Solve the system and graph the curves \begin{cases}x^{2}+3x-y+2=0 \\ y-5x=1 \end{cases}

Brittney Lord 2021-08-15 Answered

Solve the system and graph the curves
\(\begin{cases}x^{2}+3x-y+2=0 \\ y-5x=1 \end{cases}\)

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Expert Answer

averes8
Answered 2021-08-16 Author has 15043 answers

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
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