How do you simplify #\frac { 1} { ( x + 1) ( x + 2) } - \frac { 1} { ( x + 1) ( x + 3) } + \frac { 1} { ( x + 2) ( x + 3) } #?

1 Answer
Jun 9, 2017

See a solution process below:

Explanation:

To add and subtract fractions you must have each fraction over a common denominator.

For this problem the common denominator is:

#(x + 1)(x + 2)(x + 3)#

First, we can multiply each term by the appropriate form of #1# to put each fraction over the common denominator:

#((x + 3)/(x + 3) xx 1/((x + 1)(x + 2))) - ((x + 2)/(x + 2) xx 1/((x + 1)(x + 3))) + ((x + 1)/(x + 1) xx 1/((x + 2)(x + 3))) =>#

#(x + 3)/((x + 1)(x + 2)(x + 3)) - (x + 2)/((x + 1)(x + 2)(x + 3)) + (x + 1)/((x + 1)(x + 2)(x + 3))#

Next, we can add and subtract the numerators over the now common denominator:

#((x + 3) - (x + 2) + (x + 1))/((x + 1)(x + 2)(x + 3)) =>#

#(x + 3 - x - 2 + x + 1)/((x + 1)(x + 2)(x + 3)) =>#

#(x - x + x + 3 - 2 + 1)/((x + 1)(x + 2)(x + 3)) =>#

#(x + 2)/((x + 1)(x + 2)(x + 3))#

We can now cancel common terms in the numerator and denominator:

#color(red)(cancel(color(black)((x + 2))))/((x + 1)color(red)(cancel(color(black)((x + 2))))(x + 3)) =>#

#1/((x + 1)(x + 3))#

However, because we cannot divide by zero we need the following exclusions from the original expression:

Where #x != -1# and #x != -2# and #x != -3#