack two equations in two unknowns: substitution and elimination each turn that puzzle into a single equation you know how to finish. A problem with three unknowns needs a third equation, and the natural worry is that a bigger system needs a brand new method. It does not. The whole trick is to make the problem smaller. Knock one variable out of the system and three unknowns drop to two, which puts you right back on the familiar ground of the last three lessons. Solve that two-variable system, then work backward to recover the variable you set aside. One equation is a plane floating in space, and three planes in ordinary position pierce each other at a single point. That point is the one triple $(x, y, z)$ that satisfies all three equations at once.

Learning goals

Write $ax + by + cz = d$ and solve for ordered triples
Eliminate the same variable from two different pairs
Reduce to a two-variable system, then back-substitute
Explain why each round removes an unknown and terminates
Picture each equation as a plane meeting at a point
Read a false or always-true leftover as the special cases

From two unknowns to three

A linear equation in three variables can be written in the standard form

ax + by + cz = d,

where $a$ , $b$ , $c$ , and $d$ are numbers and each variable appears only to the first power. Its solutions are no longer pairs but ordered triples $(x, y, z)$ : three numbers, one for each variable, that make the equation true. A system of three linear equations stacks three such equations and asks for the triple that satisfies all three at the same time.

Take the system

x + y + z = 6, \qquad x - y + z = 2, \qquad x + y - z = 0.

The triple $(1, 2, 3)$ solves it, because $1 + 2 + 3 = 6$ , and $1 - 2 + 3 = 2$ , and $1 + 2 - 3 = 0$ are all true. A single three-variable equation, like a single two-variable equation, has far too many solutions to pin the unknowns down. It takes three independent equations to fix three unknowns, one condition for each degree of freedom.

The plan is to shrink the system one variable at a time. You already own the tool that removes a variable: elimination, from two lessons ago. Adding or subtracting two equations to cancel a variable worked no matter what the other letters were. So that same move works just as well when a third letter is riding along. The new idea in this lesson is purely organizational. You will run elimination twice, on two different pairs of the three equations, and both times you must cancel the same variable. Cancelling the same variable is what leaves you two equations sharing the same two remaining unknowns.

The reduction strategy step by step

Every three-variable system yields to the same routine.

Choose a variable to eliminate. Pick whichever one has the friendliest coefficients, often a variable that already appears with matching or opposite coefficients somewhere.
Eliminate it from one pair of equations. Combine two of the three equations, by adding or subtracting (scaling first if needed), so that variable cancels. You now have one equation in the other two unknowns.
Eliminate the same variable from a second pair. Combine a different pair, being sure to use the equation you have not touched yet, and cancel the same variable again. This gives a second equation in the same two unknowns.
Solve the two-variable system. Those two new equations form an ordinary system in two unknowns. Solve it by substitution or elimination.
Back-substitute for the third variable. Put the two values you found into any one of the three original equations and solve for the variable you eliminated.
Check the triple in all three original equations. A genuine solution satisfies every one of them, not just the two you happened to use.

The one step people get wrong is step 3. If you eliminate $z$ from the first pair but then eliminate $y$ from the second, your two leftover equations do not share the same unknowns. Without the same two unknowns, those two leftover equations cannot be solved together. Cancel the same variable both times.

The reduction in one picture. Each elimination removes one variable, so three equations in three unknowns become two in two unknowns, then one in one unknown. Solving that and back-substituting walks the answer back up the chain.

Worked example 1 Solve $x + y + z = 6$ , $2x - y + z = 3$ , and $x + 2y - z = 2$

Number the equations so the bookkeeping stays clear:

(1)\ x + y + z = 6, \qquad (2)\ 2x - y + z = 3, \qquad (3)\ x + 2y - z = 2.

Choose to eliminate $z$ . Look at the $z$ terms: equations $(1)$ and $(2)$ both have $+z$ , while equation $(3)$ has $-z$ . That makes equation $(3)$ the convenient partner, because adding it to either of the others cancels $z$ on the spot.

Eliminate $z$ from the first pair, equations $(1)$ and $(3)$ . Add them:

(x + y + z) + (x + 2y - z) = 6 + 2, \qquad 2x + 3y = 8. \quad (A)

Now eliminate the same variable $z$ from a second pair, equations $(2)$ and $(3)$ , using the equation not yet touched. Add them:

(2x - y + z) + (x + 2y - z) = 3 + 2, \qquad 3x + y = 5. \quad (B)

Equations $(A)$ and $(B)$ form an ordinary two-variable system. Solve it by substitution: equation $(B)$ gives $y = 5 - 3x$ , so replace $y$ in $(A)$ :

2x + 3(5 - 3x) = 8, \qquad 2x + 15 - 9x = 8, \qquad -7x = -7, \qquad x = 1.

Then $y = 5 - 3(1) = 2$ . Back-substitute $x = 1$ and $y = 2$ into equation $(1)$ to recover $z$ :

1 + 2 + z = 6, \qquad z = 3.

The candidate is $(1, 2, 3)$ . Check it in all three originals, not just the two pairs you combined:

1 + 2 + 3 = 6 \ \checkmark, \qquad 2(1) - 2 + 3 = 3 \ \checkmark, \qquad 1 + 2(2) - 3 = 2 \ \checkmark.

All three hold, so the solution is $(1, 2, 3)$ .

Check your understanding

You want to eliminate $z$ from the pair $x + y + z = 4$ and $x + y - z = 2$ . What do you get by adding the two equations?

2z = 2

2x + 2y = 6

x + y = 2

2x + 2y + 2z = 6

Why the reduction works and always finishes

The routine trades a big system for a smaller one and repeats. Two things need to be true for that to be trustworthy: each step must keep exactly the same set of solutions, and the shrinking must actually stop. Both follow from what you already proved for two equations.

Why eliminating a variable preserves the solution set and terminates#

A solution of a three-equation system is a triple $(x, y, z)$ that satisfies all three equations at once, and the solution set is the collection of every such triple. The reduction must not lose any of those triples or invent new ones, and it must reach an answer in finitely many moves.

Recall the guarantee proved in the elimination lesson. Suppose you keep one equation untouched. Suppose you also replace a second equation by that second equation plus a multiple of the kept one. Then the new pair has exactly the same solutions as the old pair. That guarantee holds in both directions, because the addition and multiplication properties of equality let you build the combination and then undo it. Nothing in that argument counted the variables, so it stays true word for word when the equations carry three letters instead of two.

Now run the reduction on the system, assuming the first equation contains the variable you want to remove (if it does not, reorder the equations so it does). Keep the first equation. Replace the second equation by the second minus the right multiple of the first, choosing the multiple so the target variable cancels. By the guarantee, the solution set is unchanged. Replace the third equation the same way, subtracting a multiple of the first so its copy of the target variable cancels too; again the solution set is unchanged. The system now reads: the untouched first equation, still carrying all three unknowns, together with two equations that contain only the other two unknowns. Every triple that solves the whole system must have its two-variable part solving those last two equations.

So the work splits into stages, and each stage is strictly smaller than the one before. The two leftover equations are an ordinary two-variable system, and you proved in the last lessons that such a system resolves. Eliminate again to reach one equation in one unknown. Solve that equation, then back-substitute for the second unknown. With two of the three values in hand, the untouched first equation becomes a single equation in the last unknown, which you solve directly. Each round removes one variable, so three unknowns fall to two and then to one. The process therefore cannot run forever. It halts after as many rounds as there are variables. The identical bookkeeping applies to a system of $n$ equations in $n$ unknowns. That bookkeeping drives the system down to $n - 1$ equations, then $n - 2$ , and onward to a single equation. The method is therefore not special to three unknowns; it scales to any number.

Because every step preserves the solution set, the strange endings carry a clear meaning. If a stage ever collapses to a false statement such as $0 = 4$ , then the original system had no triple satisfying it, so the system has no solution. Suppose instead a stage collapses to an always-true statement such as $0 = 0$ . That stage carried nothing new, so the triples were never narrowed down to one. Whatever equations survive then decide the outcome: infinitely many triples solve the system, or none do. With only two equations there is nothing left to decide, and the answer is infinitely many. A third equation can still contradict the others. Consider $x + y + z = 1$ paired with $2x + 2y + 2z = 2$ , which collapses to $0 = 0$ . Now add $x + y + z = 5$ to that pair: eliminating against the first equation leaves $0 = 4$ as well, so the enlarged system has no solution at all.

Check your understanding

After eliminating $z$ and solving the resulting two-variable system, you find $x = 1$ and $y = 2$ . How do you find $z$ ?

z = 0

, because it was the variable you eliminated. B Add

x

and

y

to get

z = 3

. C A fourth equation is needed to find

z

. D Substitute

x = 1

and

y = 2

into one of the original equations and solve for

z

A three-variable word problem

Three-variable systems earn their keep on problems with three unknowns and three facts. A clean example is recovering unknown prices from several receipts, where each receipt is one equation.

Worked example 2 Unknown prices from three receipts

At a snack stand the popcorn, pretzels, and drinks each have a fixed price. Three customers pay:

one popcorn, one pretzel, and one drink cost $6$ dollars,
two popcorns, one pretzel, and three drinks cost $13$ dollars,
one popcorn, two pretzels, and four drinks cost $17$ dollars.

Find the price of each item.

Name the unknowns. Let $p$ , $t$ , and $d$ be the price in dollars of a popcorn, a pretzel, and a drink. Each receipt becomes one equation:

(1)\ p + t + d = 6, \qquad (2)\ 2p + t + 3d = 13, \qquad (3)\ p + 2t + 4d = 17.

Eliminate $p$ , whose coefficient is $1$ in the first equation. Subtract twice equation $(1)$ from equation $(2)$ :

(2p + t + 3d) - 2(p + t + d) = 13 - 12, \qquad -t + d = 1. \quad (A)

Eliminate the same variable $p$ from the untouched equation $(3)$ by subtracting equation $(1)$ :

(p + 2t + 4d) - (p + t + d) = 17 - 6, \qquad t + 3d = 11. \quad (B)

Add $(A)$ and $(B)$ to cancel $t$ :

(-t + d) + (t + 3d) = 1 + 11, \qquad 4d = 12, \qquad d = 3.

From $(A)$ , $-t + 3 = 1$ , so $t = 2$ . Back-substitute into equation $(1)$ : $p + 2 + 3 = 6$ , so $p = 1$ .

The prices are $1$ dollar for popcorn, $2$ dollars for a pretzel, and $3$ dollars for a drink. Check the third receipt, the one held back the longest: $1 + 2(2) + 4(3) = 1 + 4 + 12 = 17$ , correct. The other two receipts check the same way.

The geometry: three planes in space

Two variables live on a flat page, and a linear equation there draws a line. Three variables live in space, and a linear equation $ax + by + cz = d$ draws a flat sheet stretching out forever, a plane. A system of three equations is then three planes, and a solution triple is a point lying on all three at once.

In the ordinary case the three planes tilt in different directions and meet at exactly one point, the single triple that solves the system. This is the picture behind almost every example in this lesson.

Each equation of a three-variable system is a plane in space. When the three planes tilt in general directions they cross at a single shared point, and that point is the one triple (x, y, z) solving all three equations.

Special arrangements break that single crossing, and they match the same two exceptions you met with lines. If the planes never share a common point, the system is inconsistent and has no solution. That happens when two of them are parallel, or when the three enclose a hollow like the sides of a triangular tent. If the three planes instead share a whole common line, or even coincide as one plane, then endlessly many points lie on all three. In that case the system has infinitely many solutions. The algebra flags each case for you: eliminating variables leaves a false statement in the first situation and an always-true statement in the second.

When the reduction hits no solution or infinitely many

Worked example 3 A system with no solution

Solve

(1)\ x + y + z = 6, \qquad (2)\ x + 2y + 3z = 14, \qquad (3)\ 2x + 3y + 4z = 25.

Eliminate $x$ using the first equation. Subtract equation $(1)$ from equation $(2)$ :

(x + 2y + 3z) - (x + y + z) = 14 - 6, \qquad y + 2z = 8. \quad (A)

Now clear the same variable from equation $(3)$ by subtracting $2$ times equation $(1)$ :

(2x + 3y + 4z) - 2(x + y + z) = 25 - 12, \qquad y + 2z = 13. \quad (B)

Equations $(A)$ and $(B)$ read $y + 2z = 8$ and $y + 2z = 13$ . The same expression $y + 2z$ cannot equal both $8$ and $13$ . Subtract $(A)$ from $(B)$ to see it plainly:

0 = 5.

This is false, so no triple satisfies all three equations, and the system has no solution. Geometrically the three planes fail to share any common point.

Worked example 4 A system with infinitely many solutions

Solve

(1)\ x + y + z = 6, \qquad (2)\ x + 2y + 3z = 14, \qquad (3)\ x + 3y + 5z = 22.

Eliminate $x$ again with the first equation. Subtract equation $(1)$ from equation $(2)$ :

(x + 2y + 3z) - (x + y + z) = 14 - 6, \qquad y + 2z = 8. \quad (A)

Subtract equation $(1)$ from equation $(3)$ :

(x + 3y + 5z) - (x + y + z) = 22 - 6, \qquad 2y + 4z = 16. \quad (B)

Equation $(B)$ is exactly twice equation $(A)$ , so it says nothing new. Subtract $2$ times $(A)$ from $(B)$ :

0 = 0.

The statement is always true, so the third condition was not independent, and the system has infinitely many solutions. To describe them, let $z = t$ be free. From $(A)$ , $y = 8 - 2t$ , and from equation $(1)$ , $x = 6 - y - z = 6 - (8 - 2t) - t = t - 2$ . So every triple

(x, y, z) = (t - 2,\ 8 - 2t,\ t)

solves the system, one for each value of $t$ . The three planes meet along this whole line rather than at a single point.

Check your understanding

Reducing a system of three equations in three unknowns, you eliminate a variable from two pairs and arrive at $0 = 4$ . What does this tell you?

A The system has infinitely many solutions. B The last variable must equal

4

. C The system has no solution. D You need a fourth equation to decide.

Common mistakes

Eliminating a different variable in the two pairs. If you cancel $z$ from the first pair but $y$ from the second, the two leftover equations do not share the same unknowns and cannot be solved together. Cancel the same variable both times.
Dropping one of the equations. All three equations carry information. Combine them in two different pairs so every equation is used, including the one you set aside first. Reusing the same pair twice throws away a fact and can turn a solvable system into a false “infinitely many.”
Stopping after two variables. The reduction hands you two of the three coordinates. Back-substitute to find the third, then state the full triple $(x, y, z)$ . A three-variable answer is three numbers.
Bookkeeping slips. Scaling an equation multiplies every term, and subtracting an equation flips every sign inside it. Number your equations, and keep the like terms in columns so a stray sign or a forgotten constant does not creep in.
Checking in only the equations you used. Verify the triple in all three originals. The two pairs you combined are guaranteed to hold; the equation you never revisited is the one that catches an error.
Misreading a vanished system. A leftover $0 = k$ with $k$ not zero means no solution. A leftover $0 = 0$ means only that one equation repeated another, so read what survives before answering: with three equations it can still be no solution. Neither says a variable equals zero.

Key takeaways

A linear equation in three variables is $ax + by + cz = d$ ; its solutions are ordered triples $(x, y, z)$ , and it takes three independent equations to pin three unknowns down.
Reduce, then solve. Eliminate one chosen variable from two different pairs of the equations (using all three equations, and cancelling the same variable each time) to get a two-variable system. Solve that, back-substitute for the eliminated variable, and check the triple in all three originals.
The reduction is valid because each elimination preserves the solution set, exactly as in the two-variable case, and it terminates because every round removes one unknown. The same idea scales to $n$ equations in $n$ unknowns.
Each equation is a plane in space. Three planes in general position meet at one point (one solution); planes with no common point give no solution; planes sharing a line or coinciding give infinitely many.
The algebra signals the special cases: a false statement such as $0 = 4$ means no solution. An always-true statement such as $0 = 0$ instead means that equation was redundant, leaving infinitely many or none according to what the surviving equations say.

Systems with More Variables

Learning goals

From two unknowns to three

The reduction strategy step by step

Why the reduction works and always finishes

A three-variable word problem

The geometry: three planes in space

When the reduction hits no solution or infinitely many

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Free Response Questions (FRQ)