How to solve equations with the variable in a denominator: excluded values, clearing denominators with the LCD, and checking for extraneous solutions.
The variable in a denominator
A rational equation has the variable in at least one denominator, like
\[\frac{4}{x + 1} = \frac{3}{x}\]
You solve it almost like a linear equation: clear the denominators and carry on. But there's one extra step, and it's the one that costs the most points on tests: checking that your answer doesn't make a denominator zero.
The method
Extraneous solutions
When squared terms cancel
Sometimes clearing the fractions creates \(x^2\) terms. Often they cancel, and the equation is linear after all.
Practice
Quick check
A rational equation has excluded values \(x \neq 0\) and \(x \neq 3\), and solving it gives \(x = 3\). What's the conclusion?
\(x = 3\) makes a denominator zero, so it's extraneous. If it was the only candidate, the equation has no solution.
Quick check
What are the excluded values of \(\frac{2}{x} + \frac{1}{x - 6} = 1\)?
Every denominator must be nonzero: \(x \neq 0\) and \(x - 6 \neq 0\), so \(x \neq 6\).
What comes next
That's the end of Algebra 1. You can now work with monomials and polynomials, factor, handle rational expressions and solve linear and rational equations. Next up are inequalities, systems of equations and quadratic equations, the topics of Algebra 2, which is coming soon.
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