Question 206014
Notice the edited version of your question. PLEASE, ALL QUESTIONERS TAKE NOTE: when a fraction is in vertical form, the horizontal division line acts as a grouping operator -- that means: WHEN YOU CONVERT IT INTO HORIZONTAL FORM YOU ***MUST*** USE PARENTHESES TO HAVE IT CONTINUE TO MEAN THE SAME THING.
 
So, first, the solution to what your question means as written:
{{{ 3/4=1-3x-2/x+1 }}}     subtract 2 from both sides
{{{ -5/4=-3x-2/x }}}     multiply both sides by 4x to get rid of the fractions
{{{ -5x=-12x^2-8 }}}      add {{{12x^2+8}}} to both sides
{{{ 12x^2-5x+8 = 0 }}} 

To see whether this factors, look at the discriminant (what's under the square root in the quadratic formula)
{{{b^2 - 4ac =(-5)^2 - 4(12)(8) = 25 - 384 = -359}}}  Since this number is negative, this quadratic expression is not factorable (the discriminant must be a perfect square for the quadratic to be factorable).
 
Using the quadratic formula, then, 
{{{ ax^2 + bx + c = 0 }}} --> {{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
{{{ 12x^2-5x+8 = 0 }}},  a = 12, b=-5, c = 8 (notice you can only use this when one side is 0)
{{{x = (-(12) +- sqrt( -359 ))/(2*(12)) }}} 
{{{ x = (-12 +- i*sqrt( 359 ))/24 }}}, which, since 359 is prime is as simple as it gets.
 
 
 
 
 
OK, now what the problem was probably intended to be (correct me if I'm wrong, I'm not at all certain):
 
{{{3/4 = 1 - (3x-2)/(x+1) }}}   To get rid of fractions, multiply both sides by 4(x+1)
{{{ 4(x+1)(3/4) = 4(x+1)(1 - (3x-2)/(x+1)) }}}   quick simplification of each side ( on right side 4(x-1) is multiplied by both terms)
{{{ 3(x+1) = 4(x+1) - 4(3x-2) }}}    next step of simplification: multiplying out
{{{ 3x + 3 = 4x + 4 - 12x + 8) }}}   to get x's and constants on opposite sides, add 8x - 3 to both sides
{{{ 11x = 9 }}}  so {{{ x = 9/11 }}}