Adding and Subtracting Rational Expressions

1. Factor the numerator and denominator of each fraction.

2. State all of the restrictions (denominators cannot be equal to 0).

3. If possible, simplify the common factors in each of the fractions.

4. Find a common denominator.

5. Perform addition or subtraction on the numerator.

6. If necessary, simplify the common factors again.

Consider the addition of the following rational expressions:

|\displaystyle \frac{x}{x-2} + \frac{2}{x-1}|.

1. The polynomials in the numerator and in the denominator are already factored.

2. State all of the restrictions. That is, find the values of |x| where the denominators would have a value of |0| .

   |x-2 \neq 0 \to x \neq 2|
   |x-1 \neq 0 \to x \neq 1|

3. There are no common factors to be simplified in each of the fractions.

4. Find a common denominator.

   The denominator of the first fraction is missing |(x-1)| and the denominator of the second fraction is missing |(x-2)| so that they have the same denominator. Transform the two fractions into equivalent fractions so that they have the same denominator.

   | \displaystyle \frac{x(x-1)}{(x-2)(x-1)} + \frac{2(x-2)}{(x-1)(x-2)}|

5. Add the two fractions.
   |\displaystyle \frac{x(x-1) + 2(x-2)}{(x-2)(x-1)}|
   
   |=\displaystyle \frac{x^2 - x + 2x - 4}{(x-2)(x-1)}|
   
   |=\displaystyle \frac{x^2 + x - 4}{(x-2)(x-1)}|

6. There are no common factors so the simplification stops here.

Answer : Write the simplified rational expression. Finally, do not forget to include the restrictions found initially.

|\displaystyle\frac{x}{x-2} + \frac{2}{x-1}= \frac{x^2 + x -4}{(x-2)(x-1)}| or |x \neq 1| and |x \neq 2|

Consider the addition following rational expressions:
|\displaystyle \frac{x-3}{x^2+3x+2} + \frac{x-2}{x^2-1}|

1. The two polynomials found in the denominators can be factored. First, factor |x^2+3x+2| with a trinomial method and then |x^2-1| will be factored using a difference of squares.
   |x^2+3x+2 = (x+1)(x+2)|
   |x^2-1 = (x+1)(x-1)|

   This gives the following two fractions:
   |\displaystyle \frac{x-3}{(x+1)(x+2)} + \frac{x-2}{(x+1)(x-1)}|

2. Next, state the restrictions. Find the values of |x| where the denominators would have a value of |0| .
   |x+1 \neq 0 \to x \neq -1|
   |x+2 \neq 0 \to x \neq -2|
   |x-1 \neq 0 \to x \neq 1|

3. There are no common factors to simplify in both fractions.

4. Find a common denominator.

   The denominator of the first fraction is missing |(x-1)| and the denominator of the second fraction is missing |(x+2)| so that they have the same denominator. Transform the two fractions into equivalent fractions so that they have the same denominator.
   |\displaystyle \frac{(x-3)(x-1)}{(x+1)(x+2)(x-1)} + \frac {(x-2)(x+2)}{(x+1)(x-1)(x+2)}|

5. Add the two fractions.

   |\displaystyle \frac{(x-3)(x-1) + (x-2)(x+2)}{(x+1)(x+2)(x-1)}|

   |=\displaystyle \frac{(x^2 - x - 3x + 3) + (x^2 + 2x - 2x - 4)}{(x+1)(x+2)(x-1)}|

   |=\displaystyle \frac{(x^2 - 4x + 3) + (x^2 - 4)}{(x+1)(x+2)(x-1)}|

   |=\displaystyle \frac{x^2 - 4x + 3 + x^2 - 4}{(x+1)(x+2)(x-1)}|

   |=\displaystyle \frac{2x^2 - 4x - 1}{(x+1)(x+2)(x-1)}|

6. There is no common factor so the simplification stops here.

Answer : Now, write the simplified rational expression, and do not forget to include the restrictions found initially.

|\displaystyle \frac{x-3}{x^2+3x+2} + \frac{x-2}{x^2-1} = \frac{2x^2 - 4x - 1}{(x+1)(x+2)(x-1)}| or |x \neq -2| , |x \neq -1| and |x\neq 1|

Consider the subtraction of the following rational expressions:
|\displaystyle \frac{x+1}{x^2+2x+1} - \frac{x+3}{x^2+4x+3}|

1. The denominators’ polynomials can be factored. First, factor |x^2+2x+1| with a trinomial method and then factor |x^2+4x+3| also using a trinomial method.
   |x^2+2x+1 = (x+1)(x+1)|
   |x^2+4x+3 = (x+1)(x+3)|

   This produces the following two fractions:
   |\displaystyle \frac{(x+1)}{(x+1)(x+1)} - \frac{(x+3)}{(x+3)(x+1)}|

2. State the restrictions. Find the values of |x| where the denominators would have a value of |0| .
   |x+1 \neq 0 \to x \neq -1|
   |x+3 \neq 0 \to x \neq -3|

3. Common factors can be simplified.
   
   |\displaystyle \frac{\color{red}{(x+1)}}{\color{red}{(x+1)}(x+1)} - \frac{\color{blue}{(x+3)}}{\color{blue}{(x+3)}(x+1)}|
   |=\displaystyle \frac{1}{(x+1)} - \frac{1}{(x+1)}|

4. Both fractions have the same denominator.

5. Subtract the two fractions.
   |\displaystyle \frac{1-1}{(x+1)} = \frac{0}{(x+1)} = 0|

6. There is nothing else that can be simplified.

Answer: Now, write the obtained answer and do not forget to include the restrictions found initially.
|\displaystyle \frac{x+1}{x^2+2x+1} - \frac{x+3}{x^2+4x+3}= 0| or |x\neq -1| and |x\neq -3|

Consider the subtraction of the following rational expressions:
|\displaystyle \frac{x-2}{x^2+4x+3} - \frac{2x+1}{x+3}|

1. The polynomial of the first denominator can be factored. First, factor |x^2+4x+3| with a trinomial technique.
   |x^2+4x+3 = (x+1)(x+3)|

   This gives the following two fractions:
   |\displaystyle \frac{(x-2)}{(x+1)(x+3)} - \frac{(2x+1)}{(x+3)}|

2. Next, state the restrictions. Find the values of |x| where the denominators would have a value of |0| .
   |x+1 \neq 0 \to x \neq -1|
   |x+3 \neq 0 \to x \neq -3|

3. There are no common factors.

4. Find a common denominator.

   The denominator of the second fraction is missing |(x+1)| so that the denominators are the same. Transform the two fractions into equivalent fractions so that they have the same denominator.
   |\displaystyle \frac{(x-2)}{(x+1)(x+3)} - \frac{(2x+1)(x+1)}{(x+3)(x+1)}|

5. Subtract the two fractions.
   |\displaystyle \frac{(x-2)-(2x+1)(x+1)}{(x+1)(x+3)}|
   |=\displaystyle \frac{x-2-(2x^2+2x+x+1)}{(x+1)(x+3)}|
   |=\displaystyle \frac{-2x^2-2x-3}{(x+1)(x+3)}|

6. There is nothing else that can be simplified.

Answer : Finally, write the obtained answer and do not forget to include the restrictions found initially.
|\displaystyle \frac{x-2}{x^2+4x+3} - \frac{2x+1}{x+3} = \frac{-2x^2-2x-3}{(x+1)(x+3)}| or |x\neq -3| and |x\neq -1|