Solve the following system. x^2 + y^2 = 25 2x + y = 10 The solution set

1 answer:

4 0

$\begin{cases} x^{2}+y^{2}=25 \\ 2x+y=10 \end{cases} \\ \\ \begin{cases} x^{2}+y^{2}=25 \\ y=10 -2x\end{cases} \\ \\ \begin{cases} x^{2}+(10-2x)^{2}=25 \\y=10-2x \end{cases} \\ \\ \begin{cases} x^{2}+100-40x+4x^{2}=25 \\y=10-2x \end{cases} \\ \\ \begin{cases} 5x^{2}-40x+75=0 \\y=10-2x \end{cases}$

$\begin{cases} x^{2}-8x+15=0 \\y=10-2x \end{cases} \\ \\ \begin{cases} \Delta=(-8)^{2}-4*1*15=64-60=4; \ \ \ \sqrt{\Delta} =2 \\y=10-2x \end{cases} \\ \\ \begin{cases} x_{1}= \frac{8-2}{2} =3 \ \wedge \ x_{2}= \frac{8+2}{2}=5 \\y=10-2*3=4 \ \wedge \ y=10-2*5=0 \end{cases}$

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