Question 50655
This is a fractional equation.  Start by factoring the denominators so you can find the LCD for the problem.
{{{1/(2(x-3)) + 2/(3(x-3)) = 7/(x(x-3)) }}}


The LCD = 6x(x-3), so multiply both sides of the equation by the LCD.  It's ugly at first, but in the first step all the fractions divide out!
{{{1/(2(x-3)) + 2/(3(x-3)) = 7/(x(x-3)) }}}
{{{6x(x-2)* (1/(2(x-3))) + 6x(x-3)*(2/(3(x-3))) =6x(x-3)* (7/(x(x-3))) }}}


After all the denominator factors divide out, this is what is left:
{{{3x + 4x = 42}}}
{{{7x = 42}}}
{{{x = 6}}}


One more step:  Check each denominator, and make sure that x= 6 doesn't make any denominators zero.  Also, you can check the answer by substituting x= 6 into the original equation (if you want to go to some trouble).


R^2 at SCC