Question 264983
{{{(x+1)/(x+3)=((x^2-11x)/(x^2+x-6))+(x-3)/(x-2)}}}
must make the common denominator the same to add/subtract together
just take your time and focus on just the denominators
(x^2+x-6) , (x-2) , (x+3)
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its common for books to provide the simpilar common denominator in the question.
if you can factor {{{(x^2+x-6)}}} it equals {{{((x-2)(x+3))}}}
if you arent great at factoring  then first just try{{{((x-2)(x+3))}}} to see if it = {{{(x^2+x-6)}}} since it is, you will multiply by the common denominator of{{{((x-2)(x+3))}}}
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Now heres what it should look like:
{{{((x+1)/(x+3))(((x-2)(x+3))/1)=(((x^2-11x)/(x^2+x-6))(((x-2)(x+3))/1)))}}}+{{{((x-3)/(x-2))(((x-2)(x+3))/1))}}}
Now your ready to do cross cancelations and remember{{{(x^2+x-6)=((x-2)(x+3))}}}
{{{((x+1)(x-2))=(x^2-11x)+((x-3)(x+3))}}}
Use FOIL method (First Outer Inner Last)
{{{(x^2+x-2x-2)=(x^2-11x)+(x^2-3x+3x-9)}}}
{{{x^2-x-2=x^2-11x+x^2-9}}}
{{{(x^2-x-2)=(2x^2-11x-9)}}}
{{{(x^2-x-2)-(x^2-x-2)=(2x^2-11x-9)-(x^2-x-2)}}}
{{{0=(2x^2-11x-9)-(x^2-x-2)}}}
{{{0=2x^2-11x-9-x^2+x+2)}}}
{{{0=x^2-10x-7)}}} Use the Quadratic formula: Ax^2+Bx+C=0 where A=1 B=-10 C=-7
the answer is: 10.6568542494924, -0.656854249492381
*[invoke quadratic "x", 1, -10, -7]