Looking at the coefficients of [latex]x[/latex], we can see that we can eliminate [latex]x[/latex] by adding equation (1) to equation (2).

[latex]\frac{\begin{array}{l}\text{ }x - 3y+z=4\text{ }\left(1\right)\hfill \\ -x+2y - 5z=3\text{ }\left(2\right)\hfill \end{array}}{\text{ }-y - 4z=7\text{ }\left(4\right)}[/latex]

Next, we multiply equation (1) by [latex]-5[/latex] and add it to equation (3).

[latex]\begin{array}−5x+15y−5z=−20 \hfill& \left(1\right)\text{ multiplied by }−5 \\ 5x−13y+13z=8 \hfill& \left(3\right) \\ \text{___________________} \\ 2y+8z=−12\hfill&\left(5\right) \end{array}[/latex]

Then, we multiply equation (4) by 2 and add it to equation (5).

[latex]\begin{array}−2y−8z=14 \hfill& \left(4\right)\text{ multiplied by }2 \\ 2y+8z=−12 \hfill& \left(5\right) \\ \text{___________________} \\ 0=2\end{array}[/latex]

The final equation [latex]0=2[/latex] is a contradiction, so we conclude that the system of equations in inconsistent and, therefore, has no solution.

Analysis of the Solution

In this system, each plane intersects the other two, but not at the same location. Therefore, the system is inconsistent.

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

[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 is an identity, [latex]0=0[/latex], which tells us that this system has an infinite number of solutions. There are other ways to begin to solve this system, such as multiplying equation (3) by [latex]-2[/latex], and adding it to equation (1). We then perform the same steps as above and find the same result, [latex]0=0[/latex].

When a system is dependent, we can find general expressions for the solutions. Adding equations (1) and (3), we have

[latex]\begin{array}2x+y−3z=0 \\ x−y+z=0 \\ \text{______________} \\ 3x−2z=0 \end{array}[/latex]

We then solve the resulting equation for [latex]z[/latex].

[latex]\begin{array}{l}3x - 2z=0\hfill \\ \text{ }z=\frac{3}{2}x\hfill \end{array}[/latex]

We back-substitute the expression for [latex]z[/latex] into one of the equations and solve for [latex]y[/latex].

[latex]\begin{array}{l}2x+y - 3\left(\frac{3}{2}x\right)=0\hfill \\ \text{ }2x+y-\frac{9}{2}x=0\hfill \\ \text{ }y=\frac{9}{2}x - 2x\hfill \\ \text{ }y=\frac{5}{2}x\hfill \end{array}[/latex]

So the general solution is [latex]\left(x,\frac{5}{2}x,\frac{3}{2}x\right)[/latex]. In this solution, [latex]x[/latex] can be any real number. The values of [latex]y[/latex] and [latex]z[/latex] are dependent on the value selected for [latex]x[/latex].

Analysis of the Solution

As shown 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.