We will check each equation by substituting in the values of the ordered triple for [latex]x,y[/latex], and [latex]z[/latex].

[latex]\begin{array}{ccccc}\begin{array}{r}\hfill x+y+z=2\\ \hfill \left(3\right)+\left(-2\right)+\left(1\right)=2\\ \hfill \text{True}\end{array}& & \begin{array}{r}\hfill \text{}6x - 4y+5z=31\\ \hfill 6\left(3\right)-4\left(-2\right)+5\left(1\right)=31\\ \hfill 18+8+5=31\\ \hfill \text{True}\end{array}& & \begin{array}{r}\hfill \text{}5x+2y+2z=13\\ \hfill 5\left(3\right)+2\left(-2\right)+2\left(1\right)=13\\ \hfill \text{}15 - 4+2=13\\ \hfill \text{True}\end{array}\end{array}[/latex]

The ordered triple [latex]\left(3,-2,1\right)[/latex] is indeed a solution to the system. Note that if any of the three equations was not true, it would not be a solution to the system.

It is good practice to label the equations as we go because we will need to reference them later. The key observation is that equations [latex](2)[/latex] and [latex](3)[/latex] only have the variables [latex]y[/latex] and [latex]z,[/latex] so we can solve them as a two variable system and find the solution for [latex]y[/latex] and [latex]z[/latex] immediately.

[latex]\left\{ \begin{array}{rl}-2y+z&=\text{ }6\\ 2y-2z&=\text{ }-12\end{array}\right.[/latex]

Adding the equations right away will eliminate [latex]y[/latex] and allow us to find the value of [latex]z[/latex].

[latex]\dfrac{\begin{array}{rl}\hfill \\ \:\:-2y+z&=6\hfill \\ 2y-2z&=-12\hfill \end{array}}{\text{}\text{}\text{}\text{}\text{}\:\:\:\:\:\:-z=-6}[/latex]

[latex]z=6[/latex]

Now we can substitute the value for [latex]z[/latex] that we obtained into equation [latex](2)[/latex].

[latex]\begin{array}{rl}-2y+(6)&=\;6\\-2y&=\;6-6\\-2y&=\;0\\\,\,\,\,y&=\;0\end{array}[/latex]

Be careful here not to get confused with a solution of [latex]y = 0[/latex] and an inconsistent solution. It is ok for variables to equal [latex]0.[/latex]

Now we can substitute [latex]z = 6[/latex] and [latex]y = 0[/latex] back into equation [latex](1)[/latex].

[latex]\begin{array}{rl}x-y+z&=\;5\\x-0+6&=\;5\\x+6&=\;5\\x&=\;5-6\\x&=\;-1\end{array}[/latex]

The solution is the ordered triple [latex](x,y,z)=(-1,0,6).[/latex] We leave it to you to check that this solution satisfies all three equations.

First we should identify which variable we plan to eliminate in Steps 1 and 2. Notice that the coefficients for [latex]z[/latex] are all either [latex]4[/latex] or [latex]-4.[/latex] This suggests eliminating [latex]z[/latex] in two pairs of equations.

[latex]\dfrac{\begin{array}{rll}2x+\phantom{4}y+4z=3&(1) \\ -x + 3y-4z=6&(2) \end{array}}{\phantom{2}x+4y\phantom{-24z}=9 \:\:\:\:(4)}[/latex]

[latex]\dfrac{\begin{array}{rll}-x+3y-4z=6&(2) \\ 3x -5y+4z=2&(3) \end{array}}{2x -2y\phantom{-24z}=8 \:\:\:\:(5)}[/latex]

Equations [latex](4)[/latex] and [latex](5)[/latex] form our two variable system,

[latex]\left\{ \begin{array}{rll}x+4y&=\:9&(4)\\ 2x-2y&=\:8&(5)\end{array}\right.[/latex]

We now solve this two variable system by elimination.

[latex]\begin{array}{rllrl}x+4y&=9&\xrightarrow{\cdot \; -2} & -2x-8y &=-18\\ 2x-2y&=8&\xrightarrow{\phantom{\cdot \; -2}} & 2x-2y&=8 \\ & & & \text{_______}&\text{______} \\ & & & -10y&=-10 \\ & & & y &=1 \\ \end{array}[/latex]

[latex]\begin{array}{rll}x+4y&=9&(4)\\ x+4(1)&=9& \\ x&=5 & \end{array}[/latex]

Now, substitute [latex]x=5, y=1[/latex] into equation [latex](1).[/latex]

[latex]\begin{array}{rll}2x+y+4z&=3&(1)\\ 2(5)+(1)+4z&=3&\\ 11+4z&=3&\\ 4z&=-8&\\ z&=-2 & \end{array}[/latex]

The solution is the ordered triple [latex](x,y,z)=(5,1,-2).[/latex] This solution can be checked and does satisfy all three equations.

Since [latex]x[/latex] is already eliminated from equations [latex](2)[/latex] and [latex](3),[/latex] solve them as a two variable system:

[latex]\begin{array}{rllrl}-y-4z&=\:7&\xrightarrow{\cdot \; 2} & -2y-8z &=14\\ 2y+8z&=\:-12&\xrightarrow{\phantom{\cdot \; 2}} & 2y+8z&=-12 \\ & & & \text{_______}&\text{______} \\ & & & 0 &=2 \\ \end{array}[/latex]

Recall that when we were solving systems with two variables, a contradiction such as [latex]0=2[/latex] implied that there was no solution to the system. The same is true for three variable systems.

This system has no solution. We can say the system is inconsistent.

Since the first equation is already missing [latex]x,[/latex] we use equations [latex](2)[/latex] and [latex](3),[/latex] to eliminate [latex]x[/latex] again.

[latex]\begin{array}{rllrl} -10x+2y-6z&=\;-6&\xrightarrow{\phantom{\cdot \; -2}} & -10x+2y-6z &=\;-6\\ -\phantom{5}5x-3y+\phantom{4}z&=\:-8&\xrightarrow{\cdot \; -2} & 10x+6y-2z&=\;16 \\ & & & \text{_______}&\text{______} \\ & & & 8y-8z&=\;10 \ \ \ \ (4) \\ \end{array}[/latex]

We now solve the system consisting of equations [latex](1)[/latex] and [latex](4)[/latex] by eliminating [latex]y.[/latex]

[latex]\begin{array}{rllrl} -4y+4z&=\;-5&\xrightarrow{\cdot \; 2} & -8y+8z&=\;-10\\ 8y-8z&=\;10&\xrightarrow{\phantom{\cdot \; 2}} & 8y-8z&=\;10 \\ & & & \text{_______}&\text{______} \\ & & & 0&=\;0 \\ \end{array}[/latex]

Since we reach an identity, the system is dependent and the solution set is infinite. All points along the intersection line will satisfy all three equations. In this book we are not going to describe the solution set in any more detail and will simply say there are “Infinite Solutions”.