Answer
$\{\ (\displaystyle \frac{1}{5},\frac{18}{5}), \ \ (1,2) \ \}$.
Work Step by Step
1. Solve the linear equations for one of the unknowns,
2. Substitute the expression obtained in (1) into the other equation(s).
3. Solve and back-substitute
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$1.$
$2x+y=4$
$y=4-2x$
$2.$ Substitute, solve...
$(x+1)^{2}+(y-2)^{2}=4$
$(x+1)^{2}+(4-2x-2)^{2}=4$
$(x+1)^{2}+(2-2x)^{2}=4 \quad$... expand the squares ...
$ x^{2}+2x+1+4-8x+4x^{2}-4=0\quad$...gather like terms
$5x^{2}-6x+1=0$
Factor the LHS:
Factors of 5(1)=5 that add to -6 are $-1$ and $-5$
$5x^{2}-6x+1=5x^{2}-5x-x+1=$
$=5x(x-1)-(x-1)$
$=(5x-1)(x-1)$
$(5x-1)(x-1)=0$
$5x-1=0$ or $x-1=0$
$x=\displaystyle \frac{1}{5}$ or $x=1$
3. Back-substitute...
$\left[\begin{array}{lll}
x=\frac{1}{5} & or & x=1\\
y=4-2(\frac{1}{5})=\frac{18}{5} & & y=4-2(1)=2
\end{array}\right]$
, .
The solution set is
$\{\ (\displaystyle \frac{1}{5},\frac{18}{5}), \ \ (1,2) \ \}$.
