Answer
$x=2$; $y=1$; $z=0$;
Work Step by Step
First of all we clear denominators by multiplying the first equation by $2$, the second by $4$ and the third by $6$:
We have to solve the system:
$$\begin{align*}
\begin{cases}
2x+y+z&=5\quad&\text{Equation }1\\
3x+y+6z&=7\quad&\text{Equation }2\\
2x+9y+4z&=13\quad&\text{Equation }3.
\end{cases}
\end{align*}$$
We rewrite the system as a linear system in $\textit{two}$ variables:
$$\begin{align*}
-2x-y-z&=-5\quad\text{Add }-1\text{ times Equation }1\\
3x+y+6z&=7\quad\quad\text{to Equation }2.\\
\text{___________}&\text{______}\\
x+5z&=2\quad\quad\text{New Equation }1.
\end{align*}$$
$$\begin{align*}
-18x-9y-9z&=-45\quad\text{Add }-9\text{ times Equation }1\\
2x+9y+4z&=13\quad\quad\text{to Equation }3.\\
\text{___________}&\text{______}\\
-16x-5z&=32\quad\text{New Equation }2.
\end{align*}$$
We solve the new linear system for both its variables:
$$\begin{align*}
x+5z&=2\quad\quad\text{Add new Equation }1\\
-16x-5z&=-32\quad\text{to new Equation }2.\\
\text{___________}&\text{______}\\
-15x&=-30\\
x&=2\quad\text{Solve for }x.\\
z&=0\quad\text{Substitute into new Equation }1\text{ to find }z.
\end{align*}$$
Substitute $x=2$ and $z=0$ into an original equation and solve for $y$:
$$\begin{align*}
2x+y+z&=5\quad\text{Write original Equation }1.\\
2(2)+y+0&=5\quad\text{Substitute }x=2\text{ and }z=0.\\
y&=1\quad\text{Solve for }y.\\
\end{align*}$$
The solution is $x=2, y=1,z=0$ or the ordered triple $(2,1,0)$.
