\(-x+y=5\Rightarrow y=x+5\)
Substitute \(y\) in the other equation:
\(2x^{2}-3x-(x+5)=-5\)
\(\Rightarrow 2x^{2}-3x-x-5=-5\)
\(\Rightarrow 2x^{2}-4x-5=-5\)
\(\Rightarrow 2x^{2}-4x=0\)
\(\Rightarrow 2x(x-2)=0\)
So either
\(2x=0\Rightarrow x=0\Rightarrow y=0+5=5\)
or
\(x-2=0\Rightarrow x=2\Rightarrow y=2+5=7\)
