Answer
The solution set is\[\left\{ \left( 1,4 \right) \right\}\].
Work Step by Step
Multiply \[2\]on both sides to the equation \[4x+y=8\]to get: \[8x+2y=16\].
Add the above obtained equation from both RHS and LHS as follows:
\[\underline{\begin{align}
& 3x-2y=-5 \\
& 8x+2y=16
\end{align}}\]
\[\begin{align}
& 11x\ \ \ \ \ \ \ =11 \\
& x=1
\end{align}\]
Put \[x=1\]in\[3x-2y=-5\], to get:
\[\begin{align}
& 3\left( 1 \right)-2y=-5 \\
& 3-2y=-5 \\
& -2y=-8 \\
& y=4
\end{align}\]
Put\[x=1\]and \[y=4\]in any of the given equations to check the solution:
\[\begin{align}
& 4\left( 1 \right)+4=8 \\
& 8=8
\end{align}\]
Since RHS\[=\]LHS, it implies the solution is correct.
We shall also check this solution with the equation 3x – 2y = -5
3(1) -2(4) = -5
3 – 8 = - 5
-5 = -5
So, this is the correct solution of the given system of equations.
