SOLUTION: Please help me to find the solution set(s) to this problem using Addition Method. Thank you. {{{x^2+y^2=49}}} {{{y^2-4x=49}}}

Algebra ->  Coordinate Systems and Linear Equations  -> Lessons  -> Linear Equations Lesson -> SOLUTION: Please help me to find the solution set(s) to this problem using Addition Method. Thank you. {{{x^2+y^2=49}}} {{{y^2-4x=49}}}      Log On


   

Question 434221: Please help me to find the solution set(s) to this problem using Addition Method. Thank you.
x%5E2%2By%5E2=49
y%5E2-4x=49
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E2%2By%5E2=49
y%5E2-4x=49
Rearrange it to
x%5E2%2By%5E2=49
-4x%2By%5E2=49
Multiply the 2nd eq by -1
x%5E2%2By%5E2=49
4x-y%5E2=-49
------------------addition method eliminates y^2, we are left with:
x^2 + 4x = 0
Factor out x
x(x+4) = 0
Two solutions
x = 0
x + 4 = 0
x = -4
:
It's obvious that x = 0 will work, find y when x = -4
Replace x with - 4 in the 2nd original equation
-4(-4) + y^2 = 49
16 + y^2 = 49
y^2 = 49-16
y^2 = 33
y = sqrt%2833%29
:
You can check the solutions in the 1st original equation