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Section 7-2 : Linear Systems with Three Variables

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

\[\begin{align*}3x - 9z & = 33\\ 7x - 4y - z & = - 15\\ 4x + 6y + 5z & = - 6\end{align*}\]

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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 we can easily solve the first equation for \(x\) and get an equation involving only \(z\) which we can in turn plug in the second and third equation.

Here is the first equation solved for \(x\).

\[\begin{align*}3x - 9z & = 33\\ 3x & = 9z + 33\hspace{0.25in} \to \hspace{0.25in}x = 3z + 11\end{align*}\] Show Step 2

Plugging the equation we found above into the second and third equations and doing some simplification gives,

\[\begin{array}{*{20}{c}}{7\left( {3z + 11} \right) - 4y - z = - 15}\\{4\left( {3z + 11} \right) + 6y + 5z = - 6}\end{array}\hspace{0.25in} \to \hspace{0.25in}\begin{array}{*{20}{c}}{ - 4y + 20z = - 92}\\{6y + 17z = - 50}\end{array}\] Show Step 3

Now, notice that if we multiply the first equation above by 3 and the second equation above by 2 we can cancel the \(y\)’s when we add the results. Here is that work.

\[\begin{align*} -4y+20z & =-92 & \underrightarrow{\times \,\,3} & & -12y+60z=-276 \\ 6y+17z & =-50 & \underrightarrow{\times \,\,2} & & \underline{12y+34z=-100} \\ & & & & 94z=-376 \end{align*}\] Show Step 4

From the equation above we can see that we must have \(z = - 4\).

Show Step 5

We can plug \(z = - 4\) into either of the equations we got in Step 2 and solve for \(y\). We’ll use the second equation for this propose.

\[6y + 17\left( { - 4} \right) = - 50\hspace{0.25in} \to \hspace{0.25in}y = 3\] Show Step 6

Finally, plug \(z = - 4\) we got in Step 1 to determine the value of \(x\).

\[x = 3\left( { - 4} \right) + 11 = - 1\]

The solution to the system is then : \(\require{bbox} \bbox[2pt,border:1px solid black]{{x = - 1,\,\,y = 3,\,\,z = - 4}}\).