Fractional Equations

What this move does

An equation with the unknown in a denominator is called a fractional equation. What we know how to solve are polynomial equations, so we “multiply away” the denominators:

\frac{x}{x - 1} = \frac{5 x - 3}{x ^{2} - 1} \times (x^{2} - 1) x (x + 1) = 5 x - 3

Find the lowest common denominator, and while you are at it, write down the “forbidden zone”: the values of $x$ that make a denominator $0$ .
Multiply both sides by the common denominator. Multiply every term, including the ones with no denominator.
Solve the polynomial equation.
Check: substitute each root into the common denominator. Any root that makes it $0$ is an extraneous root and must be rejected.

Why do extraneous roots appear? Because step 2 is not “free”. From $A = B$ you can always get $A \cdot M = B \cdot M$ . But going back, from $A \cdot M = B \cdot M$ to $A = B$ , needs $M \neq = 0$ . If $x$ happens to make $M = 0$ , then $A \cdot M = B \cdot M$ becomes $0 = 0$ , which is of course true, even if $A$ and $B$ mean nothing at all. An extraneous root is a zero of the very expression you multiplied by.

In the picture above, the dashed curve is the polynomial after clearing denominators. It crosses the horizontal axis once more than the solid curve, and that extra crossing lands exactly on the red vertical line.

Predict first

Solve $\frac{1}{x - 2} + 3 = \frac{x - 1}{x - 2}$ . Clearing denominators gives $1 + 3 (x - 2) = x - 1$ , so $x = 2$ . What, then, is the solution of the original equation?

x = 2

No solutionEvery number except

2

Choose an answer first; then you can see the result.

Work through one

Example: solve x/(x−2) = 8/(x²−4)

Step 1 / 6

Find the common denominator and write down the forbidden zone
$x^{2} - 4 = (x - 2) (x + 2)$ , so the lowest common denominator is $x^{2} - 4$ . Forbidden zone: $x \neq = 2$ and $x \neq = - 2$ , shown as two red vertical lines.

$\frac{x}{x - 2} = \frac{8}{( x - 2 ) ( x + 2 )}$
Multiply both sides by the common denominator
Solve the polynomial equation
Check x = 2
Check x = −4
Write the conclusion

Extraneous roots made to order

Extraneous roots are not just bad luck. They appear under one precise condition: a root of the polynomial equation equals a number in the forbidden zone. So problems with a letter in them often ask it the other way round: for which value of the letter does the equation have an extraneous root?

The equation below has a parameter $k$ . Clearing denominators gives $x = 2 (x - 3) + k$ , that is, $x = 6 - k$ . Drag $k$ and this root slides along the horizontal axis.

Predict first

For which value of $k$ does the equation $\frac{x}{x - 3} = 2 + \frac{k}{x - 3}$ in $x$ have no solution?

Choose an answer first; then you can see the result.

Messy denominators? Substitute first

Some fractional equations become high-degree once you clear the denominators, but one piece keeps coming back. Treat that piece as a single letter and the equation gets simple at once.

Example: solve (x/(x−1))² − 5·x/(x−1) + 6 = 0

Step 1 / 4

Find the piece that keeps coming back
$\frac{x}{x - 1}$ appears twice. Let $t = \frac{x}{x - 1}$ . Do not forget the forbidden zone: $x \neq = 1$ .

$(\frac{x}{x - 1})^{2} - 5 \cdot \frac{x}{x - 1} + 6 = 0$
Solve the equation in t
Go back to x
Check

When things go wrong

Break it

Counterexample lab: remove one condition

Conclusion: the root of the polynomial equation is a root of the original equation

$\frac{3}{x - 1} = \frac{2}{x + 1}$ : clearing denominators gives $3 (x + 1) = 2 (x - 1)$ , so $x = - 5$ . Click to check it: neither denominator is $0$ , so it is a genuine root.

Practice

Exercise 1Basic

Solve $\frac{2}{x - 3} = \frac{1}{x + 1}$ .

Exercise 2Basic

Solve $\frac{x}{x - 4} - 3 = \frac{4}{x - 4}$ . (If there is a solution, enter it. If there is none, type “no solution”.)

Exercise 3Advanced

Solve $\frac{x}{x + 1} = \frac{2}{x ^{2} - 1}$ .

Exercise 4Challenge

The equation $\frac{1}{x - 3} + \frac{k x}{x ^{2} - 9} = \frac{2}{x + 3}$ in $x$ has an extraneous root. Find all possible values of $k$ . (Separate them with commas.)

What this move does

Work through one

Example: solve x/(x−2) = 8/(x²−4)

Extraneous roots made to order

Messy denominators? Substitute first

Example: solve (x/(x−1))² − 5·x/(x−1) + 6 = 0

When things go wrong

Counterexample lab: remove one condition

Practice

Mistake Museum · Exhibits from this stop

Stopping once you have a root, without checking

Forgetting to multiply a term that has no denominator