Question 528248
<pre>
{{{(2x+1)/(x+3)}}} + {{{(3x)/(x-2)}}} + 2x

Write the term 2x as {{{2x/1}}}, so all terms will be fractions:

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

The LCD is {x+3)(x-2).  We must get all three fractions to have
the LCD for the denominator by multiplying by unit fractions.

The denominator of the first fraction needs to be multiplied by 
(x-2) to become the LCD, so we multiply the first fraction by 
the unit fraction {{{red((x-2)/(x-2))}}}

The denominator of the second fraction needs to be multiplied by 
(x+3) to become the LCD, so we multiply the second fraction by 
the unit fraction {{{red((x+3)/(x+3))}}}

The denominator of the third fraction needs to be multiplied by 
(x+3)(x-2) to become the LCD, so we multiply the first fraction by 
the unit fraction {{{red(((x+3)(x-2))/((x+3)(x-2)))}}}


{{{red((x-2)/(x-2))}}}·{{{(2x+1)/(x+3)}}} + {{{red((x+3)/(x+3))}}}·{{{(3x)/(x-2)}}} + {{{red(((x+3)(x-2))/((x+3)(x-2)))}}}·{{{2x/1}}}


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

Multiply the tops out but not the bottoms:

{{{(2x^2-3x-2)/((x-2)(x+3))}}} + {{{(3x^2+9x)/((x+3)(x-2))}}} + {{{((x^2+x-6)(2x))/((x+3)(x+2))}}}

{{{(2x^2-3x-2)/((x-2)(x+3))}}} + {{{(3x^2+9x)/((x+3)(x-2))}}} + {{{(2x^3+2x^2-12x)/((x+3)(x+2))}}}

Since all the deminators are equal to the LCD, we write the sum of the
numerators over the LCD:

{{{((2x^2-3x-2)+(3x^2+9x)+(2x^3+2x^2-12x))/((x+2)(x+3))}}}

Remove the parentheses on top:

{{{(2x^2-3x-2+3x^2+9x+2x^3+2x^2-12x)/((x+2)(x+3))}}}

Collect like terms in the top and arrange in descending order:

{{{(2x^3+7x^2-6x-2)/((x+2)(x+3))}}}

Edwin</pre>