SOLUTION: Please note that graphing is not allowed to solve this. If x^2+y^2=36 and (x^2/16)+(y^2/64)=1, then what are the values of x and y?

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Please note that graphing is not allowed to solve this. If x^2+y^2=36 and (x^2/16)+(y^2/64)=1, then what are the values of x and y?      Log On


   

Question 458999: Please note that graphing is not allowed to solve this.
If x^2+y^2=36 and (x^2/16)+(y^2/64)=1, then what are the values of x and y?
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
If x^2 + y^2 = 36 and (x^2/16)+(y^2/64)=1, then what are the values of x and y?
Get rid of the annoying fraction in the 2nd equation
%28x%5E2%2F16%29%2B%28y%5E2%2F64%29=1
multiply by 64 and we have
4(x^2) + y^2 = 64
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Use elimination with the 1st equation just as it is
4x^2 + y^2 = 64
x^2 + y^2 = 36
----------------subtraction eliminates y^2, find x^2
3x^2 = 28
x^2 = 28%2F3
x = sqrt%2828%2F3%29 = 2sqrt%287%2F3%29
:
Find y, replace x^2 with 28%2F3 in the 1st equation
28%2F3 + y^2 = 36
y^2 = 36 - 28%2F3
y^2 = 108%2F3 - 28%2F3
y = 80%2F3
y = sqrt%2880%2F3%29 = 4sqrt%285%2F3%29
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