Question

In the system shown below, what are the coordinates of the solution that lies in quadrant IV?
x^2+y^2=25
x-y^2=-5

Answer 1

x2−y2=25

(x+y)(x−y)=25

x+y=25

x−y=1
add to get
2x=26

x=13
then solve for y

Answer 2

help please

Answer 3

answer is (13,12)
(4,3) does not work because 4 + 3 = 7, not 25

Answer 4

If you solve the system by elimination, we note that adding both equations together gives you \(\large x^2+x = 20\implies x^2+x-20=0\). This equation can be factored, so we see that \(\large x^2+x-20=-\implies (x+5)(x-4)=0\). Therefore, \(\large x=-5\) or \(\large x=4\). In order for the point of intersection to be in the fourth quadrant, the x value needs to be positive, but the y value needs to be negative. Therefore, if \(\large x=4\), plugging this into the first equation gives us \(\large 4^2+y^2 = 25\implies y^2 = 9 \implies y=\pm 3\). But as I mentioned before, we need y to be negative in order to get the point in the fourth quadrant. Hence, the solution that we want is \(\large (4,-3)\).

I've attached a graph of the two equations, so you can clearly see that this point is in fact the solution we're after.

I hope this makes sense! :-)