Question 547678
Okay, If I'm understanding properly we have:


Solve for y in 
{{{x+2y=2x-1}}}   
 {{{x+2y-x=2x-1-x}}}
{{{2y=x-1}}}  
{{{2y/2=(x-1)/2}}}
{{{y=(x-1)/2}}} 

Hopefully I understood what you were trying to ask, just let me know if you still do not understand:)