# Solve {(x-3)^2+(y+1)^2=5; x-3y=7}

Solve $$\begin{cases}(x-3)^2+(y+1)^2=5\\x-3y=7\end{cases}$$

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liingliing8

Let:
$$\displaystyle{\left({x}−{3}\right)}^{{2}}+{\left({y}+{1}\right)}^{{2}}={5}{\left({1}\right)}$$
$$x−3y=7$$
Use substitution. Solve for x using (2) to obtain (3):
$$x=3y+7(3)$$
Substitute (3) to (1) and solve for y:
$$\displaystyle{\left({3}{y}+{7}−{3}\right)}^{{2}}+{\left({y}+{1}\right)}^{{2}}={5}$$
$$\displaystyle{\left({3}{y}+{4}\right)}^{{2}}+{\left({y}+{1}\right)}^{{2}}={5}$$
$$\displaystyle{\left({9}{y}^{{2}}+{24}{y}+{16}\right)}+{\left({y}^{{2}}+{2}{y}+{1}\right)}={5}$$
$$\displaystyle{10}{y}^{{2}}+{26}{y}+{17}={5}$$
$$\displaystyle{10}{y}^{{2}}+{26}{y}+{12}={0}$$
Factor:
$$\displaystyle{2}{\left({5}{y}^{{2}}+{13}{y}+{6}\right)}={0}$$
$$\displaystyle{2}{\left({y}+{2}\right)}{\left({5}{y}+{3}\right)}={0}$$
By zero product property,
$$y=−2,−35$$
Solve for the corresponding $$xx$$ values using (3).
When $$y=−2y=−2,$$
$$x=3(−2)+7$$
$$x=1$$
When $$y=−\frac{3}{5}$$,
$$x=3(−\frac{3}{5})+7$$
$$x=\frac{26}{5}$$
So, the solutions are:
$$(1,−2)$$ and $$(\frac{26}{5},−\frac{3}{5})$$

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