Finding Solutions for Systems of Quadratic Equations

[b_ball_chic]

OMG! This question is annoying me to death...

x^2 + y^2 = 25
x^2 + y = -5

I understand you substitute, and I did...
x^2 = -5 - y

So I plugged it in...
(-5 - y) + y^2 = 25
subtract 25) y^2 - y - 30 = 0

I tried the quadratic formula AND did the whole (y + 1)(y - 30)=0,
but they would not equal the answers: (3,4)(-3,4)(-5,0)

Please help! I can't understand why I can't get this answer, but yet I can get the others. :

 

[vincentchan]

I tried the quadratic formula AND did the whole (y + 1)(y - 30)=0,

wrong...
(y+1)(y-30)= y^2 -29y -30
do it again pls

 

[b_ball_chic]

okay...i see...
(y-6)(y+5)=0
So I plugged in:
6) x^2 = -5 - (6)
x^2 = -11
I also plugged in -5) x ^2 = -5 - (-5) = 0
That got me the (-5,0), but I still do not understand how to get the other two points...

 

[vincentchan]

see if you copy your question correctly,
x^2 + y^2 = 25
x^2 + y = -5
has one solution only , which is (0,-5), if you don't believe me, you can try plug in your ANSWER in the equations above, you will see LHS is NOT equal RHS

 

[b_ball_chic]

ack...

yea..i copied the question wrong...it was:
x^2 + y ^2 = 25
y - x^2 = -5
Got the answers, thnx SO much!

I STILL can't believe I copied the wrong question...sorry I put you through the trouble with that, too...