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Let (x, y) be an ordered pair of real numbers that satisfies the equation \(x^{2}+y^{2}=14x+48y \). What is the minimum value of x?

 Jan 14, 2020
 #1
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Nvm, I solved it

 
 Jan 15, 2020
 #2
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2(7x+24y)=x^2+y^2

x^2-14x=48y-y^2

x(x-14)=y(48-y) 

(x,y)=(0,0) , (14,48) , (0,48),(14,0) etc.. 

 
 Jan 15, 2020

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