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how do i solve this?

 Sep 29, 2018
 #1
avatar+2345 
+1

\((2x-y)^2\) 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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