Answer
$\left(\dfrac{8}{a},-\dfrac{3}{b},-\dfrac{5}{c}\right)$
Work Step by Step
We are given the system:
$\begin{cases}
ax-by-2cz=21\\
ax+by+cz=0\\
2ax-by+cz=14
\end{cases}$
We will use the addition method. Add Equation 2 to Equation 1 and Equation 3 to eliminate $y$:
$\begin{cases}
ax-by-2cz+ax+by+cz=21+0\\
2ax-by+cz+ax+by+cz=14+0
\end{cases}$
$\begin{cases}
2ax-cz=21\\
3ax+2cz=14
\end{cases}$
Multiply Equation 1 by 2 and add it to Equation 2 to eliminate $z$ and determine $x$:
$\begin{cases}
2(2ax-cz)=2(21)\\
3ax+2cz=14
\end{cases}$
$\begin{cases}
4ax-2cz=42\\
3ax+2cz=14
\end{cases}$
$4ax-2cz+3ax+2cz=42+14$
$7ax=56$
$x=\dfrac{56}{7a}$
$x=\dfrac{8}{a}$
Substitute the value of $x$ in the Equation $2ax-cz=21$ to determine $z$:
$2a\left(\dfrac{8}{a}\right)-cz=21$
$16-cz=21$
$cz=-5$
$z=-\dfrac{5}{c}$
Substitute the values of $x, z$ is Equation 2 of the given system to find $y$:
$ax+by+cz=0$
$a\left(\dfrac{8}{a}\right)+by+c\left(-\dfrac{5}{c}\right)=0$
$8+by-5=0$
$by=-3$
$y=-\dfrac{3}{b}$
The system's solution is:
$\left(\dfrac{8}{a},-\dfrac{3}{b},-\dfrac{5}{c}\right)$
