Lesson 8.6: Solve Rational Equations

New Concept

To solve rational equations, we eliminate fractions by multiplying by the LCD. This simplifies the problem, but we must check for extraneous solutions—values that make a denominator zero in the original equation.

What’s next

Let's put this into practice. You'll work through interactive examples, followed by a series of practice cards to master solving these equations.

Property

A rational equation is two rational expressions connected by an equal sign. You must make sure to know the difference between rational expressions and rational equations. The equation contains an equal sign.

Examples

• The equation $\frac{x}{4} + \frac{1}{2} = \frac{7}{4}$ is a rational equation because two rational expressions are set equal.
• $\frac{5}{z+2} = \frac{2}{z}$ is a rational equation where the variable appears in the denominators.
• The statement $\frac{a}{a-5} + 3 = \frac{5}{a-5}$ is an equation because of the equal sign, which we can solve for $a$.

Property

An extraneous solution to a rational equation is an algebraic solution that would cause any of the expressions in the original equation to be undefined. We note any possible extraneous solutions, $c$, by writing $x \neq c$ next to the equation.

Examples

• For the equation $\frac{2x}{x-5} = \frac{10}{x-5}$, solving gives $2x=10$, so $x=5$. However, $x=5$ is an extraneous solution because it makes the denominator zero.
• In $\frac{y^2}{y-3} = \frac{9}{y-3}$, we get $y^2 = 9$, which yields $y=3$ and $y=-3$. The solution $y=3$ is extraneous, but $y=-3$ is a valid solution.
• The equation $\frac{x+7}{(x-2)(x-1)} = \frac{8}{x-2} - \frac{6}{x-1}$ leads to the algebraic solution $x=2$. Since $x=2$ makes the original denominators zero, it is an extraneous solution, and the equation has no solution.

Explanation

An extraneous solution is a 'fake' answer. It looks correct after simplifying, but it breaks the original equation by making a denominator zero. This is why you must always check your solutions against the initial restrictions.

Property

To solve equations with rational expressions:

1. Note any value of the variable that would make any denominator zero.
2. Find the least common denominator of all denominators in the equation.
3. Clear the fractions by multiplying both sides of the equation by the LCD.
4. Solve the resulting equation.
5. Check. If any values found in Step 1 are algebraic solutions, discard them. Check any remaining solutions in the original equation.

Examples

• To solve $\frac{1}{y} + \frac{1}{4} = \frac{5}{8}$, we note $y \neq 0$. The LCD is $8y$. Multiplying gives $8 + 2y = 5y$, so $8 = 3y$, and $y = \frac{8}{3}$.
• Solve $1 - \frac{3}{x} = \frac{10}{x^2}$. With $x \neq 0$ and LCD $x^2$, we get $x^2 - 3x = 10$, or $x^2 - 3x - 10 = 0$. Factoring gives $(x-5)(x+2)=0$, so $x=5$ or $x=-2$.
• Solve $\frac{4}{p+2} + \frac{1}{p-2} = \frac{8}{p^2-4}$. Note $p \neq \pm 2$. The LCD is $(p+2)(p-2)$. This simplifies to $4(p-2) + 1(p+2) = 8$, which gives $5p - 6 = 8$, so $p = \frac{14}{5}$.

Explanation

The main strategy is to 'clear the fractions' by multiplying the entire equation by the Least Common Denominator (LCD). This transforms the rational equation into a simpler polynomial equation that you already know how to solve.

Property

Many formulas used in business, science, economics, and other fields use rational equations to model the relation between two or more variables.
We will now see how to solve a rational equation for a specific variable.
The same strategy applies: identify restrictions, find the LCD, clear fractions, and then use algebraic properties to isolate the desired variable.

Examples

• To solve the physics formula $I = \frac{V}{R}$ for $R$, note $R \neq 0$. Multiply by the LCD, $R$, to get $IR = V$. Then, divide by $I$ to isolate $R$: $R = \frac{V}{I}$.
• Solve the acceleration formula $a = \frac{v-u}{t}$ for $t$. Note $t \neq 0$. Multiply by $t$ to get $at = v-u$. Then divide by $a$ to find $t = \frac{v-u}{a}$.
• To solve $\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}$ for $R$, the LCD is $R R_1 R_2$. Multiplying yields $R_1 R_2 = R R_2 + R R_1$. Factor out $R$ to get $R_1 R_2 = R(R_2 + R_1)$, so $R = \frac{R_1 R_2}{R_1 + R_2}$.

Explanation

This is like rearranging a formula. Instead of finding a number, you're isolating one variable in terms of the others. The process of clearing fractions by multiplying by the LCD still works perfectly, even with multiple variables.