Step 1

Given:

\(\begin{cases}x^{2}+3x-y+2=0 & \rightarrow\ (1) \\ y-5x=1 & \rightarrow\ (2) \end{cases}\)

From equation (2), we get

\(\displaystyle{y}={1}+{5}{x}\)

Now put \(\displaystyle{y}={1}+{5}{x}\) in equation (1), we get

\(\displaystyle\Rightarrow\ {x}^{{{2}}}+{3}{x}-{\left({1}+{5}{x}\right)}+{2}={0}\)

\(\displaystyle\Rightarrow\ {x}^{{{2}}}+{3}{x}-{1}-{5}{x}+{2}={0}\)

\(\displaystyle\Rightarrow\ {x}^{{{2}}}-{2}{x}+{1}={0}\)

\(\displaystyle\Rightarrow{\left({x}-{1}\right)}^{{{2}}}={0}\)

\(\displaystyle\Rightarrow\ {x}-{1}={0}\)

\(\displaystyle\Rightarrow\ {x}={1}\)

Now put \(\displaystyle{x}={1}\) in equation (2), we get

\(\displaystyle{y}-{5}{\left({1}\right)}={1}\)

\(\displaystyle\Rightarrow\ {y}-{5}={1}\)

\(\displaystyle\Rightarrow\ {y}={6}\)

Therefore the solution is,

\(\displaystyle{x}={1}\) and \(\displaystyle{y}={6}\)

Step 2