Answer
The solution of the system is $\left( 0.5,3,-2 \right)$.
Work Step by Step
It is given that:
$7z-3=2\left( x-3y \right)$ ...... (I)
$5y-3z-7=4x$ ...... (II)
$4+5z=3\left( 2x-y \right)$ ...... (III)
From equation (I) $7z-3=2\left( x-3y \right)$:
$2x-6y-7z=-3$ ...... (IV)
From equation (II) $5y-3z-7=4x$:
$4x-5y-3z=-7$ ...... (V)
From equation (III) $4+5z=3\left( 2x-y \right)$:
$6x-3y-5z=4$ ...... (VI)
Now, multiply the equation (IV) by 2 and subtract from equation (V):
$\begin{align}
& 4x-12y-14z=-6 \\
& 4x-5y-3z=-7 \\
\end{align}$
Solve and we get:
$-7y-11z=1$ ...... (VII)
Multiply equation (IV) by 3 and subtract from equation (VI):
$\begin{align}
& 6x-18y-21z=-9 \\
& 6x-3y-5z=4 \\
\end{align}$
Solve and we get:
$-15y-16z=-13$ ...... (VIII)
From equation (VIII) and, putting the value of z in equation (VII):
$\begin{align}
& -7y-11\left( \frac{13-15y}{16} \right)=1 \\
& -7y-\frac{143}{16}+\frac{165y}{16}=1 \\
& -112y-143+165y=16 \\
& 53y=159
\end{align}$
So,
$\begin{align}
& y=\frac{159}{53} \\
& y=3 \\
\end{align}$
Putting the value of $y=3$ in equation (VII):
$\begin{align}
& -7\left( 3 \right)-11z=1 \\
& -21-11z=1 \\
& z=-2
\end{align}$
Putting the values of $y=3$ and $z=-2$ in equation (I):
$\begin{align}
& 2x-6\left( 3 \right)-7\left( -2 \right)=-3 \\
& x=0.5
\end{align}$
Thus, the ordered triple $\left( 0.5,3,-2 \right)$ satisfies the three systems.
