Solving Rational Equations

A civil engineer calculating water flow rates uses equations where the flow speed appears in the denominator. When she sets two flow expressions equal and solves, she gets two solutions. One of them, when plugged back in, makes a denominator equal zero — which is physically meaningless. She throws it out. The other solution is the answer. That discarded solution has a name: an extraneous solution. And the technique she used to solve the equation in the first place — multiplying both sides by the LCD — is exactly what this lesson is about.

A rational equation is any equation containing at least one rational expression, meaning a fraction with a variable in the denominator. The strategy for solving them is to eliminate every denominator at once by multiplying both sides of the equation by the least common denominator. This turns a rational equation into a polynomial equation you already know how to solve.

Here is the critical rule: after you solve, you must substitute every solution back into the original equation. If a solution makes any denominator equal to zero, it is extraneous — it is not a real solution, and you must discard it. An extraneous solution is not a mistake in your algebra. It arises because multiplying both sides by an expression containing a variable can introduce solutions that were never valid in the first place.

Start with a straightforward example. Solve:

The denominators are $x$ and $2$, so the LCD is $2x$. Multiply every term by $2x$.

Now here is where rational equations get interesting. Sometimes the algebra produces a solution that looks fine until you check it. Solve:

Both fractions share the denominator $x - 3$, so the LCD is $x - 3$.

Wait — the variable disappeared entirely and left a contradiction. That means there is no solution. But suppose the algebra had instead produced a specific value like $x = 3$. Plugging $x = 3$ into the original makes both denominators $3 - 3 = 0$, which is undefined. Any such value must be excluded before you even start: write the restriction $x \neq 3$ first, and if your algebra later produces 3, discard it immediately.

Here is a fuller example where an extraneous solution actually appears. Solve:

The denominators are $x + 2$ and $3$. The LCD is $3(x+2)$. The restriction is $x \neq -2$.

The check here is confirming that neither solution equals $-2$. When the quadratic factors cleanly and one root equals the restricted value, that root is extraneous and must be discarded. The other root, if it passes the check, is the only solution.

To visualize what is happening, the graph below shows the left side and right side of the original equation in that last example as two separate functions. Their intersections are the solutions.

Notice the vertical asymptote at $x = -2$. The graph confirms two intersection points, consistent with the two solutions the algebra produced.