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Posted by **Princess** on Tuesday, December 11, 2012 at 11:07am.

x - 2y = -3

- Math -
**Steve**, Tuesday, December 11, 2012 at 11:18amx = 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)

- Math -
**Knights**, Tuesday, December 11, 2012 at 11:20amfor 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

- Math -
**Reiny**, Tuesday, December 11, 2012 at 11:26amI 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)

- Math -
**Reiny**, Tuesday, December 11, 2012 at 11:28amarrgghhh!

for first point:

y = 2, then x = 4-3 = 1 , just like Steve had above

point is (1,2)

- Math -
**Steve**, Tuesday, December 11, 2012 at 11:29amso, is this a typo fest, or what?

Good job, Knights.

Hope you can dig out the correct solutions, Princess!

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