Systems in Three Variables

Intuition

Take three equations in three unknowns, such as $x + y + z = 4$ , $x + 2y + 3z = 5$ , and $2x + 3y + z = 12$ . A solution is a triple $(x, y, z)$ that makes all three equations true at once, and the goal is to find every such triple. You can already eliminate a variable to drop to a smaller system, solve that, then substitute back; this lesson sharpens the question. What does the whole set of solutions look like, and why does the arithmetic always land on one of a short list of answers? In two variables one equation drew a line; in three it draws a flat sheet, a plane. A system of three such equations is three planes in space, and a solution is a point that lies on all three. Three planes can cross at one point, share a whole line, coincide on a single sheet, or miss each other entirely. The elimination routine below finds the solution and tells you, from the arithmetic alone, which of those four outcomes you are in.

Learning goals

Picture each equation as a plane in three-dimensional space
Eliminate one variable at a time to reach triangular form, then back-substitute
Tell a false row (no solution) apart from a vanishing row (a free variable)
Sort the outcome into a point, a line, a plane, or no solution
Parameterize a line or a plane with one letter per free variable

From a line to a plane

A linear equation in three variables has the standard form

ax + by + cz = d,

with numbers $a$ , $b$ , $c$ , $d$ and at least one of $a$ , $b$ , $c$ nonzero, and each variable to the first power only. Its solutions are ordered triples $(x, y, z)$ , three numbers at once. Compare the counting. In two variables an equation $ax + by = c$ leaves one degree of freedom: solve for whichever variable has a nonzero coefficient, and the other is free to be anything, so the solutions sweep out a line. (If $b \neq 0$ , that means picking $x$ freely and solving for $y$ .) In three variables an equation $ax + by + cz = d$ leaves two degrees of freedom: solve for one variable using whichever coefficient is nonzero, and the other two are free, so the solutions sweep out a two-dimensional sheet, a plane. (If $c \neq 0$ , that means picking $x$ and $y$ freely and solving for $z$ .) Going up one variable lifts the picture up one dimension.

One equation, one dimension up. In two variables the solutions of ax + by = c fill a line in the plane; in three variables the solutions of ax + by + cz = d fill a flat plane in space. Each extra variable adds a degree of freedom to the solution set.

A system of three linear equations stacks three of these and asks for the triples satisfying all three at once. Geometrically it stacks three planes and asks which points lie on every one. That is why solving a three-variable system is the study of how planes can intersect.

The elimination algorithm: reduce to triangular form

You met elimination as a way to cancel a variable by adding or subtracting equations. The previous lessons also proved the key fact we lean on: replacing one equation by itself plus a multiple of another changes nothing about the solution set. The upgrade here is to stop combining pairs at random and instead aim at a target shape. Use the first equation to remove $x$ from the other two, then use the new second equation to remove $y$ from the third. This only works if the equation you are eliminating with actually contains the variable you are clearing; if it does not, for instance the first equation happens to have no $x$ term, swap it with a lower equation that does. Swapping the order of two equations changes nothing about the solution set, so it is always allowed. What remains is a triangular (staircase) system: the first equation still carries all three unknowns, the second carries only two, the third carries only one.

The target shape. Forward elimination clears the lower-left entries so each equation begins one variable later than the one above it, leaving a staircase. The bottom row has a single unknown; back-substitution then climbs the stairs, solving for z, then y, then x.

The bottom equation has one unknown, so you solve it outright. Carry that value up into the second equation and solve for its second unknown. Carry both up into the first equation and solve for the last. This forward sweep to triangular form, then the climb back up, is a fixed procedure you can run without inventing anything. Two lessons from now you will drop the letters and record only the coefficients in a rectangular array called a matrix, and this identical sweep becomes a purely mechanical routine. For now keep the letters, so the reasoning stays visible.

Worked example 1 Solve $x + y + z = 4$ , $x + 2y + 3z = 5$ , $2x + 3y + z = 12$

Number the equations to keep the bookkeeping honest:

(1)\ x + y + z = 4, \qquad (2)\ x + 2y + 3z = 5, \qquad (3)\ 2x + 3y + z = 12.

Use equation $(1)$ to clear $x$ from the other two. Subtract $(1)$ from $(2)$ , and subtract $2$ times $(1)$ from $(3)$ :

(2) - (1): \ y + 2z = 1, \qquad (3) - 2(1): \ y - z = 4.

Now clear $y$ from the second of these by subtracting it from the first, which leaves a single equation in $z$ :

(y + 2z) - (y - z) = 1 - 4, \qquad 3z = -3.

The system is now triangular, one variable dropping away at each step:

\begin{array}{rrrcr} x & +\, y & +\, z & = & 4 \\[2pt] & y & +\, 2z & = & 1 \\[2pt] & & 3z & = & -3 \end{array}

Climb from the bottom. The last row gives $z = -1$ . Put that into the middle row:

y + 2(-1) = 1 \ \Rightarrow \ y = 3.

Put $y = 3$ and $z = -1$ into the top row:

x + 3 + (-1) = 4 \ \Rightarrow \ x = 2.

The candidate is $(2, 3, -1)$ . Check it in all three originals, including the one you touched last:

2 + 3 - 1 = 4 \ \checkmark, \qquad 2 + 6 - 3 = 5 \ \checkmark, \qquad 4 + 9 - 1 = 12 \ \checkmark.

The three planes meet at the single point $(2, 3, -1)$ .

Check your understanding

A system has been reduced to the triangular form $x - y + 2z = 1$ , $y + z = 5$ , and $4z = 8$ . What is $x$ ?

0

2

3

-3

A word problem in three unknowns

Three unknowns generally need three independent facts to pin down one value for each, and each fact becomes one equation. Independent just means each fact adds real information: repeating something you already know does not count as a fourth fact. The elimination routine solves the resulting system the same way, whether the third fact is a full weighted total or a plain relationship between the unknowns.

Worked example 2 A three-way coffee blend

A roaster builds a $12$ -pound blend from three coffees. The three coffees are a house roast at $6$ dollars per pound, a dark roast at $9$ dollars per pound, and a decaf at $12$ dollars per pound. The blend costs $102$ dollars in all, and the roaster uses as much house roast as dark and decaf combined. How many pounds of each go in?

Let $h$ , $d$ , and $k$ be the pounds of house, dark, and decaf. The three facts become

(1)\ h + d + k = 12, \qquad (2)\ 6h + 9d + 12k = 102, \qquad (3)\ h = d + k.

Divide equation $(2)$ by $3$ to shrink the numbers, and rewrite equation $(3)$ with all terms on one side:

(2')\ 2h + 3d + 4k = 34, \qquad (3')\ h - d - k = 0.

Use $(3')$ to clear $h$ . Subtract $(3')$ from $(1)$ , and subtract $2$ times $(3')$ from $(2')$ :

(1) - (3'): \ 2d + 2k = 12 \ \Rightarrow \ d + k = 6, \qquad (2') - 2(3'): \ 5d + 6k = 34.

Those two equations are an ordinary two-variable system. Subtract $5$ times the first from the second to strip out $d$ :

(5d + 6k) - 5(d + k) = 34 - 30, \qquad k = 4.

Then $d = 6 - k = 2$ , and $h = d + k = 6$ . Check the cost, the fact held back longest:

6(6) + 9(2) + 12(4) = 36 + 18 + 48 = 102 \ \checkmark.

The blend takes $6$ pounds of house roast, $2$ pounds of dark, and $4$ pounds of decaf.

The four possible outcomes

Every example so far ended at one point, but that is only the most common ending. Recall the two-variable result from the previous lesson: a system there is consistent (has a solution) or inconsistent (none). A consistent system is in turn either independent (exactly one solution) or dependent (infinitely many). The infinite case had a single shape, a whole line where two equations described the same line. Space adds a dimension, and with it a new possibility. A dependent system in three variables can leave the solutions filling a line or filling an entire plane, so the catalog grows from three entries to four.

One example of each of the four possible solution sets: crossing at a single point, sharing one line, coinciding as a single plane (both infinitely many solutions), or sharing no point at all (no solution, shown here as parallel planes, though planes can also fail to share a point without being parallel). Every three-equation system lands in exactly one of these four outcomes.

How do you tell which of the four you are looking at, just from the arithmetic? Read the triangular form. The next three examples work out a line, a false row, and a plane in turn; after that we can explain why these four outcomes, and no others, are the only ones possible.

Degenerate systems and how to read them

When the sweep ends with anything other than one value pinned down for each variable, the arithmetic tells you which shape you have. A false row ( $0 = k$ with $k \neq 0$ ) means no solution, and it overrides everything else. If no row is false, then a vanishing row ( $0 = 0$ ) is a redundant constraint, so a variable is free and you parameterize. To parameterize, name each free variable with a letter and solve for the rest in terms of it.

Worked example 3 A vanishing row: the solutions form a line

Solve

(1)\ x + 3y + z = 3, \qquad (2)\ -x + y + z = 1, \qquad (3)\ -x + 5y + 3z = 5.

Clear $x$ using equation $(1)$ . Add $(1)$ to each of the others, since their $x$ terms are $-x$ :

\begin{aligned} (1) + (2): \ 4y + 2z &= 4 \ \Rightarrow \ 2y + z = 2, \\ (1) + (3): \ 8y + 4z &= 8 \ \Rightarrow \ 2y + z = 2. \end{aligned}

Both reductions give the identical equation $2y + z = 2$ , so subtracting one from the other leaves $0 = 0$ . The third equation carried no new information. Only two real constraints survive, equation $(1)$ and $2y + z = 2$ , so one variable is free: the solutions form a line.

Parameterize by naming the free variable. Set $z = t$ ; then $2y + z = 2$ gives

2y + t = 2 \ \Rightarrow \ y = 1 - \frac{t}{2}.

Recover $x$ from equation $(2)$ , $-x + y + z = 1$ :

-x + \left(1 - \frac{t}{2}\right) + t = 1 \ \Rightarrow \ -x + 1 + \frac{t}{2} = 1 \ \Rightarrow \ x = \frac{t}{2}.

Every solution has the form

(x, y, z) = \left(\frac{t}{2},\ 1 - \frac{t}{2},\ t\right),

one triple for each value of $t$ . (You could set $z = 2t$ instead to clear the fraction out of $y$ , giving $(t,\ 1 - t,\ 2t)$ : the same line, just labeled with a differently scaled letter.) Check the held-back equation $(3)$ :

-\frac{t}{2} + 5\left(1 - \frac{t}{2}\right) + 3t = -\frac{t}{2} + 5 - \frac{5t}{2} + 3t = 5

for every $t$ . The three planes share this whole line.

Worked example 4 A false row: no solution

Solve

(1)\ x + y + 2z = 4, \qquad (2)\ 2x + 3y + z = 5, \qquad (3)\ 3x + 4y + 3z = 12.

Clear $x$ with equation $(1)$ . Subtract $2$ times $(1)$ from $(2)$ , and $3$ times $(1)$ from $(3)$ :

(2) - 2(1): \ y - 3z = -3, \qquad (3) - 3(1): \ y - 3z = 0.

The same expression $y - 3z$ cannot equal both $-3$ and $0$ . Subtract the first from the second to see it:

0 = 3.

This row is false, so no triple satisfies all three equations. The system is inconsistent, and the three planes enclose space without ever sharing a common point.

Check your understanding

Elimination on a three-equation system produces the rows $x + y + z = 2$ , $y - z = 1$ , and $0 = 5$ . What is the solution set?

A empty (no solution) B a single point C a line (infinitely many solutions) D a plane (infinitely many solutions)

Worked example 5 Two vanishing rows: the solutions form a plane

Solve

\begin{aligned} (1)\ 2x - y + z &= 3, \\ (2)\ -4x + 2y - 2z &= -6, \\ (3)\ 6x - 3y + 3z &= 9. \end{aligned}

Look before computing. Equation $(2)$ is exactly $-2$ times equation $(1)$ , and equation $(3)$ is exactly $3$ times equation $(1)$ . Elimination confirms it: adding $2$ times $(1)$ to $(2)$ gives $0 = 0$ , and subtracting $3$ times $(1)$ from $(3)$ gives $0 = 0$ . Both extra rows vanish, so only one real constraint survives. With two variables left undetermined, the solution set is a whole plane, namely the single plane $2x - y + z = 3$ that all three equations describe.

Parameterize with two letters. Solve for $y$ , whose coefficient is $-1$ so no fractions appear, and let the other two run free as $x = s$ and $z = t$ :

2x - y + z = 3 \ \Rightarrow \ y = 2x + z - 3 = 2s + t - 3.

Every solution has the form

(x, y, z) = (s,\ 2s + t - 3,\ t),

one triple for each pair $(s, t)$ . Two free letters sweep out a two-dimensional sheet, which is exactly what it means for the answer to be a plane.

Check your understanding

A system reduces to the single surviving equation $x - y + z = 2$ , with the other two rows vanishing to $0 = 0$ . Writing $y = s$ and $z = t$ , which parameterization gives every solution?

(2 + s - t,\ s,\ t)

(2 - s + t,\ s,\ t)

(s,\ 2 + s - t,\ t)

(2 + s - t,\ t,\ s)

Why exactly these four?

Why exactly these four outcomes, and never a stray answer like two points? The triangular form settles it. The count that matters is how many rows out of the triangular system still carry real information, that is, how many did not vanish into $0 = 0$ .

Why the solution set is a point, a line, a plane, or nothing#

Run the forward elimination until the system is triangular, exactly as in the examples above. Each step replaces one equation by itself plus a multiple of another, and the earlier lessons proved that such a step neither loses a solution nor invents one. So the triangular system has exactly the solutions the original did. Read it from the bottom up.

If any row comes out false ( $0 = k$ with $k \neq 0$ ), no triple can satisfy it, so the solution set is empty. That case is done; assume from here that no row is false.

Otherwise, count the surviving equations, the rows that did not vanish into $0 = 0$ . Each surviving equation pins down one variable once the variables below it are known, the same way back-substitution worked in every worked example above. Every variable that no surviving equation pins down is free: you may give it any value you like, and each choice produces a different solution.

Surviving equations	Free variables	Solution set
3	0	a single point
2	1	a line (one parameter)
1	2	a plane (two parameters)

A count of $0$ surviving equations would need all three original equations to vanish, which cannot happen for a genuine system of three equations. So a consistent system always lands in one of the three rows above, and together with the empty case, the complete catalog is point, line, plane, or nothing. It can never be exactly two solutions, or any other finite count past one, because as soon as one variable is free, running it through infinitely many values produces infinitely many solutions.

Common mistakes

Eliminating a different variable in the two pairs. Aim the whole forward sweep at one variable first. If you clear $x$ from the second equation but $y$ from the third, the two rows that are left carry different pairs of unknowns, so they are not yet the matching two-variable system the next step needs.
Reading $0 = 0$ as no solution, or $0 = k$ as infinitely many. A vanishing row ( $0 = 0$ ) is a redundant equation: provided no row is false, it frees a variable and the system has infinitely many solutions. A false row ( $0 = k$ with $k \neq 0$ ) means none, and it wins even if another row vanished. Keep the two straight.
Confusing a line of solutions with a plane of solutions. One vanishing row leaves two constraints and one free variable, a line. Two vanishing rows leave one constraint and two free variables, a plane. Count the free variables, not just whether the answer is “infinitely many.”
Reporting a plane or line as if it were a single triple. When the answer is infinite, you must parameterize: introduce $t$ (or $s$ and $t$ ) and give the full family like $(t/2,\ 1 - t/2,\ t)$ , not one sample point.
Stopping at two coordinates. The sweep hands you the bottom variable, then the middle one. Climb all the way to the top equation and solve for the last unknown before stating the triple.
Checking in only the equation you set aside. A slip can occur in any step, not just the one you set aside last. Checking that one equation, as most examples above do, catches many arithmetic errors, but substituting into all three original equations, as Worked Example 1 does, is the surest way to catch a slip anywhere in the sweep. For a parameterized line or plane, confirm algebraically that each original equation holds for every value of the parameter.

Key takeaways

A linear equation $ax + by + cz = d$ has a plane of solutions, so a three-equation system is three planes and a solution is a point on all three.
Solve by forward elimination to triangular form, then back-substitute. Clear $x$ from the lower two equations, clear $y$ from the bottom one, solve the single remaining equation, and climb back up.
The solution set is exactly one of four shapes: a point (three surviving equations), a line (two surviving equations, one free variable), a plane (one surviving equation, two free variables), or empty. It can never be two points or any finite count past one.
The algebra announces the shape. Unique values mean a point, and a false row $0 = k$ means empty. If no row is false, a vanishing row $0 = 0$ is a redundant equation that frees a variable, so you parameterize the line or plane.
To parameterize, name each free variable with a letter and solve each surviving equation for one variable, giving a family like $(t/2,\ 1 - t/2,\ t)$ for a line or $(s,\ 2s + t - 3,\ t)$ for a plane.

Systems in Three Variables

Learning goals

From a line to a plane

The elimination algorithm: reduce to triangular form

A word problem in three unknowns

The four possible outcomes

Degenerate systems and how to read them

Why exactly these four?

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice