Answer
\[x=1,\ y=-2\]
Work Step by Step
Multiply \[3\]on both sides to the equation \[3x-4y=11\]to get: \[9x-12y=33\].
Multiply \[4\]on both sides to the equation \[2x+3y=-4\]to get: \[8x+12y=-16\]
Add the above obtained equation from both RHS and LHS as follows:
\[\underline{\begin{align}
& 9x-12y=33 \\
& 8x+12y=-16
\end{align}}\]
\[\begin{align}
& 17x\ \ \ \ \ \ \ \ \ =17 \\
& x=1
\end{align}\]
Put \[x=1\]in\[3x-4y=11\], to get:
\[\begin{align}
& 3\left( 1 \right)-4y=11 \\
& 3-4y=11 \\
& -4y=8 \\
& y=-2
\end{align}\]
Put\[x=1\]and \[y=-2\]in any of the given equations to check the solution:
\[\begin{align}
& 2\left( 1 \right)+3\left( -2 \right)=-4 \\
& 2-6=-4 \\
& -4=-4
\end{align}\]
Since RHS\[=\]LHS, it implies the solution is correct.
Now check its solution by putting x = 1 and y = -2 in 3x -4y = 11 too.
3(1) -4(-2) = 11
3 +8 = 11
11 = 11
So LHS = RHS
It means this is a solution to the above system of equations.
