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Adding Rational Expressions

The process for adding two rational expressions is almost identical to the process for adding two fractions.

  1. Reduce all of the fractions. You don't have to do this but it'll make your life easier later on.
  2. Find the least common denominator.
  3. Make each rational expression into an equivalent expression with the least common denominator.
  4. Make a new fraction by adding the numerators and keeping the least common denominator.
  5. Reduce the fraction from step four.

Example 1

Simplify $$\frac{x - 1}{x - 3}+\frac{x}{x + 4}$$

Both of those fractions are already reduced (none of the parts can be reduced further and there aren't any factors that can be canceled) so we can go right to finding the least common denominator. That's going to be $(x - 3)(x + 4)$. To get each fraction with the same denominator we have to multiply the numerator and denominator of the first expression by $x + 4$ and the second one by $x - 3$.

$$\frac{(x + 4)(x - 1)}{(x + 4)(x - 3)}+\frac{x(x - 3)}{(x + 4)(x - 3)}$$ $$\frac{x^2 + 3x - 4}{(x + 4)(x - 3)}+\frac{x^2 - 3x}{(x + 4)(x - 3)}$$ $$\frac{x^2 + 3x - 4 + x^2 - 3x}{(x + 4)(x - 3)}$$ $$\frac{2x^2 - 4}{(x + 4)(x - 3)}$$

Notice how I didn't multiply out the denominator. The next step is to reduce the expression which means first factoring the numerator and the denominator. Since I need the denominator factored to do the reducing, there was no point in multiplying it out. In this case, the numerator can't be factored so there's nothing for us to cancel, i.e. the expression can't be reduced any further. Since that's the case, I'll multiply out the denominator to get our final answer.

$$\frac{2x^2 - 4}{x^2 + 4x - 12}$$

Example 3

Simplify $$\frac{x^2 + x - 2}{2x^2 + 5x + 2}+\frac{x^2 - 5x - 6}{2x^2 - 7x - 4}$$

First we need to factor all of the polynomials in both expressions.

$$\frac{(x + 2)(x - 1)}{(x + 2)(2x + 1)}+\frac{(x - 6)(x + 1)}{(2x + 1)(x - 4)}$$

Since the numerator and denominator of the first expression both have a $x + 2$ we can reduce the expression by canceling them both.

$$\frac{x - 1}{2x + 1}+\frac{(x - 6)(x + 1)}{(2x + 1)(x - 4)}$$

Now, looking at the factors in the denominator, the least common denominator is $(2x + 1)(x - 4)$. The second expression already has that as its denominator so all we need to do is multiply the numerator and denominator of the first expression by $x - 4$.

$$\frac{(x - 4)(x - 1)}{(x - 4)(2x + 1)}+\frac{(x - 6)(x + 1)}{(2x + 1)(x - 4)}$$ $$\frac{x^2 - 5x + 4}{(x - 4)(2x + 1)}+\frac{x^2 - 5x - 6}{(2x + 1)(x - 4)}$$ $$\frac{x^2 - 5x + 4 + x^2 - 5x - 6}{(x - 4)(2x + 1)}$$ $$\frac{2x^2 - 10x - 2}{(x - 4)(2x + 1)}$$

We can factor a 2 out of the numerator which would leave us with $2(x^2 - 5x - 1)$. That can't be factored any further so we can get the final answer by multiplying out the denominator.

$$\frac{2x^2 - 10x - 2}{2x^2 - 7x - 4}$$

Example 2

Simplify $$\frac{x + 1}{x^2 - 11x + 30}+\frac{3x + 6}{x^2 - 5x - 6}$$

In this case, both of the denominators and the second numerator can be factored so that's going to be our first step.

$$\frac{x + 1}{(x - 6)(x - 5)}+\frac{3(x + 2)}{(x - 6)(x + 1)}$$

None of the numerators and denominators have any terms in common so we can't cancel anything. Dont' be tempted by the two $x + 1$ terms. Since they're in different fractions and we're adding, they can't be canceled. (If we were multiplying the two fractions then we could cancel them.) Since the expressions can't be reduced any further, we need to move on to finding the least common denominator.

Finding the least common denominator here is a lot like it is with fractions. We need to take all of the common factors but we don't need to include duplicates. That means, in this case, our least common denominator will be $(x - 6)(x + 1)(x - 5)$. Notice how I only included the $(x - 6)$ once even though it appears in both denominators. That's what I meant by not including duplicates - we only need to add it once. Now we need to multiply the numerator and denominator of the first fraction by $(x + 1)$ and the second by fraction by $(x - 5)$.

$$\frac{(x + 1)(x + 1)}{(x + 1)(x - 6)(x - 5)}+\frac{3(x - 5)(x + 2)}{(x - 5)(x - 6)(x + 1)}$$ $$\frac{x^2 + 2x + 1}{(x + 1)(x - 6)(x - 5)}+\frac{3x^2 - 9x - 30}{(x - 5)(x - 6)(x + 1)}$$ $$\frac{x^2 + 2x + 1 + 3x^2 - 9x - 30}{(x + 1)(x - 6)(x - 5)}$$ $$\frac{4x^2 - 7x - 29}{(x + 1)(x - 6)(x - 5)}$$

The numerator can't be factored any further so we won't be able to cancel any factors with the denominator to reduce the expression. To get the final answer, we just need to multiply out the expression in the denominator.

$$\frac{4x^2 - 7x - 29}{(x^2 - 5x - 6)(x - 5)}$$ $$\frac{4x^2 - 7x - 29}{(x^3 - 10x^2 + 19x + 30)}$$

Example 4

Simplify $$\frac{3}{x^2 - 2x + 1}+\frac{2}{x^2 + x - 2}$$

The first step is to factor all of the polynomials

$$\frac{3}{(x - 1)(x - 1)}+\frac{2}{(x - 1)(x + 2)}$$ $$\frac{3}{(x - 1)^2}+\frac{2}{(x - 1)(x + 2)}$$

None of the numerators and denominators have any terms in common so we can't cancel anything which brings us to finding the least common denominator. In this case, that's going to be $(x - 1)^2(x + 2)$. Notice how I kept the exponent of the $x - 1$ term. Just like when you're finding the least common multiple of a number, you take the highest exponent of each of the factors. Now that we've got that, we can multiply each of the two expressions so that they have the same denominator.

$$\frac{3(x + 2)}{(x + 2)(x - 1)(x - 1)}+\frac{2(x - 1)}{(x - 1)(x + 2)(x - 1)}$$ $$\frac{3x + 6}{(x + 2)(x - 1)(x - 1)}+\frac{2x - 2}{(x - 1)(x + 2)(x - 1)}{ = }$$ $$\frac{3x + 6 + 2x - 2}{(x + 2)(x - 1)(x - 1)}$$ $$\frac{5x + 4}{(x + 2)(x - 1)(x - 1)}$$

The numerator can't be factored any further and doesn't match any of the factors in the denominator so we can't reduce the expression any further. To get the final answer, we just need to multiply out the expression in the denominator.

$$\frac{5x + 4}{x^3 - 3x + 2}$$
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