For the system shown below, what are the coordinates of the solution that lies in quadrant I

3y^2 - 5x^2 = 7
3y^2 - x^2 = 23

I keep getting (-2,3) but that's in quadrant II.

There are 4 solutions, one in each quadrant: (± 2, ± 3)

Q I : (2, 3)
Q II : (−2, 3)
Q III : (−2, −3)
Q IV : (2, −3)

Since this is a conic section centered at the origin, the signs of the numbers don't matter. If this is a coordinate, (a, b), then other coordinates of this conic section are (-a, b), (a, -b), (-a, -b). This is only if the conic section is centered at the origin, which in this case is.
The solution in quadrant I is (2, 3).

1) 3y^2 - 5x^2 = 7
2) 3y^2 - x^2 = 23
Subtract 2) from 1) -4x^2 = - 16
x^2 = 4 so x = + or - 2
for x = -2 substituting in 1) 3y^2 -20 = 7
3y^2 = 27
y^2 = 9 so y = + or - 3

x = +2 and y = + 3 are both in the first quadrant

subtract the bottom equation from the top equation to get:

0 - 4x^2 = -16
x^2 = 4
x = 2

plug x=2 into any equation to get
3y^2 - 4 = 23
3y^2 = 27
y^2 = 9
y = 3
(2,3)

by subtracting eq. ! frm @ we get
3y2 - 3y^2 - x^2 + 5x^2=23- 7
->4x^2=16
-> x^2= 4
->3y^2=23+4
->y^2=9
so the soln are
x=+-2,y=+-3
so in first quad we get (2,3)
guess U forgot the (+-) part...
hope it helped......

-3y^2+5x^2=-7
3y^2-x^2=23
4x^2=16
4x^2-16=0
(2x+4)(2x-4)=0
x=-2 or x=2

3y^2-5(2)^2=7
3y^2-5(4)=7
3y^2-20=7
3y^2-27=0
3(y^2-9)=0
3(y+3)(y-3)=0
y=-3 or y=3

(2,3)

Another solution is (2, 3) in Q1

3y² - 5x² = 7
3y² - x² = 23

4x² = 16
x² = 4
x = ±2

3y² = 27
y² = 9
y = ±3

(2, 3)