Question 63398
{{{ 1/(4x+12) - 1/(x^2-9) = 5/(x-3) }}}
Factor the denominators
{{{ 1/(4(x+3)) - 1/((x+3)(x-3)) = 5/(x-3) }}}
find common denominators ... in this case 4(x+3)(x-3)
If a fraction is missing any of these multiply the numerator and denominator by that quantity
{{{ (1(x-3))/(4(x+3)(x-3)) - 1(4)/(4(x+3)(x-3)) = (5(4)(x+3))/(4(x+3)(x-3)) }}}
multiply out the numerators.
{{{ (x-3)/(4(x+3)(x-3)) - 4/(4(x+3)(x-3)) = (20x+60)/(4(x+3)(x-3)) }}}
Since the denominators are all the same, and this is an equation .... DROP THEM
{{{ x-3 - 4 = 20x+60 }}}
Combine like terms
{{{ x - 7 = 20x + 60 }}}
subtract x from both sides
{{{ - 7 = 19x +60 }}}
subtract 60 from both sides
{{{ -67 = 19x }}}
divide by 19
{{{ -67/19 =x }}}