Answer
The solution of the system is $\left( \frac{1}{2},\frac{1}{3},-1 \right)$.
Work Step by Step
It is known that:
$3\left( 2x+y \right)+5z=-1$ ...... (I)
$2\left( x-3y+4z \right)=-9$ ...... (II)
$4\left( 1+x \right)=-3\left( z-3y \right)$ ...... (III)
From equation (I):
$3\left( 2x+y \right)+5z=-1$
$6x+3y+5z=-1$ ...... (IV)
From equation (II):
$2\left( x-3y+4z \right)=-9$
$2x-6y+8z=-9$ ...... (V)
From equation (III):
$\begin{align}
& 4\left( 1+x \right)=-3\left( z-3y \right) \\
& 4+4x=-3z+9y
\end{align}$
$4x-9y+3z=-4$ ...... (VI)
Multiply equation (V) by 3 and subtract from equation (IV):
$\begin{align}
& 6x+3y+5z=-1 \\
& 6x-18y+24z=-27 \\
\end{align}$
$21y-192=26$ ...... (VII)
Now multiply equation (V) by 7 and subtract from equation (VII):
$\begin{align}
& 21y-19z=26 \\
& 21y-91z=98 \\
\end{align}$
$3y-13z=14$ ...... (VIII)
Multiply equation (VIII) by 7 and subtract from equation (VII):
$\begin{align}
& 21y-19z=26 \\
& 21y-91z=98 \\
\end{align}$
Solve and get:
$\begin{align}
& 72z=-72 \\
& z=-1
\end{align}$
Put the value of z in equation (VII)
$\begin{align}
& 21y+19=26 \\
& 21y=7 \\
& y=\frac{1}{3}
\end{align}$
Put the value of $y=\frac{1}{3}$ and $z=-1$ in equation (IV):
$\begin{align}
& 6x+3\left( \frac{1}{3} \right)-5=-1 \\
& x=\frac{1}{2}
\end{align}$
Hence, the ordered triple $\left( \frac{1}{2},\frac{1}{3},-1 \right)$ satisfies the three systems.
