SOLUTION: I know how to determine where these graphs intersect by graphing them but I have to determine algebraically where these graphs intersect. {{{ 4x^2 +16y^2 =64 }}} {{{ 2x-y^2=-4

Algebra ->  Systems-of-equations -> SOLUTION: I know how to determine where these graphs intersect by graphing them but I have to determine algebraically where these graphs intersect. {{{ 4x^2 +16y^2 =64 }}} {{{ 2x-y^2=-4      Log On


   



Question 1045863: I know how to determine where these graphs intersect by graphing them but I have to determine algebraically where these graphs intersect.
+4x%5E2+%2B16y%5E2+=64+
+2x-y%5E2=-4+

Found 2 solutions by solver91311, MathLover1:
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


Multiply the second equation by 16



Add the two equations, term by term, to get:



Solve for

Substitute the roots for x in either equation and then solve for y. Discard one of the x-values because you get a complex number result for y. The remaining value of x paired with either of the two real roots for y are your two intersection points.

John

My calculator said it, I believe it, that settles it


Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
+4x%5E2+%2B16y%5E2+=64+.............(1)
+2x-y%5E2=-4+.............(2).........both sides multiply by 16
-------------------------------------------
+4x%5E2+%2B16y%5E2+=64+.............(1)
+32x-16y%5E2=+-64+.............(2)
--------------------------------------------------------add
+4x%5E2+%2B16y%5E2+%2B32x-16y%5E2=64+-64
+4x%5E2+%2Bcross%2816y%5E2%29+%2B32x-cross%2816y%5E2%29=0
+4x%5E2++%2B32x=0................both sides divide by 4
+x%5E2++%2B8x=0...........factor

+x%28x++%2B8%29=0
solutions:
+x=0
if +x++%2B8=0->x=-8
now find y
go to +2x-y%5E2=-4+.............(2) substitute x
if +x=0 we have +2%2A0-y%5E2=-4+-> +-y%5E2=-4+%7D%7D-%3E+%7B%7B%7B+4=y%5E2+%7D%7D-%3E%7B%7B%7By=2 or y=-2
so, if x=0 there are two intersection points :
(0,2) and (0,-2)
if +x=-8 we have +2%2A%28-8%29-y%5E2=-4+-> +-16-y%5E2=-4+-> +-16%2B4=y%5E2+->+-12=y%5E2+->y=sqrt%28-12%29 or y=-sqrt%28-12%29 -> complex solutions
y=2i%2Asqrt%283%29 or y=-2i%2Asqrt%283%29