$\displaystyle \displaystyle \frac {2x+1} {3x-7}= 2 \Rightarrow
\frac {(2x+1)\cdot (3x-7)} {(3x-7) }= 2\cdot (3x-7 ) \Rightarrow 2x+1= 2\cdot (3x-7) $
so you will lose fraction after dividing $\displaystyle \displaystyle \frac {(2x+1)\cdot (3x-7)}{(3x-7)} = (2x+1) $
do you understand ?
P.S. when said multiply all .... means that you multiply and left and right side of equation !!!
let's look at this this way ...
$\displaystyle 2x = 2 \Rightarrow x = 2/2 = 1 $
multiply both sides of equation with 2 and you will have
$\displaystyle 4x= 4 \Rightarrow x = 4/4 = 1 $
do you see that nothing is changed ?
now with fractions ....
$\displaystyle \frac {2x}{2} = 1 $
because of that under fraction is 2 (can be anything) to lose that fraction you multiply both sides with that whatever is under fraction ....
in this case multiply with 2
$\displaystyle \frac {2x}{2}\cdot 2 = 1 \cdot 2 \Rightarrow \frac {2x \cdot 2 }{2} = 1\cdot 2 \Rightarrow 2x = 2 \Rightarrow x =2/2 = 1 $
you don't multiply by two in your problem there... i just do that with my example to show you that nothing changes if you multiply both sides (or add or divide or ... ) Whatever is "down" in fraction (can't figure out how to write it on English ) you multiply both sides with THAT in your case it's 3x-7
so you will be free of the fraction
P.S. when multiplying fraction with number you don't multiply down expression
$\displaystyle \displaystyle \frac {x+y}{a+b} \cdot 10 = \frac {10x +10y}{a+b} $
but only if you are multiplying fraction with fraction ...
$\displaystyle \displaystyle \frac {x+y}{a+b} \cdot \frac {10}{3} = \frac {10x+10y}{3a+3b} $
you have solution in my first post (#2)
it's not problem for you with English ... it's mine
P.S. you need to get familiar with basic operations with fractions, and on fractions (as multiplying , divide, add , subtract .... ) so you will not have problem piking up stuff like this