SOLUTION: Longjonsilver had walked me through the following problem, but didn't actually solve it for me. I would like to see if the answer I came up with is correct. Thank you. x^2+y

Algebra ->  Systems-of-equations -> SOLUTION: Longjonsilver had walked me through the following problem, but didn't actually solve it for me. I would like to see if the answer I came up with is correct. Thank you. x^2+y      Log On


   

Question 6923: Longjonsilver had walked me through the following problem, but didn't actually solve it for me. I would like to see if the answer I came up with is correct. Thank you.

x^2+y^2=13 = 2x^2+2y^2=26
y=2x^2-5 = 2x^2=y+5
y+5+2y^2=26
2y^2+y-21=0
(2y+7)(y-3)
2y+7=0 y-3=0
y=-7/2 or y=3

x^2-7/2^2=13
x^2+49/4=13
x^2=52/4-49/4
x^2=3/4
x=+ or - sqrt3/4

x^2+3^2=13
x^2+9=13
x^2=4
x=+ or - 2
ORDERED PAIRS ARE (-7/2, SQRT3/4) (-7/2,-SQRT3/4) (3,2) (3,-2)

THANK YOU IN ADVANCE!!!
Answer by longjonsilver(2297) About Me  (Show Source):
You can put this solution on YOUR website!
nearly perfect :-)

the only "wrong" thing is that you have written the ordered pairs as (y,x)...we usually write them as (x, y), so correct the order within all 4 and then you have perfection.

Well done
Jon.