Answer
The solution of the system is $\left( -1,-2,3 \right)$.
Work Step by Step
Now, it is given that
$2x+3y+7z=13$ ...... (I)
$3x+2y-5z=-2z$ ...... (II)
$5x+7y-3z=-28$ ...... (III)
Multiply equation (I) by 3 and equation (II) by 2 and subtract:
$\begin{align}
& 6x+9y+21z=39 \\
& 6x+4y-10z=-44 \\
\end{align}$
Solve and we get:
$5y+31z=83$ ...... (IV)
Now, multiply equation (II) by 5 and equation (III) by 3 and subtract:
$\begin{align}
& 15x+10y-25z=-110 \\
& 15x+21y-9z=-84 \\
& -11y-16z=-26
\end{align}$
Solve and we get:
$11y+16z=26$ ...... (V)
Multiply equation (IV) by 11 and equation (V) by 5 and subtract:
$\begin{align}
& 55y+341z=913 \\
& 55y+60z=130 \\
& 261z=783
\end{align}$
Solve and we get:
$z=3$
Putting the value of $z=3$ in equation (V):
$\begin{align}
& 11y+16\left( 3 \right)=26 \\
& 11y+48=26 \\
& 11y=-22 \\
& y=-2
\end{align}$
Putting the value of $z=3$ and $y=-2$ in equation (I):
$\begin{align}
& 2x-6+21=13 \\
& 2x=13-15 \\
& x=-1
\end{align}$
Thus, the ordered triple $\left( -1,-2,3 \right)$ satisfies the three systems.
