Example

Solve the following system.

$\begin{array}{ll}\text{ }x - 3y+z=4\,\,\,\,\,\,\,\,\,\,\,\,\left(1\right)\\ \,\,\,\,\,\,-y-4z=7\,\,\,\,\,\,\,\,\,\,\,\,\,\left(2\right)\,\,\,\,\\\,\,\,\,\,\,\,\,2y+8z=-12\,\,\,\,\,\,\,(3)\end{array}$

Answer: If you multiply equation $(2)$ by $2$ and add $(2)$ and $(3)$ together, you can eliminate y and solve for $z$. First, multiply both sides of equation $(2)$ by $2$.

[latex]\begin{array}\,\,\,\,\,\,2(-y-4z)=2(7)\\-2y-8z=14\end{array}[/latex]

Next, add equations [latex](2)[/latex] and [latex](3)[/latex] together to eliminate y and solve for [latex]z[/latex].

[latex]\begin{array}\,\,\,\,\,\,-2y-8z=14\\\underline{2y+8z=-12}\\0+0=2\\0=2\end{array}[/latex]

Recall that when we were solving systems with two variables, a non-true solution such as [latex]0=2[/latex] implied that there was no solution to the system.

This can happen with a system of three variables as well.

This system has no solution.

Example

Find the solution to the given system of three equations in three variables.

[latex]\begin{array}{rr}\hfill \text{ }2x+y - 3z=0& \hfill \left(1\right)\\ \hfill 4x+2y - 6z=0& \hfill \left(2\right)\\ \hfill \text{ }x-y+z=0& \hfill \left(3\right)\end{array}[/latex]

Answer: First, we can multiply equation [latex](1)[/latex] by [latex]-2[/latex] and add it to equation [latex](2)[/latex].

[latex]\begin{array}\,-4x−2y+6z=0 \hfill& \text{equation }\left(1\right)\text{multiplied by }−2 \\ \,4x+2y−6z=0\hfill&\left(2\right) \end{array}[/latex]

We do not need to proceed any further. The result we get when adding the two equations is an identity, [latex]0=0[/latex] which tells us that this system has an infinite number of solutions. As shown in the figure below, two of the planes are the same and they intersect the third plane on a line. The solution set is infinite, as all points along the intersection line will satisfy all three equations.