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Subtracting Rational Expressions

Fundamentally, the process for subtracting two rational expressions follows the same steps as the process for adding only with subtraction in step four. We've found that students often have challenges with the subtraction process so we've included it as a separate section to give us a chance to look at it in more detail.

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 subtracting 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 so we can go right to finding the least common denominator. In this case, 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 $x - 3$.

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

If you aren't clear on how I got the first numerator on the second line, take a look at our page on subtracting polynomials.

The numerator factors to 2(3x - 2). Since there's neither a 2 nor a $3x - 2$ in the denominator, there's nothing we can cancel here. This makes our final answer:

$$\frac{6x - 4}{x^2 + 4x - 12}$$

Example 2

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 $x - 4$.

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

The numerator can't be factored any further and there's nothing that goes into 10 evenly in the denominator so that expression is our final answer.

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