Question 135446
Start with the given system


{{{x^2+y^2=16}}}
{{{x^2-2y=8}}} 



{{{x^2=8+2y}}} Isolate {{{x^2}}} for the second equation


{{{8+2y+y^2=16}}} Plug in {{{x^2=8+2y}}} into the first equation



{{{8+2y+y^2-16=0}}} Subtract 16 from both sides



{{{y^2+2y-8=0}}} Combine like terms



{{{(y+4)(y-2)=0}}} Factor the left side (note: if you need help with factoring, check out this <a href=http://www.algebra.com/algebra/homework/playground/change-this-name4450.solver>solver</a>)




Now set each factor equal to zero:

{{{y+4=0}}} or  {{{y-2=0}}} 


{{{y=-4}}} or  {{{y=2}}}    Now solve for y in each case



So our y values are

 {{{y=-4}}} or  {{{y=2}}} 




{{{x^2=8+2x}}} Start with the given equation


{{{x^2=8+2(-4)}}} Plug in {{{y=-4}}}



{{{x^2=0}}} Simplify


{{{x=0}}} Take the square root of both sides



So our first ordered pair is (0,-4)




---------------------------





{{{x^2=8+2x}}} Start with the given equation


{{{x^2=8+2(2)}}} Plug in {{{y=2}}}



{{{x^2=12}}} Simplify


{{{x=0+-sqrt(12)}}} Take the square root of both sides


{{{x=2*sqrt(3)}}} or {{{x=-2*sqrt(3)}}}



So our next ordered pairs are ({{{2*sqrt(3)}}},2) or ({{{-2*sqrt(3)}}},2) 




--------------------------

Answer:


So the solutions are 


 (0,-4), ({{{2*sqrt(3)}}},2), or ({{{-2*sqrt(3)}}},2)