Linear Systems in Several Unknowns

What this move does

Three unknowns, three equations, and the method is one word: eliminate.

3 unknowns eliminate one 2 unknowns eliminate another 1 unknown

Each step is the elimination by adding or subtracting from the previous stop, just done a few more times:

Pick the unknown that is easiest to eliminate (its coefficients are $\pm 1$ , or it is already missing from some equation).
Use the same equation to eliminate this unknown from the other two, giving two equations with one unknown fewer.
Eliminate once more between those two new equations, so only one unknown is left, and solve.
Substitute back: put the value you found back in, layer by layer, to find the other unknowns. Finally check in the original equations.

The workbench above writes each equation as a row of coefficients. A coefficient that has become $0$ shows as “·”. When the elimination is done right, the “·” cells line up like the steps of a staircase.

Predict first

There are only two equations: $x + y + z = 6$ and $x - y + z = 2$ . Of the three unknowns, which one already has a definite value?

x

y

z

none of them

Choose an answer first; then you can see the result.

Work through one

Example: solve 2x + y − z = 3, x − y + 2z = 5, 3x + 2y + z = 10

Step 1 / 7

Pick an unknown first
The coefficient of $y$ is $1$ in ① and $- 1$ in ②, so eliminating $y$ is least work. Decision made: eliminate $y$ both times, both times using ①.

$⎩ ⎨ ⎧ 2 x + y - z = 3 x - y + 2 z = 5 3 x + 2 y + z = 10$
② + ①
③ − 2 × ①
Eliminate x between ② and ③
Substitute z back
Substitute x back
Check in the original equations

Look at the structure first

Following the steps always works, but a quick look before you start can sometimes save more than half the effort.

One problem, several solutions

Solve the system $x + y = 5$ (①), $y + z = 7$ (②), $z + x = 6$ (③).

Step by step

① − ② eliminates $y$ : $x - z = - 2$ .

Add ③: $2 x = 4$ , $x = 2$ .

Substitute back: ① gives $y = 3$ and ③ gives $z = 4$ .

CostNo thinking needed, but more steps.

Add all three

Each of $x$ , $y$ and $z$ appears twice across the three equations. Add them all:

2 (x + y + z) = 18, x + y + z = 9

Subtract ①, ② and ③ from it in turn: $z = 4$ , $x = 2$ , $y = 3$ .

CostYou first have to notice that the three equations “look alike”.

For a system where the three letters play the same role (a cyclic system), finding the block $x + y + z$ first is often fastest.

When it fails

Break it

Counterexample lab: remove one condition

Conclusion: a system of three linear equations in three unknowns has exactly one solution

$x + y + z = 6$ , $2 x - y + z = 3$ , $x + 2 y - z = 2$ : the three equations are independent. Eliminate all the way and every unknown is determined.

Practice

Exercise 1Basic

Given $x + y + z = 12$ , $x - y = 1$ and $y - z = 1$ , find $y$ .

Exercise 2Basic

Solve the system $x + 2 y + z = 8$ , $2 x - y + z = 3$ , $x + y - z = 0$ . (Write $x$ , $y$ , $z$ in order, separated by commas.)

Exercise 3Advanced

Solve the system $x + y = 11$ , $y + z = 14$ , $z + x = 13$ . (Write $x$ , $y$ , $z$ in order.)

Exercise 4Challenge

Given $2 x + 3 y + 4 z = 20$ and $3 x + 4 y + 5 z = 26$ , find $x + y + z$ .

What this move does

Work through one

Example: solve 2x + y − z = 3, x − y + 2z = 5, 3x + 2y + z = 10

Look at the structure first

Step by step

Add all three

When it fails

Counterexample lab: remove one condition

Practice

Mistake Museum · Exhibits from this stop

Eliminating a different unknown each time

Using the same pair of equations twice