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Section 7-2 : Linear Systems with Three Variables

1. Find the solution to the following system of equations.

\[\begin{align*}2x + 5y + 2z & = - 38\\ 3x - 2y + 4z & = 17\\ - 6x + y - 7z & = - 12\end{align*}\]

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Start Solution

Before we get started with the solution process for this system we need to make it clear that there is no “one correct solution path”. There are lots of solution paths that we can take to find the solution to this system. All are correct and all will end up with the same solution to the system (provided the work has been done correctly of course…).

Okay, let’s get started on the solution to this system.

For this system it looks like if we multiply the first equation by 3 and the second equation by 2 both of these equations will have \(x\) coefficients of 6 which we can then eliminate if we add the third equation to each of them.

So, let’s first do the multiplication.

\[\begin{align*} 2x+5y+2z &=-38 & \underrightarrow{\times \,\,3} & \hspace{0.25in} & 6x+15y+6z & =-114 \\ 3x-2y+4z & =17 & \underrightarrow{\times \,\,2} & \hspace{0.25in} & 6x-4y+8z & =34 \\ -6x+y-7z & =-12 & \underrightarrow{\text{same}} & \hspace{0.25in} & -6x+y-7z & =-12 \\ \end{align*}\] Show Step 2

Okay, we’ll now replace the first equation with the sum of the first and third equation and we’ll replace the second equation with the sum of the second and third equation. Here is the result from doing those operations.

\[\begin{align*}16y - z & = - 126\\ - 3y + z & = 22\\ - 6x + y - 7z & = - 12\end{align*}\] Show Step 3

Next notice that we can eliminate \(z\) from the first equation simply by replacing it with the sum of the first and second equation. Here is the result from that operation.

\[\begin{align*}13y\,\,\,\,\,\,\,\,\, & = - 104\\ - 3y + z & = 22\\ - 6x + y - 7z & = - 12\end{align*}\] Show Step 4

Okay, from the first equation we can see that we must have \(y = - 8\).

Show Step 5

We can plug \(y = - 8\) into the second equation and solve that for \(z\).

\[ - 3\left( { - 8} \right) + z = 22\hspace{0.25in} \to \hspace{0.25in}z = - 2\] Show Step 6

Finally, plug \(y = - 8\) and \(z = - 2\) into the third equation and solve for \(x\).

\[\begin{align*} - 6x + \left( { - 8} \right) - 7\left( { - 2} \right) & = - 12\\ - 6x + 6 & = - 12\hspace{0.25in} \to \hspace{0.25in}x = 3\end{align*}\]

The solution to the system is then : \(\require{bbox} \bbox[2pt,border:1px solid black]{{x = 3,\,\,y = - 8,\,\,z = - 2}}\).