Linear Systems with Three Variables

Main idea

With three unknowns, eliminate the same variable twice and you're left with a two-variable system you already know how to solve. In pictures, each equation is a flat plane, and the solution is the point where all three meet.

\begin{aligned} a_1x + b_1y + c_1{\htmlData{sym=0}{z}} &= {\htmlData{sym=1}{d_1}} \\ a_2x + b_2y + c_2{\htmlData{sym=0}{z}} &= d_2 \\ a_3x + b_3y + c_3{\htmlData{sym=0}{z}} &= d_3 \end{aligned}

What the symbols mean

$z$
Third unknown
Each equation is now a plane in 3D space.
$d_1$
Constant
The first equation's right side.

Explanation

A three-variable system is simply three equations in three unknowns, $x$ , $y$ and $z$ , that all have to hold at once. It's the same question as last lesson with one more unknown, and you need it whenever three things are unknown, like fitting a parabola through three points. The block of equations looks intimidating, but each row is just $x$ , $y$ and $z$ (the unknowns) times some numbers (the $a$ 's, $b$ 's and $c$ 's), adding up to a target (the $d$ ). Same idea as before, just one size bigger. Bear with it.

Strategy. The trick from two variables still works: knock out one variable, and the problem gets smaller. You need to do it twice, using two different pairs of equations, so that you end up with two equations in the same two unknowns.

Pick a variable to eliminate. Use equations 1 and 2 to remove it, then equations 1 and 3 (or 2 and 3).
Solve the resulting two-variable system.
Substitute back to find the third variable.

What can happen. Each equation in three unknowns is a plane in space, so it's the same story as with lines, one dimension up. Three planes can meet in a single point, in no point at all (the system is inconsistent), or along a whole line (infinitely many solutions). There's an interactive picture further down where you can slide along the line where two planes meet and find where it hits the third.

See it

Each of these three equations is a plane, drawn here by its outline. The first two, x + y + z = 6 and 2x − y + z = 3, meet in a line, and that line is what eliminating a variable leaves you with. Slide along it and watch the gap to the third plane shrink to zero at (1, 2, 3), the one point on all three planes.

Worked examples

Example 1

Solve $x + y + z = 6$ , $2x - y + z = 3$ , $x + 2y - z = 2$ .

1
Eliminate $z$ , since its coefficients are all $\pm1$ . Eq. 2 minus Eq. 1 gives $x - 2y = -3$ , and Eq. 1 plus Eq. 3 gives $2x + 3y = 8$ .

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

Solve $x + y + z = 6$ , $2x - y + z = 3$ , $x + 2y - z = 2$ . Enter $x$ , $y$ , $z$ .

Problem 2+5 XP

Solve $x + y + z = 4$ , $2x - y + 3z = 14$ , $x + 2y - z = -3$ .

Problem 3+5 XP

Three numbers sum to 24. The first is twice the second, and the third is 4 more than the first. Find them.

Problem 4+5 XP

Find $a$ , $b$ , $c$ so that $y = ax^2 + bx + c$ passes through $(1, 2)$ , $(2, 3)$ and $(3, 6)$ .

Problem 5+5 XP

What does it mean if elimination produces $0 = 7$ ?

Problem 6+5 XP

A theater sold 400 tickets for 3600 in all: adults paid 10, students 6 and children 8. There were twice as many adult tickets as child tickets. How many of each? Enter adults, students, children.

Problem 7+5 XP

A ball's height $h = at^2 + bt + c$ is measured as 25 m at $t = 1$ s, 30 m at $t = 2$ s and 25 m at $t = 3$ s. Find $a$ , $b$ , $c$ .

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

Find the circle $x^2 + y^2 + Dx + Ey + F = 0$ through $(1, 0)$ , $(0, 1)$ and $(2, 3)$ . Enter $D$ , $E$ , $F$ .

Where it's used

Balancing chemical equations, fitting a parabola through three measurements, and nutrition planning with three foods are three-variable systems.

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