# PSAT Math: Rational Expressions and Equations

A rational expression is simply a ratio (or fraction) of polynomials. In other words, it is a fraction with a polynomial as the numerator and another polynomial as the denominator. The rules that govern fractions and polynomials also govern rational expressions, so if you know these well, you’ll be in good shape when you encounter one on Test Day.

There are a few important tidbits to remember about rational expressions; these are summarized here. They are also true for rational equations.

• For an expression to be rational, the numerator and denominator must both be polynomials.
• Like polynomials, rational expressions are also designated certain degrees based on the term with the highest variable exponent sum. For instance, the expression $\frac{1\phantom{\rule{.2em}{0ex}}-\phantom{\rule{.2em}{0ex}}2x}{3{x}^{2}+\phantom{\rule{.2em}{0ex}}3}$ has a first-degree numerator and a second-degree denominator.
• Because rational expressions by definition can have polynomial denominators, they will often be undefined for certain values. For example, the expression $\frac{x\phantom{\rule{.2em}{0ex}}-\phantom{\rule{.2em}{0ex}}4}{x\phantom{\rule{.2em}{0ex}}+\phantom{\rule{.2em}{0ex}}2}$ is defined for all values of x except –2. This is because when x = –2, the denominator of the expression is 0, which would make the expression undefined.
• Factors in a rational expression can be cancelled when simplifying, but under no circumstances can you do the same with individual terms. Consider, for instance, the expression $\frac{{x}^{2}-\phantom{\rule{.2em}{0ex}}x\phantom{\rule{.2em}{0ex}}-\phantom{\rule{.2em}{0ex}}6}{{x}^{2}+\phantom{\rule{.2em}{0ex}}5x\phantom{\rule{.2em}{0ex}}+\phantom{\rule{.2em}{0ex}}6}$. Many students will attempt to cancel the x 2, x, and 6 terms to give$\frac{1\phantom{\rule{.2em}{0ex}}-\phantom{\rule{.2em}{0ex}}1\phantom{\rule{.2em}{0ex}}-\phantom{\rule{.2em}{0ex}}1}{1\phantom{\rule{.2em}{0ex}}+\phantom{\rule{.2em}{0ex}}5\phantom{\rule{.2em}{0ex}}+\phantom{\rule{.2em}{0ex}}1}=\frac{–1}{7}$, which is never correct. Don’t even think about trying this on Test Day.
• Like fractions, rational expressions can be proper or improper. A proper rational expression has a lower-degree numerator than denominator $\left(\text{e}.\text{g}.,\frac{1\phantom{\rule{.2em}{0ex}}-\phantom{\rule{.2em}{0ex}}x}{{x}^{2}+\phantom{\rule{.2em}{0ex}}3}\right)$, and an improper one has a higher-degree numerator than denominator $\left(\text{e}.\text{g}.,\frac{{x}^{2}+\phantom{\rule{.2em}{0ex}}3}{1\phantom{\rule{.2em}{0ex}}-\phantom{\rule{.2em}{0ex}}x}\right)$. The latter can be simplified using polynomial long division.

### PSAT Math Practice: Solving Rational Equations

Rational equations are just like rational expressions except for one difference: They have an equal sign. They follow the same rules as rational expressions. The steps you take to solve the more friendly-looking linear equations apply to rational equations as well.

When solving rational equations, beware of extraneous solutions, solutions derived that don’t satisfy the original equation. This happens when the derived solution causes 0 in the denominator of any of the terms in the equation (because division by 0 is not possible).

Take the equation $\frac{1}{x\phantom{\rule{.2em}{0ex}}+\phantom{\rule{.2em}{0ex}}4}+\frac{1}{x\phantom{\rule{.2em}{0ex}}-\phantom{\rule{.2em}{0ex}}4}=\frac{8}{\left(x\phantom{\rule{.2em}{0ex}}+\phantom{\rule{.2em}{0ex}}4\right)\left(x\phantom{\rule{.2em}{0ex}}-\phantom{\rule{.2em}{0ex}}4\right)}$, for instance.

After multiplying both sides by the common denominator (x + 4)(x − 4), you have (x − 4) + (x + 4) = 8. Solving for x yields 2x = 8 which simplifies to x = 4. However, when 4 is substituted for x, you get 0 in the denominator of both the second and third terms of the equation, so 4 is an extraneous solution. Therefore, this equation is said to have no solution.

### Note

Whenever you encounter an equation with variables in a denominator or under a radical, make sure you check the solutions by plugging the values back into the original equation.

1. Which value of x satisfies the equation $\frac{4}{x}+\frac{2}{x\phantom{\rule{.2em}{0ex}}-\phantom{\rule{.2em}{0ex}}8}=\frac{–2}{{x}^{2}\phantom{\rule{.2em}{0ex}}-\phantom{\rule{.2em}{0ex}}8x}$ ?

A. –1
B. 1
C. 4
D. 5

Work through the Kaplan Method for Math step-by-step to solve this question. The following table shows Kaplan’s strategic thinking on the left, along with suggested math scratchwork on the right.

Strategic Thinking Math Scratchwork

Step 1: Read the question, identifying and organizing important information as you go

You’re asked to determine which value of x satisfies the equation.

Step 2: Choose the best strategy to answer the question

Start by identifying a common denominator. By factoring x out of the denominator of the term on the right side of the equation, you’ll see that the common denominator across all three terms is x(x − 8). Multiply the entire equation by this expression, then distribute properly and solve for x.

${x}^{2}\phantom{\rule{.2em}{0ex}}\text{–}\phantom{\rule{.2em}{0ex}}8x\phantom{\rule{.2em}{0ex}}\to \phantom{\rule{.2em}{0ex}}x\left(x\phantom{\rule{.2em}{0ex}}–\phantom{\rule{.2em}{0ex}}8\right)$

$x\left(x\phantom{\rule{.2em}{0ex}}–\phantom{\rule{.2em}{0ex}}8\right)\left(\frac{4}{x}\text{+}\frac{2}{x\phantom{\rule{.2em}{0ex}}–\phantom{\rule{.2em}{0ex}}8}\text{=}\frac{–2}{x\left(x\phantom{\rule{.2em}{0ex}}–\phantom{\rule{.2em}{0ex}}8\right)}\right)$
$\begin{array}{rcl}4\left(x\phantom{\rule{.2em}{0ex}}–\phantom{\rule{.2em}{0ex}}8\right)\phantom{\rule{.2em}{0ex}}\text{+}\phantom{\rule{.2em}{0ex}}2x& =& –2\\ 4x\phantom{\rule{.2em}{0ex}}–\phantom{\rule{.2em}{0ex}}32\phantom{\rule{.2em}{0ex}}\text{+}\phantom{\rule{.2em}{0ex}}2x& =& –2\\ 6x& =& 30\\ x& =& 5\end{array}$

Step 3: Check that you answered the right question

Because “no solution” is not a choice, you can be confident that 5 is not an extraneous solution. Choice (D) is therefore correct.