Question 1151158
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Great job so far getting the proper factorizations. 


Examine the denominators
(x+2)(x+3) and (x+2)(x+1)


We have the unique factors: (x+2), (x+3) and (x+1)
These unique factors form the Lowest Common Denominator (LCD).
The LCD is (x+1)(x+2)(x+3). 


The goal is to get each fraction to have the LCD. For the first fraction we have (x+2)(x+3). What's missing to have this turn into the LCD? That would be (x+1).


So we must multiply top and bottom of the first fraction by (x+1) so we get to the LCD.
{{{(x)/((x+2)(x+3)) = (x*highlight((x+1)))/((x+2)(x+3)highlight((x+1))) = (x^2+x)/((x+1)(x+2)(x+3))}}}


The first fraction {{{(x)/((x+2)(x+3))}}} turns into {{{(x^2+x)/((x+1)(x+2)(x+3))}}} after performing this step.


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Now turn to the second fraction 
{{{(2)/((x+2)(x+1))}}}
which is the same as 
{{{(2)/((x+1)(x+2))}}}



What's missing from the denominator to have it equal to the LCD? That would be (x+3). 


Multiply top and bottom of this fraction by (x+3) to get...
{{{(2)/((x+1)(x+2)) = (2*highlight((x+3)))/((x+1)(x+2)*highlight((x+3))) = (2x+6)/((x+1)(x+2)(x+3))}}}


So {{{(2)/((x+1)(x+2))}}} turns into {{{(2x+6)/((x+1)(x+2)(x+3))}}}

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To summarize so far:
{{{(x)/((x+2)(x+3))}}} turns into {{{(x^2+x)/((x+1)(x+2)(x+3))}}}


{{{(2)/((x+1)(x+2))}}} turns into {{{(2x+6)/((x+1)(x+2)(x+3))}}}



From here, we can subtract the fractions as the denominators are both the same.
{{{(x)/((x+2)(x+3))-(2)/((x+1)(x+2)) = (x^2+x)/((x+1)(x+2)(x+3))-(2x+6)/((x+1)(x+2)(x+3))}}}


{{{(x)/((x+2)(x+3))-(2)/((x+1)(x+2)) = (x^2+x-(2x+6))/((x+1)(x+2)(x+3))}}}


{{{(x)/((x+2)(x+3))-(2)/((x+1)(x+2)) = (x^2+x-2x-6)/((x+1)(x+2)(x+3))}}}


{{{(x)/((x+2)(x+3))-(2)/((x+1)(x+2)) = (x^2-x-6)/((x+1)(x+2)(x+3))}}}


{{{(x)/((x+2)(x+3))-(2)/((x+1)(x+2)) = ((x-3)(x+2))/((x+1)(x+2)(x+3))}}}


{{{(x)/((x+2)(x+3))-(2)/((x+1)(x+2)) = ((x-3)*highlight((x+2)))/((x+1)*highlight((x+2))(x+3))}}}


{{{(x)/((x+2)(x+3))-(2)/((x+1)(x+2)) = ((x-3)*cross((x+2)))/((x+1)*cross((x+2))(x+3))}}}


{{{(x)/((x+2)(x+3))-(2)/((x+1)(x+2)) = (x-3)/((x+1)(x+3))}}}


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


{{{(x)/((x+2)(x+3))-(2)/((x+1)(x+2)) = (x-3)/(x^2+4x+3)}}}


Your answer would either be {{{(x-3)/((x+1)(x+3))}}} or {{{(x-3)/(x^2+4x+3)}}} depending on if your teacher wants you keep the denominator factored or not.
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