Lesson 7.4: Solve Rational Equations

New Concept

We will solve rational equations—equations with variables in the denominator. Our main strategy is to multiply by the LCD to clear the fractions, solve the resulting equation, and then check for extraneous solutions.

What’s next

Next, you will walk through interactive examples and then apply these skills in a series of practice problems and challenge exercises.

Property

A rational equation is an equation that contains a rational expression. The equation contains an equal sign, which distinguishes it from a rational expression.

Examples

• The equation $\frac{x}{4} + \frac{1}{2} = 2$ is a rational equation. We can solve it to find that $x=6$.

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

• Solve $\frac{x^2}{x-2} = \frac{4}{x-2}$. Multiplying by $x-2$ gives $x^2 = 4$, so $x=2$ or $x=-2$. Since $x=2$ makes the denominator zero, it is an extraneous solution. The only valid solution is $x=-2$.

• Solve $\frac{1}{x-4} - \frac{1}{x^2-16} = \frac{2}{x+4}$. The LCD is $(x-4)(x+4)$. The equation becomes $(x+4)-1 = 2(x-4)$, which gives $x+3 = 2x-8$, so $x=11$. There are no extraneous solutions.

Property

To solve equations with rational expressions:
Step 1. Note any value of the variable that would make any denominator zero.
Step 2. Find the least common denominator (LCD) of all denominators in the equation.
Step 3. Clear the fractions by multiplying both sides of the equation by the LCD.
Step 4. Solve the resulting equation.
Step 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 $\frac{1}{x} + \frac{1}{4} = \frac{5}{8}$, note $x \neq 0$. The LCD is $8x$. Multiply through: $8x(\frac{1}{x}) + 8x(\frac{1}{4}) = 8x(\frac{5}{8})$. This gives $8 + 2x = 5x$, so $3x=8$, and $x=\frac{8}{3}$.

• To solve $\frac{1}{y} + \frac{2}{5} = \frac{1}{3}$, note $y \neq 0$. The LCD is $15y$. Multiply through: $15 + 6y = 5y$. This gives $y = -15$.

Property

After clearing the fractions by multiplying by the LCD, the resulting equation may be a quadratic equation. To solve it, first write the quadratic equation in standard form ($ax^2 + bx + c = 0$). Then, solve the equation by factoring, using the square root property, or the quadratic formula. Finally, check for any extraneous solutions.

Examples

• Solve $1 - \frac{3}{x} = \frac{10}{x^2}$. With LCD $x^2$, we get $x^2 - 3x = 10$. In standard form, $x^2 - 3x - 10 = 0$. Factoring gives $(x-5)(x+2)=0$, so the solutions are $x=5$ and $x=-2$.

• Solve $1 + \frac{5}{y} = \frac{14}{y^2}$. The LCD is $y^2$, giving $y^2 + 5y = 14$. This becomes $y^2 + 5y - 14 = 0$, which factors to $(y+7)(y-2)=0$. The solutions are $y=-7$ and $y=2$.

Property

To solve a rational equation for a specific variable, follow these steps:

1. Note any value of the variables that would make any denominator zero.
2. Clear the fractions by multiplying both sides of the equation by the LCD.
3. Simplify and collect all terms with the target variable on one side of the equation.
4. If necessary, factor out the target variable.
5. Isolate the target variable by dividing both sides by the remaining terms or factors.

Examples

• Solve $m = \frac{y-5}{x-4}$ for $y$. Multiply both sides by $x-4$ to get $m(x-4) = y-5$. Then, add 5 to both sides to isolate $y$: $y = m(x-4) + 5$.

• Solve $\frac{1}{a} + \frac{1}{b} = c$ for $a$. The LCD is $ab$. Multiplying gives $b + a = abc$. To isolate $a$, get all $a$ terms on one side: $b = abc - a$. Factor out $a$: $b = a(bc-1)$. Divide to solve: $a = \frac{b}{bc-1}$.