Question 88232
this is just a fraction problem. We treat it just like numerical fractions:


{{{ (6/(3x-2)) + (3/2x) }}}


if we had {{{ 2/3 + 1/4 }}} we would have to find the LCM of both 3 and 4 --> 3*4 which we can evaluate --> 12.
We do the same with (3x-2) and 2x. Their LCM is (3x-2)*2x which we cant evaluate so we leave it like that.


Next step is to multiply both fractions by 1 (so they remain unchanged) but write the 1 as a fraction:


{{{ (2/3)(4/4) + (1/4)(3/3) }}}
{{{ (8/12) + (3/12) }}}
{{{ (11/12) }}}


Algebraically we cannot do all these steps. We can do the first one:
{{{ (6/(3x-2))*((2x)/(2x)) + (3/2x)*((3x-2)/(3x-2)) }}}

We can multiply the numerators together and the denominators too:
{{{ (6(2x))/((2x)(3x-2)) + (3(3x-2))/(2x(3x-2)) }}}

And now add them together since they have the same denominator:
{{{ (6(2x) + 3(3x-2))/((2x)(3x-2)) }}}


So that is the fractions done. The rest is just manipulation of the algebra to try to simplify it:
{{{ (12x + 3(3x-2))/((2x)(3x-2)) }}}
{{{ (12x + 9x-6)/((2x)(3x-2)) }}}
{{{ (21x-6)/((2x)(3x-2)) }}}


or perhaps multiply out the denominator:
{{{ (21x-6)/(6x^2-4x) }}}


but we tend not to do this - we like things factorised. The answer is best quoted as {{{ (21x-6)/((2x)(3x-2)) }}}


hope this helps

cheers
Jon.