Find the intersection or the solution to the following system of equations: [latex]3x+2y-z=-1, -2x-2y+3z=5,[/latex] and [latex]5x+2y-z=3.[/latex]

As we did with a set of two equations, first line up the equations to choose the variable that we wish to eliminate:

[latex]\left\{ \begin{array}{rrrrrrr} 3x&+&2y&-&z&=&-1 \\ -2x&-&2y&+&3z&=&5 \\ 5x&+&2y&-&z&=&3 \right. \end{array}[/latex]

For these equations, it looks easiest to eliminate the [latex]y[/latex]-variable. To do this, add the first and second equations together and then add the second and third equations together:

[latex]\begin{array}{rr} \begin{array}{rrrrrrrr} &3x&+&2y&-&z&=&-1 \\ +&-2x&-&2y&+&3z&=&5 \\ \hline &&&x&+&2z&=&4 \end{array} &\hspace{0.25in} \begin{array}{rrrrrrrr} &-2x&-&2y&+&3z&=&5 \\ +&5x&+&2y&-&z&=&3 \\ \hline &&&3x&+&2z&=&8 \end{array} \end{array}[/latex]

Now, you are left with [latex]x+2z=4[/latex] and [latex]3x + 2z = 8.[/latex] We now solve these as done previously with a set of two equations:

[latex]\left\{ \begin{array}{rrrrr} x&+&2z&=&4 \\ 3x&+&2z&=&8 \right. \end{array}[/latex]

Multiply either the top or the bottom equation by −1 to eliminate the [latex]z[/latex]-variable.

[latex]\begin{array}{rrrrrrr} &(x&+&2z&=&4)&(-1) \\ &3x&+&2z&=&8& \\ \\ &-x&-&2z&=&-4& \\ +&3x&+&2z&=&8& \\ \hline &&&\dfrac{2x}{2}&=&\dfrac{4}{2}& \\ \\ &&&x&=&2& \end{array}[/latex]

Next, find [latex]z[/latex] using one of [latex]x+2z=4[/latex] or [latex]3x+2z=8[/latex] and the solution [latex]x= 2.[/latex] [latex]x+2z=4[/latex] looks to be the easiest to work with.

[latex]\begin{array}{rrrrr} x&+&2z&=&4 \\ 2&+&2z&=&4 \\ -2&&&&-2 \\ \hline &&\dfrac{2z}{2}&=&\dfrac{2}{2} \\ \\ &&z&=&1 \end{array}[/latex]

Finally,  find [latex]y[/latex] using one of the original three equations:

[latex]\begin{array}{rrrrrrr} 3x&+&2y&-&z&=&-1 \\ 3(2)&+&2y&-&(1)&=&-1 \\ 6&+&2y&-&1&=&-1 \\ &&5&+&2y&=&-1 \\ &&-5&&&=&-5 \\ \hline &&&&\dfrac{2y}{2}&=&\dfrac{-6}{2} \\ \\ &&&&y&=&-3 \end{array}[/latex]

These planes intersect at the point [latex]x = 2,[/latex] [latex]y = -3,[/latex] and [latex]z = 1[/latex], or the coordinate [latex](2, -3, 1).[/latex]

Find the intersection or the solution to the following system of equations: [latex]x+2y-z=0, 3x-2y=-2,[/latex] and [latex]y+z=3.[/latex]

First, line up the equations to choose the variable that we wish to eliminate:

[latex]\left\{ \begin{array}{rrrrrrr} x&+&2y&-&z&=&0 \\ 3x&-&2y&&&=&-2 \\ &&y&+&z&=&3 \right. \end{array}[/latex]

In this example, adding the first and last equations eliminates the variable [latex]z,[/latex] without having to modify any of the equations:

[latex]\begin{array}{rrrrrrrr} &x&+&2y&-&z&=&0 \\ +&&&y&+&z&=&3 \\ \hline &&&x&+&3y&=&3 \\ \end{array}[/latex]

Now, there are two equations left:

[latex]\left\{ \begin{array}{rrrrr} 3x&-&2y&=&-2 \\ x&+&3y&=&3 \right. \end{array}[/latex]

First multiply the bottom equation by −3, then add it to the top equation, to eliminate the variable [latex]x[/latex]:

[latex]\begin{array}{rrrrrrr} &(x&+&3y&=&3)&(-3) \\ \\ &3x&-&2y&=&-2& \\ +&-3x&-&9y&=&-9& \\ \hline &&&-11y&=&-11& \\ &&&y&=&1& \\ \end{array}[/latex]

Now choose one of the two remaining equations, [latex]3x-2y=-2[/latex] or [latex]x+3y=3,[/latex] to find the variable [latex]x.[/latex] Choosing [latex]x+3y= 3,[/latex] leaves:

[latex]\begin{array}{rrrrr} x&+&3(1)&=&3 \\ x&+&3&=&3 \\ &-&3&=&-3 \\ \hline &&x&=&0 \end{array}[/latex]

Finally, to find the third variable, use one of the original three equations: [latex]x+2y-z=0, 3x-2y=-2,[/latex]  or [latex]y+z=3.[/latex] Choosing [latex]y + z = 3,[/latex] gives:

[latex]\begin{array}{rrrrr} (1)&+&z&=&3 \\ &&z&=&2 \end{array}[/latex]

These planes intersect at the point [latex]x = 0, y = 1,[/latex] and [latex]z = 2,[/latex] or the coordinate [latex](0, 1, 2).[/latex]