2.16 - Linear equations in three unknowns

One triple must satisfy every equation

Can three different-looking equations still fail to determine three numbers? Consider $x+y+z=6$, $2x+2y+2z=12$ and $3x+3y+3z=18$. There are three equations on the page, but the last two simply repeat the first at a different scale. Both $(1,2,3)$ and $(0,0,6)$ work, and infinitely many other triples do too.

The surprise is that counting equations is not enough: they must provide new information. In the systems you will solve here, that information lets you pin down one common triple. The method is an extension of solving two simultaneous equations, rather than a completely new piece of algebra.

With three unknowns, the answer contains three values. Their labels matter: swapping the values of $x$ and $z$ usually gives a different proposed solution.

For instance, $(x,y,z)=(2,1,3)$ satisfies $x+y+z=6$ and $2x-y+z=6$. It also satisfies $x+2y-z=1$. Checking only the first equation would leave many wrong triples undetected.

The following check focuses on what “simultaneous” requires, before any calculation becomes lengthy.

Eliminating the same unknown twice

Choose one unknown to remove from two different pairs of equations, leaving two equations in the same remaining unknowns.

Label the original equations so that each subtraction can be followed. If you remove $x$ once but $y$ next, your new equations may still contain all three unknowns between them. Remove the same unknown twice instead. Multiplying and combining equations is not a procedure supplied on the formulae sheet: learn the method.

Worked example 2 changes only the right side of the third equation, from $1$ to $8$. The elimination pattern stays the same, but the solution changes.

Subtraction affects every term and the constant on the other side. The labels prevent an innocent-looking sign change from becoming hard to trace.