SOLUTION: algebraically find the x-coordinates where the line y=2x-3 intersects the circle (x-3)^2+(y+1)^2=68.

Question 1064246: algebraically find the x-coordinates where the line y=2x-3 intersects the circle (x-3)^2+(y+1)^2=68.
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
(x-3)^2+(2x-3+1)^2=68, by substitution
x^2-6x+9+4x^2-8x+4=68
5x^2-14x-55=0
(5x+11)(x-5)=0
x=-11/5,5
y=7, so (5,7). That would be 2^2+8^2=68
x=-11/5, so y=-37/5 and (-11/5, -37/5). That would be 676/25+1024/25=1700/25=68
(5,7) and (-11/5, -37/5)

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