Answer
$\left(-\dfrac{9}{a},\dfrac{5}{b},\dfrac{5}{c}\right)$
Work Step by Step
We are given the system:
$\begin{cases}
ax-by+2cz=-4\\
ax+3by-cz=1\\
2ax+by+3cz=2
\end{cases}$
We will use the addition method. Multiply Equation 1 by 3 and add it to Equation 2 to eliminate $y$. Add Equation 1 to Equation 3 to eliminate $y$:
$\begin{cases}
ax+3by-cz+3(ax-by+2cz)=1+3(-4)\\
2ax+by+3cz+ax-by+2cz=2+(-4)
\end{cases}$
$\begin{cases}
ax+3by-cz+3ax-3by+6cz=1-12\\
3ax+5cz=-2
\end{cases}$
$\begin{cases}
4ax+5cz=-11\\
3ax+5cz=-2
\end{cases}$
Multiply Equation 2 by -1 and add it to Equation 1 to eliminate $z$ and determine $x$:
$\begin{cases}
4ax+5cz=-11\\
-3ax-5cz=-(-2)
\end{cases}$
$4ax+5cz-3ax-5cz=-11+2$
$ax=-9$
$x=-\dfrac{9}{a}$
Substitute the value of $x$ in the Equation $3ax+5cz=-2$ to determine $z$:
$3a\left(-\dfrac{9}{a}\right)+5cz=-2$
$-27+5cz=-2$
$5cz=25$
$z=-\dfrac{25}{5c}$
$z=\dfrac{5}{c}$
Substitute the values of $x, z$ is Equation 1 of the given system to find $y$:
$ax-by+2cz=-4$
$a\left(-\dfrac{9}{a}\right)-by+2c\left(\dfrac{5}{c}\right)=-4$
$-9-by+10=-4$
$by=5$
$y=\dfrac{5}{b}$
The system's solution is:
$\left(-\dfrac{9}{a},\dfrac{5}{b},\dfrac{5}{c}\right)$
