SOLUTION: Solve by any method - Addition or Substitution a){{{x+y^2=4}}} {{{x^2+y^2=16}}} b) (x+1)Square + (y+1)Square=5 2x-y=3

Algebra ->  Systems-of-equations -> SOLUTION: Solve by any method - Addition or Substitution a){{{x+y^2=4}}} {{{x^2+y^2=16}}} b) (x+1)Square + (y+1)Square=5 2x-y=3      Log On


   

Question 730715: Solve by any method - Addition or Substitution
a)x%2By%5E2=4
x%5E2%2By%5E2=16



b) (x+1)Square + (y+1)Square=5
2x-y=3
Answer by tommyt3rd(5050) About Me  (Show Source):
You can put this solution on YOUR website!
I will do a)



x%2By%5E2=4
x%5E2%2By%5E2=16

y%5E2=4-x
therefore
x%5E2%2B4-x=16
so that
x%5E2-x-12=0

x=4 or x=-3

x=-3 leads to y=++sqrt%287%29 or y=+-+sqrt%287%29
x=4 leads to y=0