Question 178160
{{{x/(x^2-2x-3)-3x/(x^2-9)}}} Start with the given expression



{{{x/((x+1)(x-3))-3x/(x^2-9)}}} Factor the first denominator



{{{x/((x+1)(x-3))-3x/((x+3)(x-3))}}} Factor the second denominator




Take note that the LCD is {{{(x+1)(x+3)(x-3)}}} (the LCD is comprised of the unique factors of both denominators). So the goal is to get EVERY denominator equal to the LCD



{{{(x*highlight((x+3)))/((x+1)*highlight((x+3))(x-3))-3x/((x+3)(x-3))}}} Multiply both the numerator and denominator of the first fraction by {{{x+3}}} to get the denominator equal to the LCD



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



{{{(x^2+3x)/((x+1)(x+3)(x-3))-(3x*highlight((x+1)))/(highlight((x+1))(x+3)(x-3))}}} Multiply both the numerator and denominator of the second fraction by {{{x+1}}} to get the denominator equal to the LCD



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



Now that EVERY denominator is equal to the LCD, we can subtract the fractions:



{{{(x^2+3x-(3x^2+3x))/((x+1)(x+3)(x-3))}}} Combine the fractions by subtracting the numerators over the common denominator.



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



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



{{{(-2x^2)/(x^3+x^2-9x-9)}}} Multiply the terms out in the denominator (this step is optional)



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Answer:


So 


{{{x/(x^2-2x-3)-3x/(x^2-9)=(-2x^2)/((x+1)(x+3)(x-3))}}} where {{{x<>-3}}}, {{{x<>-1}}}, or {{{x<>0}}}



Or..


{{{x/(x^2-2x-3)-3x/(x^2-9)=(-2x^2)/(x^3+x^2-9x-9)}}} where {{{x<>-3}}}, {{{x<>-1}}}, or {{{x<>0}}}



Note: both answers are the same, but the second one is if you decided to expand the denominator