Question 548292
{{{2/(x-3)+2/x=1}}}
I don't know about you... but I do not like having fractions... lets get rid of them!
To do this, we will first multiply every term by x-3 to get
{{{(x-3)*(2/(x-3))+(x-3)*(2/x)=(x-3)*1}}}
{{{2+(2x-6/x)=(x-3)}}}
Now let's multiply every term by x
{{{x*2+x*(2x-6/x)=x*(x-3)}}}
{{{2x+2x-6=x^2-3x}}}
{{{4x-6=x^2-3x}}}

This is just a quadratic, so lets put it in the form ax^2+bx+c=0
{{{4x-6-4x+6=x^2-3x-4x+6}}}
{{{0=x^2-7x+6}}}
Now we need to think of two numbers that multiply to our c value (6) and add to our b value (-7) ... if you think hard enough you will find that -1 and -6 does the trick.  
So our equation factors to 
{{{(x-1)(x-6)=0}}}

Now if two things have a product of 0, then at least one of them must be equal to 0 
This means
{{{x-1=0}}}
{{{x=1}}}
or 
{{{x-6=0}}}
{{{x=6}}}


And there is the two solutions:)