Question
In Exercises 35 - 48, solve the system graphically.$ \left\{\begin{array}{l} \hspace{1cm} \hspace{1cm} -x + y = 3\\x^2 - 6x - 27 + y^2 = 0\end{array}\right. $
Step 1
The first equation, $-x + y = 3$, is a linear equation and can be graphed as a straight line. The second equation, $x^2 - 6x - 27 + y^2 = 0$, is a circle equation and can be graphed as a circle. Show more…
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