Lesson: February 10, 2025. Quiz: February 13, 2025.

  • Solve rational equations in one variable. Identify extraneous solutions to rational equations and give examples of how they arise.

Assignment

Solving a Rational Equation

The general strategy is to eliminate your denominators so that you are back in polynomial territory.

\[\begin{align} \frac{1}{x+4} &= 2 \\[1em] 1 &= 2(x+4) \\ 0 &= 2x + 7 \\ x &= -\frac{7}{2} \end{align}\]

The catch is that you have to watch out for extraneous solutions. Your original equation will have restrictions on the domain, so any solution that falls outside the domain is not actually a solution, despite appearances.

\[\begin{align} \frac{1}{x-5} + \frac{x}{x-3} &= \frac{2}{x^2-8x+15} \\[1em] \frac{(x-3)+(x^2-5x)}{(x-5)(x-3)} &= \frac{2}{(x-5)(x-3)} \\[1em] x^2 - 4x - 3 &= 2 \\ x^2 - 4x - 5 &= 0 (x-5)(x+1) &= 0 \\ x &=\{-1,5\} \end{align}\]

Based on the original domain, $x\neq5$ since it’s an extraneous solution and we are left only with $x=-1$.

Here’s another example from the book (4B on page 227), but worked out a different way. Pay careful attention to what happens from $(12)$ to $(13)$.

\[\begin{align} \frac{3}{x-3} &= \frac{x}{x-3} - \frac{x}{4} \\[1em] \frac{3}{x-3} - \frac{x}{x-3} &= - \frac{x}{4} \\[1em] \frac{3 - x}{x-3} &= -\frac{x}{4} \\[1em] -\frac{x-3}{x-3} &= -\frac{x}{4} \\[1em] -1 &= -\frac{x}{4} \\[1em] x&=4 \end{align}\]

The way that problem was worked out above didn’t lead to any extraneous solutions, but will if you don’t move your expressions with like denominators to the same side.