Answer
\[x=\frac{14}{5},\ y=\frac{1}{5}\]
Work Step by Step
The equation \[-4x+y=-11\]can be written as follows:
\[\begin{align}
& -4x+y=-11 \\
& y=-11+4x \\
& y=4x-11
\end{align}\]
Substitute\[y=4x-11\]in the equation\[2x-3y=5\], to get:
\[\begin{align}
& 2x-3\left( 4x-11 \right)=5 \\
& 2x-12x+33=5 \\
& -10x=-28 \\
& x=\frac{14}{5}
\end{align}\]
Substitute \[x=\frac{14}{5}\]in\[y=4x-11\], to get:
\[\begin{align}
& y=4\left( \frac{14}{5} \right)-11 \\
& =\frac{56}{5}-11 \\
& =\frac{56-55}{5} \\
& =\frac{1}{5}
\end{align}\]
Put\[x=\frac{14}{4}\]and \[y=\frac{1}{5}\]in any of the given equations to check the solution:
\[\begin{align}
& 2\left( \frac{14}{5} \right)-3\left( \frac{1}{5} \right)=5 \\
& \frac{28}{5}-\frac{3}{5}=5 \\
& \frac{25}{5}=5 \\
& 5=5
\end{align}\]
Since RHS\[=\]LHS, it implies the solution is correct.
Check its solution with - 4x + y = -11 too.
-4(14/5) + 1/5 = - 11
-56/5 + 1/5 = - 11
-55/5 = -11
-11 = -11
So,\[\left( \frac{14}{5},\frac{1}{5} \right)\] is a solution to this system of equations.
