Answer
The required values of x, y and z are:
$ x=-\frac{9}{a}$ ; $ y=\frac{5}{b}$ ; $ z=\frac{5}{c}$.
Work Step by Step
Mark the given system of equations as shown below:
$ ax-by+2cz=-4$ (I)
$ ax+3by-cz=1$ (II)
$2ax+by+3cz=2$ (III)
Add equations (I) and (II) in order to eliminate a,
$\begin{align}
& \text{ }ax-\text{ }by+2cz=-4 \\
& \underline{-ax-3by+\text{ }cz=-\text{ }1} \\
& \text{ }-4by+3cz=-5\text{ }\left( \text{IV} \right) \\
\end{align}$
By multiplying equation (II) by -2 and adding to eliminate a from equations (I) and (III), we get
$\begin{align}
& \text{ 2}ax+\text{ }by+3cz=2 \\
& \underline{-2ax-6by+2cz=2} \\
& \text{ }-5by\text{ +5}cz=0\text{ }\left( \text{V} \right) \\
\end{align}$
By multiplying equation (IV) by 5, equation (V) by 4, and adding equations (IV) and (V), we get the value of z,
$\begin{align}
& -20by+15cz=-\text{25} \\
& \underline{-20by-20cz=\text{ }0} \\
& \text{ }-5cz=-25 \\
\end{align}$
Divide both sides by -5c,
$\begin{align}
& \frac{-5cz}{-5c}=\frac{-25}{-5c} \\
& \text{ }z=\frac{5}{c} \\
\end{align}$
Substitute the value of z in equation (IV) to get the value of y,
$\begin{align}
& -4by+3c\left( \frac{5}{c} \right)=-5 \\
& -4by+15=-5
\end{align}$
Now, subtract 15 from both sides to get,
$\begin{align}
& -4by+15-15=-5-15 \\
& -4by=-20
\end{align}$
By dividing both sides by -4b to get,
$\begin{align}
& \frac{-4by}{-4b}=\frac{-20}{-4b} \\
& \text{ }y=\frac{5}{b} \\
\end{align}$
Substitute the values of y and z in equation (I) to get the value of x,
$\begin{align}
& ax-b\left( \frac{5}{b} \right)+2c\left( \frac{5}{c} \right)=-4 \\
& ax-5+10=-4 \\
& ax+5=-4
\end{align}$
Now subtract 5 from both sides, we get,
$\begin{align}
& ax+5-5=-4-5 \\
& ax=-9
\end{align}$
Now, dividing both sides by a to get,
$\begin{align}
& \frac{ax}{a}=-\frac{9}{a} \\
& \text{ }x=-\frac{9}{a} \\
\end{align}$
Hence, the values are $ x=-\frac{9}{a}$, $ y=\frac{5}{b}$, and $ z=\frac{5}{c}$.
