Answer
\[x=0,\ y=2\]
Work Step by Step
The equation \[5x=4y-8\]can be written as \[5x-4y=-8\].
Multiply \[7\]on both sides to the equation \[5x-4y=-8\]to get: \[35x-28y=-56\].
Multiply \[4\]on both sides to the equation \[3x+7y=14\]to get: \[12x+28y=56\].
Add the above obtained equation from both RHS and LHS as follows:
\[\underline{\begin{align}
& 35x-28y=-56 \\
& 12x+28y=56
\end{align}}\]
\[\begin{align}
& 47x\ \ \ \ \ \ \ \ \ \ =0 \\
& x=0
\end{align}\]
Put \[x=0\]in\[5x-4y=-8\] to get:
\[\begin{align}
& 5\left( 0 \right)-4y=-8 \\
& -4y=-8 \\
& y=2
\end{align}\]
Put\[x=0\]and \[y=2\]in any of the given equations to check the solution:
\[\begin{align}
& 3\left( 0 \right)+7\left( 2 \right)=14 \\
& 14=14
\end{align}\]
Since RHS\[=\]LHS, it implies the solution is correct.
Now we shall check with 5x = 4y -8 too.
5(0) = 4(2) – 8
0 = 8-8
0=0
LHS = RHS
So, this (0, 2) is a solution of the above system of equations.