how do i solve this?

 Sep 29, 2018

\((2x-y)^2\) represents \(18^2\) . Therefore, in order for there to be a valid value for and y\(2x-y=18\)


Do x=20 and y=2 fit this category? Well, let's see. 


\(x=20,y=2;\\ 2*20-2\stackrel{?}{=}18\\ 38\neq 18\)


No, it doesn't. Let's try the next, x=2 and y=20.


\(x=2,y=20;\\ 2*2-20\stackrel{?}{=}18\\ -16\neq 18\)


Ok, x=2 and y=20 are not it. Let's try the next pair. 


\(x=10,y=2;\\ 2*10-2\stackrel{?}{=}18\\ 18= 18 \checkmark\)


There we go! We found the pair that works, x=2 and y=20.

 Sep 29, 2018

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