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

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

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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