We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
260
1
avatar+569 

 

how do i solve this?

 Sep 29, 2018
 #1
avatar+2340 
+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

9 Online Users

avatar