SOLUTION: Solve by substitution {{{x^2+y^2=16}}} {{{x^2-2y=8}}}

Algebra ->  Graphs -> SOLUTION: Solve by substitution {{{x^2+y^2=16}}} {{{x^2-2y=8}}}       Log On


   

Question 135446: Solve by substitution
x%5E2%2By%5E2=16
x%5E2-2y=8
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Start with the given system

x%5E2%2By%5E2=16
x%5E2-2y=8


x%5E2=8%2B2y Isolate x%5E2 for the second equation

8%2B2y%2By%5E2=16 Plug in x%5E2=8%2B2y into the first equation


8%2B2y%2By%5E2-16=0 Subtract 16 from both sides


y%5E2%2B2y-8=0 Combine like terms


%28y%2B4%29%28y-2%29=0 Factor the left side (note: if you need help with factoring, check out this solver)



Now set each factor equal to zero:
y%2B4=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%5E2=8%2B2x Start with the given equation

x%5E2=8%2B2%28-4%29 Plug in y=-4


x%5E2=0 Simplify

x=0 Take the square root of both sides


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



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




x%5E2=8%2B2x Start with the given equation

x%5E2=8%2B2%282%29 Plug in y=2


x%5E2=12 Simplify

x=0%2B-sqrt%2812%29 Take the square root of both sides

x=2%2Asqrt%283%29 or x=-2%2Asqrt%283%29


So our next ordered pairs are (2%2Asqrt%283%29,2) or (-2%2Asqrt%283%29,2)



--------------------------
Answer:

So the solutions are

(0,-4), (2%2Asqrt%283%29,2), or (-2%2Asqrt%283%29,2)