At what point does the graph of 3x+4y=15 intersect the graph of x^2+y^2=25?
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\( \frac{15-3x}{4}=\sqrt{25-x^2}\\ 225-90x+9x^2=16\cdot 25-16x^2\\ 25x^2-90x-175=0\\x^2-\frac{18}{5}x-7=0\\ x= \frac{9}{5}\pm \sqrt{\frac{81}{25}+7}\\ x=\frac{9}{5}\pm \frac{16}{5}\)
\(x\in \{-\frac{7}{5},5\}\)
\(y\in \{\ \frac{24}{5},\ 0\}\)
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