Question 191386
{{{(x+1)/(x+2)+1/(x-3)-(x^2-2)/(x^2-x-6)}}} Start with the given expression.



{{{(x+1)/(x+2)+1/(x-3)-(x^2-2)/((x+2)(x-3))}}} Factor the last denominator



Take note that the LCD is {{{(x+2)(x-3)}}}



{{{((x+1)(x-3))/((x+2)(x-3))+1/(x-3)-(x^2-2)/((x+2)(x-3))}}} Multiply the first fraction by {{{(x-3)/(x-3)}}} (to get this denominator equal to the LCD)



{{{(x^2-2x-3)/((x+2)(x-3))+1/(x-3)-(x^2-2)/((x+2)(x-3))}}} FOIL



{{{(x^2-2x-3)/((x+2)(x-3))+(1(x+2))/((x+2)(x-3))-(x^2-2)/((x+2)(x-3))}}} Multiply the second fraction by {{{(x+2)/(x+2)}}} (to get this denominator equal to the LCD)



{{{(x^2-2x-3)/((x+2)(x-3))+(x+2)/((x+2)(x-3))-(x^2-2)/((x+2)(x-3))}}} Distribute



Now that the denominators are all equal, we can combine the fractions



{{{(x^2-2x-3+(x+2)-(x^2-2))/((x+2)(x-3))}}} Combine the fractions



{{{(x^2-2x-3+x+2-x^2+2)/((x+2)(x-3))}}} Distribute



{{{(-x+1)/((x+2)(x-3))}}} Combine like terms.
 


{{{(-x+1)/(x^2-x-6)}}} FOIL the denominator



So {{{(x+1)/(x+2)+1/(x-3)-(x^2-2)/(x^2-x-6)=(-x+1)/(x^2-x-6)}}} where {{{x<>-2}}} or {{{x<>3}}}