Systems with More Variables

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

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. Each equation now holds three letters instead of two, and a solution is three numbers instead of two, one for each letter.

An equation like these, written in the standard form

ax + by + cz = d,

is a linear equation in three variables: $a$ , $b$ , $c$ , and $d$ are numbers, $a$ , $b$ , and $c$ are not all zero, and each variable appears only to the first power. Its solutions are ordered triples $(x, y, z)$ . A system of three linear equations stacks three such equations and asks for the triple that satisfies all three at the same time. A single three-variable equation, like a single two-variable equation, has far too many solutions to pin the unknowns down. Fixing three unknowns takes three equations that each contribute a genuinely new condition, not one that just repeats another: three independent equations, one condition for each unknown.

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. Canceling the same variable is what leaves you two equations sharing the same two remaining unknowns.

Check your understanding

Which of these is written in the standard form $ax + by + cz = d$ for a linear equation in three variables?

4x + 2y - z = 9

4x + 2yz = 9

4x^2 - y + z = 9

4x + 2y - z

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

Check your understanding

Which ordered triple $(x, y, z)$ solves all three equations $x + y + z = 6$ , $x - y + z = 4$ , and $2x + y - z = 5$ ?

(3, 1, 2)

(2, 1, 3)

(4, 0, 2)

(1, 2, 3)

Why the reduction works and always finishes

The routine trades a big system for a smaller one and repeats, and that is trustworthy only if each step keeps exactly the same solutions and the shrinking actually stops. Both follow from what you already proved for two equations.

Why eliminating a variable preserves the solution set, and why two rounds finish it#

A solution of the system is a triple $(x, y, z)$ that satisfies all three equations at once, and the reduction must not lose or invent one. Recall the guarantee from the elimination lesson: keep one equation untouched, replace a second equation by that equation plus a multiple of the kept one, and the new pair has exactly the same solutions as the old pair, because you can undo the step by subtracting the same multiple back. Nothing in that argument counted the letters, so it holds just as well with three variables as with two.

Apply it twice, first reordering the equations if needed so the kept equation’s coefficient on the target variable is not zero. Keep the first equation. Replace the second equation with the second minus the right multiple of the first, chosen so the target variable cancels; 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; the solution set is unchanged again. What is left is the untouched first equation, still carrying all three unknowns, plus two equations that hold only the other two unknowns. Those two form an ordinary two-variable system, which you already know how to solve; back-substituting into the untouched first equation then finds the last unknown. Two rounds of elimination take three unknowns down to two, then to one, and solving that last equation directly finishes the job: the process cannot run forever.

Check your understanding

Why does the reduction always come to an end, instead of running forever, no matter which three-variable system you start with?

A Each round of elimination removes one unknown, and a system starts with only finitely many unknowns, so eventually none are left to eliminate. B Because you are only ever allowed to eliminate the variable

z

. C Because a three-variable system always has exactly one solution. D Because elimination only works on the first two equations you write down.

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 step added no new condition, so it did nothing to narrow the triples down. Whatever equations survive still have to be checked: infinitely many triples solve the system, or none do. In the two-variable case, with only two equations to begin with, hitting $0 = 0$ leaves nothing standing in the way, so the answer there is always infinitely many. A third equation changes that: it 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 $7$ dollars,
two popcorns, one pretzel, and three drinks cost $11$ dollars,
one popcorn, two pretzels, and four drinks cost $14$ 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 = 7, \qquad (2)\ 2p + t + 3d = 11, \qquad (3)\ p + 2t + 4d = 14.

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

(2p + t + 3d) - 2(p + t + d) = 11 - 14, \qquad -t + d = -3. \quad (A)

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

(p + 2t + 4d) - (p + t + d) = 14 - 7, \qquad t + 3d = 7. \quad (B)

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

(-t + d) + (t + 3d) = -3 + 7, \qquad 4d = 4, \qquad d = 1.

From $(A)$ , $-t + 1 = -3$ , so $t = 4$ . Back-substitute into equation $(1)$ : $p + 4 + 1 = 7$ , so $p = 2$ .

The prices are $2$ dollars for popcorn, $4$ dollars for a pretzel, and $1$ dollar for a drink. Check the third receipt, the one held back the longest: $2 + 2(4) + 4(1) = 2 + 8 + 4 = 14$ , 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.

Check your understanding

In the ordinary case, three planes in space, one for each equation of a three-variable system, cross at exactly one shared point. What does that point represent?

A The one ordered triple

(x, y, z)

that satisfies all three equations at once. B The planes must be parallel to each other. C The largest of the three unknowns. D One of infinitely many solutions to the system.

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 all three cross each other in pairs without ever sharing one point in common. 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 equation added no new condition beyond the first two, and the system has infinitely many solutions: pick any value for $z$ , and $(A)$ gives $y$ , then equation $(1)$ gives $x$ . Try a few values and see the pattern:

$z$	$y = 8 - 2z$	$x = 6 - y - z$	Triple
$0$	$8$	$-2$	$(-2, 8, 0)$
$1$	$6$	$-1$	$(-1, 6, 1)$
$2$	$4$	$0$	$(0, 4, 2)$

Every row solves all three original equations; check the first one: $-2 + 8 + 0 = 6$ , $-2 + 2(8) + 3(0) = 14$ , and $-2 + 3(8) + 5(0) = 22$ , all true. A different choice of $z$ gives a different triple on the same line, and there is no limit to how many you could list.

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 $y + z = 5$ and $2y + 2z = 10$ . What does this tell you?

A The system has no solution. B The second equation added no new condition, so the system has infinitely many solutions. C

y

and

z

must each equal

5

. D A fourth equation is needed to decide.

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

Poor organization: the wrong variable, or a dropped equation. Cancel the same variable from both 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. And use all three original equations, combined in two different pairs: you cannot tell by eye whether an equation is essential or redundant, and 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 added no new condition, 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 canceling 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 each round removes one unknown.
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.

Practice

Multiple Choice Questions (MCQ)

Progressively harder sets of questions. Each opens on its own page.

Level 1 Foundational 12 questions Start → Level 2 Intermediate 12 questions Start → Level 3 Challenge 12 questions Start →

Core practice

Practice problems at the level of the course, to be worked out on paper. Hints one at a time, then the answer or the full worked solution, with your progress kept in this browser.

More practice (optional)

Extra sets, as hard as the Challenge set. Each one opens on its own page.

More resources (optional)

Other explanations of this lesson, if you want a second take.

Intermediate Algebra: solve systems of equations with three variables (free textbook) (OpenStax)
Systems of linear equations (Khan Academy)
Systems of linear equations: definitions (Purplemath)
Systems of linear equations (any number of variables) (Math is Fun)

Go deeper (optional)

You can skip this and keep going. Read it if you want to know more.

Why elimination works and finishes for any number of equations, not just three

The lesson proved it for three equations in three unknowns. Nothing in that argument depended on the number three, so the same reasoning runs on any system of $n$ equations in $n$ unknowns.

Why elimination preserves solutions and terminates for a system of any size#

Keep the first equation, reordering the equations first if needed so its coefficient on the target variable is not zero, and eliminate that variable from every other equation using the same move: replace an equation with that equation minus the right multiple of the kept one. Each replacement preserves the solution set for the same reason it did with two or three equations, because the step can always be undone. After one round, the untouched first equation still carries every variable, and the other $n - 1$ equations carry only the remaining $n - 1$ variables. Those $n - 1$ equations form a smaller system of exactly the same kind, one equation and one unknown lighter. Run the same round on it, and the next, and the next: each round strips away one equation and one unknown together, so after $n - 1$ rounds a single equation in a single unknown remains, or a round collapses to a false or always-true statement first. Those endings mean exactly what they meant for three equations: a false statement rules out a solution, and an always-true one means that equation was redundant, leaving the surviving equations to decide the outcome. When a single equation does remain, solve it, then climb back up the chain, substituting each newly known value into the equation from the round before, until every unknown is found. The chain has only $n - 1$ rounds to climb, so it always finishes, one way or the other.

Writing every solution of an infinite family with one parameter

The lesson showed a few triples on the line of solutions for $x + y + z = 6$ , $x + 2y + 3z = 14$ , and $x + 3y + 5z = 22$ : pick a value for $z$ , and $x$ and $y$ follow. Writing that pattern once, for every value at once, uses a parameter, a letter standing for “whichever number you pick.”

Let $z = t$ , where $t$ can be any number. From $(A)$ , $y + 2t = 8$ , so $y = 8 - 2t$ . 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 triple for each value of $t$ . Setting $t = 0, 1, 2$ reproduces the same three triples found by hand above, and every other value of $t$ gives another point on the same line.

A bit of history (optional)

Why would anyone want a system with eight unknowns? Here is one answer, from $1845$ .

Electricity was new, and the current inside a wire could not be measured directly. A circuit with several branches was a real puzzle. The current splits at every junction, and how it splits depends on the whole network, not on any single piece.

Gustav Kirchhoff was a student of twenty-one in Prussia, a kingdom in what is now Germany. He wrote down two rules that settled the matter. At any junction, the current arriving equals the current leaving. Around any closed loop, the voltage gained and the voltage lost cancel out. Each junction and each loop gives one equation, and each branch of the circuit is one unknown.

That is a system, and it is exactly the kind you have been reducing. A circuit with eight branches yields eight unknowns and eight equations. Nobody solves such a thing by eye. You knock out one unknown, then another, until a single one is left. Then you climb back up and pick up the rest. That is this lesson’s method, run a few more rounds.

Kirchhoff’s rules are still how engineers describe a circuit, and the tools that solve those systems still run elimination underneath. Three unknowns give you three planes meeting at a point, and that is the last picture you can draw. Past three, the geometry runs out of room and the algebra carries on without it.