Systems in Three Variables

Intuition

You already know how to solve three equations in three unknowns: knock out one variable to drop to a two-variable system, solve that, then back-substitute. This lesson is not that lesson again. It asks a sharper question. What does the set of all solutions actually look like, and why does the algebra 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. So a system of three such equations is three planes hanging in space, and a solution is a point resting on all three. Three planes can cross at one point, or lean on a shared line, or fall onto the same sheet, or share no single point among all three. In that last case the planes either miss entirely or meet only in pairs, and those four pictures are the whole story. The other half of the lesson turns the solving into a fixed routine. The routine is a forward sweep that leaves the system in a neat triangular shape you can read from the bottom up. Two lessons from now that routine becomes a machine; here you learn why it works and how to hear each of the four shapes speaking through the arithmetic.

Learning goals

Picture each equation as a plane in space
Eliminate forward to triangular form, then back-substitute
Sort the outcome into point, line, plane or empty
Read a false row as empty and a zero row as free
Parameterize a line or plane with a letter per free variable
Note the routine is mechanical, which is why a matrix can run it

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 $a$ , $b$ , $c$ not all zero) 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: pick $x$ and $y$ is forced, so the solutions sweep out a line. In three variables an equation $ax + by + cz = d$ leaves two degrees of freedom: pick $x$ and $y$ freely and $z$ is forced, so the solutions sweep out a two-dimensional sheet, a plane. 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. 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 need three independent facts, and each fact becomes one equation. 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 shape of the solution set

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.

What three planes can do. They cross at one point (a unique solution), or lean on a shared line, or coincide as a single sheet (both give infinitely many solutions), or fail to share any point at all (no solution). These four are the only possibilities.

Why exactly these four, and never a stray answer like two points? The triangular form settles it. The count that matters is how many equations survive the forward sweep as genuine constraints.

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

Run the forward elimination until the system is triangular. 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.

First, a row may collapse to $0 = k$ with $k \neq 0$ . No triple can make $0$ equal a nonzero number, so the system has no solution and its solution set is empty. Set that case aside.

Otherwise every surviving row genuinely constrains the unknowns. Call a variable a pivot when it is the first variable still present in some row of the triangular system. That row pins its pivot down as soon as the variables after it are known. A variable that is never a pivot is free, meaning you may hand it any value you like. Suppose $r$ of the rows survive as real constraints, so there are $r$ pivots and $3 - r$ free variables. Back-substitution then determines each pivot uniquely from whatever values the free variables were given. So the solutions are exactly the triples you sweep out by running the free variables through all their choices.

Now count. If $r = 3$ , no variable is free, back-substitution forces one value for each unknown, and the solution set is a single point. If $r = 2$ , one variable is free; letting it run traces a one-parameter family of triples, which is a line. If $r = 1$ , two variables are free, and the two-parameter family they generate fills a plane. A count of $r = 0$ would need all three equations to vanish, which cannot happen for a genuine system, and no other value of $r$ exists. So a consistent three-variable system has one solution, a line of them, or a plane of them. Such a system never has exactly two solutions, or any other finite count past one, because a single free variable already delivers infinitely many. Together with the empty case, the complete catalog is point, line, plane, or nothing.

Check your understanding

Elimination turns a three-equation system into $x + y + z = 2$ and $y - z = 1$ , with the third equation reduced to $0 = 0$ . What is the solution set?

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

Degenerate systems and how to read them

When the sweep ends anywhere but three clean pivots, 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 $r = 2$ and one variable is free: the solutions form a line.

Parameterize. In $2y + z = 2$ , set $z = 2t$ so the division by $2$ comes out clean; as $t$ runs over all numbers, $z = 2t$ still reaches every value. Then

2y + 2t = 2 \ \Rightarrow \ y = 1 - t.

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

-x + (1 - t) + 2t = 1 \ \Rightarrow \ -x + 1 + t = 1 \ \Rightarrow \ x = t.

Every solution has the form

(x, y, z) = (t,\ 1 - t,\ 2t),

one triple for each value of $t$ . Check the held-back equation $(3)$ : $-t + 5(1 - t) + 3(2t) = -t + 5 - 5t

6t = 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.

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 $r = 1$ constraint survives. With $3 - 1 = 2$ free variables, 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 $x + z = 3$ and $y - z = -1$ , with the third equation vanishing. Writing $z = t$ , which parameterization gives all the solutions?

(3 - t,\ t - 1,\ t)

(3 + t,\ -1 - t,\ t)

(t - 3,\ t - 1,\ t)

(3 - t,\ -1 - t,\ t)

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 leftover equations share no common pair of unknowns and cannot be solved together.
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 $(t,\ 1 - t,\ 2t)$ , 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 equations you combined. The pair you added is guaranteed to hold. Verify the triple in the equation you set aside, because that is where a slip in arithmetic shows itself.

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 pivots), a line (two pivots, one free variable), a plane (one pivot, 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 $(t,\ 1 - t,\ 2t)$ for a line or $(s,\ 2s + t - 3,\ t)$ for a plane.
This forward-then-back routine is fixed and mechanical, which is why the matrices lesson can hand it to a grid of coefficients and let a machine run it.

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 shape of the solution set

Degenerate systems and how to read them

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Free Response Questions (FRQ)