Answer
$x=-\dfrac{1}{2}$; $y=1$; $z=1$;
Work Step by Step
First of all we clear denominators by multiplying the first equation by $6$, the second by $12$ and the third by $6$:
We have to solve the system:
$$\begin{align*}
\begin{cases}
2x+5y+4z&=8\quad&\text{Equation }1\\
2x+8y+3z&=10\quad&\text{Equation }2\\
4x+y+9z&=8\quad&\text{Equation }3.
\end{cases}
\end{align*}$$
We rewrite the system as a linear system in $\textit{two}$ variables:
$$\begin{align*}
-2x-5y-4z&=-8\quad\quad\text{Add }-1\text{ times Equation }1\\
2x+8y+3z&=10\quad\quad\text{to Equation }2.\\
\text{___________}&\text{______}\\
3y-z&=2\quad\quad\quad\text{New Equation }1.
\end{align*}$$
$$\begin{align*}
-4x-10y-8z&=-16\quad\text{Add }-2\text{ times Equation }1\\
4x+y+9z&=8\quad\quad\text{to Equation }3.\\
\text{___________}&\text{______}\\
-9y+z&=-8\quad\quad\text{New Equation }2.
\end{align*}$$
We solve the new linear system for both its variables:
$$\begin{align*}
3y-z&=2\quad\quad\text{Add new Equation }1\\
-9y+z&=-8\quad\text{to new Equation }2.\\
\text{___________}&\text{______}\\
-6y&=-6\\
y&=1\quad\text{Solve for }y.\\
z&=1\quad\text{Substitute into new Equation }1\text{ to find }z.
\end{align*}$$
Substitute $y=1$ and $z=1$ into an original equation and solve for $x$:
$$\begin{align*}
2x+5y+4z&=8\quad\text{Write original Equation }1.\\
2x+5(1)+4(1)&=8\quad\text{Substitute }y=1\text{ and }z=1.\\
x&=-\dfrac{1}{2}\quad\text{Solve for }x.\\
\end{align*}$$
The solution is $x=-\dfrac{1}{2}, y=1,z=1$ or the ordered triple $\left(-\dfrac{1}{2},1,1\right)$.
