Posted by Princess on Tuesday, December 11, 2012 at 11:07am.
x = 2y-3, so we have
(2y-3)^2 + 9y^2 = 37
4y^2 - 12y + 9 + 9y^2 = 37
13y^2 - 12y - 28 = 0
(y-2)(13y+14) = 0
so, solutions are (1,2) and (-67/3,-14/13)
for 2nd equation, we get x = 2y-3
thus plugging into first equation gets
(2y-3)^2+9y^2 = 37.
4y^2-12y+9+9y^2 = 13y^2-12y+9=37 = 13y^2-12y-28 = 0.
Solving with the quadratic formula gives us: 2,-14/13
I assume you are solving.
from the 2nd: x = 2y-3
sub into the 1st
(2y-3)^2 + y^2 - 37 = 0
4y^2 - 12y + 9 + 9y^2 - 37 = 0
13y^2 - 12y - 28 = 0
(y-2)(13y + 14) = 0
y = 2 or y = -14/13
if y = 2, then x = 2-3 = -1
if y = -14/13, x = -28/13 - 3 = -67/13
points of intersection:
(-1,2) and (-67/13 , -14/13)
arrgghhh!
for first point:
y = 2, then x = 4-3 = 1 , just like Steve had above
point is (1,2)
so, is this a typo fest, or what?
Good job, Knights.
Hope you can dig out the correct solutions, Princess!
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