Question 151293
{{{(8)/(x+3)+(9)/(x-3)}}} Start with the given expression


Now let's find the LCM of the denominators {{{(x+3)}}} and {{{(x-3)}}}. It turns out that the LCM of these denominators is {{{(x+3)(x-3)}}}. So the goal is to get both denominators to the LCD {{{(x+3)(x-3)}}}



{{{((x-3)/(x-3))((8)/((x+3)))+((x+3)/(x+3))((9)/((x-3)))}}} Multiply the 1st fraction by {{{((x-3))/((x-3))}}}




{{{(8(x-3))/((x+3)(x-3))+((x+3)/(x+3))((9)/((x-3)))}}} Combine the fractions.



{{{(8(x-3))/((x+3)(x-3))+((x+3)/(x+3))((9)/((x-3)))}}} Multiply the 2nd fraction by {{{((x+3))/((x+3))}}}



{{{(8(x-3))/((x+3)(x-3))+(9(x+3))/((x+3)(x-3))}}} Combine the fractions.



{{{(8x-24)/((x+3)(x-3))+(9x+27)/((x+3)(x-3))}}} Distribute



{{{(8x-24+9x+27)/((x+3)(x-3))}}} Add the fractions. This is now possible since both denominators are equal.



{{{(17x+3)/((x+3)(x-3))}}} Combine like terms.



{{{(17x+3)/(x^2-9)}}} FOIL



So {{{(8)/(x+3)+(9)/(x-3)}}} simplifies to {{{(17x+3)/(x^2-9)}}}



In other words, {{{(8)/(x+3)+(9)/(x-3)=(17x+3)/(x^2-9)}}} where {{{x<>-3}}} or {{{x<>3}}}