Solving Rational Equations

Intuition

$\frac{1}{x} + \frac{1}{2} = \frac{5}{2x}$

The variable is stuck in a denominator, but you already know how to clear a fraction: multiply both sides by something every denominator divides into. Do that here and the equation becomes $2 + x = 5$ , which you can solve on sight.

There is one catch. In the first lesson of this course you proved that multiplying both sides of an equation by an expression that can equal zero is not reversible, so it can hand you a root the original equation never had, an extraneous root. That is real here too, since $x$ sits in a denominator and denominators can be zero. What this lesson proves is that for a rational equation the danger is completely predictable: the only root that can ever be extraneous is one that was excluded from the start. Once you know that, checking a candidate stops being a guessing game and becomes a single question with a definite answer.

Learning goals

List the excluded values before any algebra
Multiply every term by the LCD, then treat the roots as candidates, not answers
Reject an extraneous root however correctly it was derived
Distinguish no solution from an identity over the domain
Apply a second filter in a word problem, and name it

What a rational equation is

A rational equation is an equation whose sides are built from rational expressions:

\frac{1}{x} + \frac{1}{2} = \frac{5}{2x}, \qquad\qquad \frac{1}{x-2} + \frac{x}{x+2} = \frac{8}{x^2 - 4}.

The case worth all the care, and the one this lesson is about, is when the variable appears in a denominator. A constant denominator, like the $2$ in $\frac{x}{2}$ , is never zero and rules nothing out, which is why an equation such as $\frac{x}{2} + 1 = 4$ behaves like any linear equation. Put the variable underneath and the denominator can vanish, and where it vanishes the equation stops saying anything at all.

That is the thread of this chapter. A rational expression is undefined wherever its denominator is zero, so an equation built from rational expressions carries a domain: the real numbers that make none of its denominators zero. The rest are its excluded values, and you find them exactly as you did when simplifying, by factoring each denominator and setting each factor to zero:

In $\frac{1}{x} + \frac{1}{2} = \frac{5}{2x}$ the denominators are $x$ , $2$ , and $2x$ , so $x = 0$ is excluded and nothing else is.
In $\frac{1}{x-2} + \frac{x}{x+2} = \frac{8}{x^2-4}$ the last denominator factors as $x^2 - 4 = (x-2)(x+2)$ , so $x = 2$ and $x = -2$ are both excluded.

An excluded value is not a value where the equation is false. It is a value the equation does not speak about at all. Asking whether $x = 2$ solves the second equation is not a question with an answer: at $x = 2$ one of the expressions in it does not exist. Write the excluded values down first, before you touch the algebra, because in a moment the algebra will erase every trace of them.

Check your understanding

Which values are excluded from $\dfrac{3}{x-5} + \dfrac{x}{x+1} = 2$ ?

x = 5

and

x = -1

x = -5

and

x = 1

x = 5

only D

x = 2

Clearing the denominators

The least common denominator is the least common multiple of all the denominators in the equation. Multiply both sides by it, term by term, and every denominator divides into it and cancels. What is left is a polynomial equation, which you know how to solve.

Take the first equation above. Its denominators are $x$ , $2$ , and $2x$ , so the LCD is $2x$ :

2x \cdot \frac{1}{x} \;+\; 2x \cdot \frac{1}{2} \;=\; 2x \cdot \frac{5}{2x} \qquad\Longrightarrow\qquad 2 + x = 5 \qquad\Longrightarrow\qquad x = 3.

Two habits are worth building right here. Multiply every term, including any term that has no denominator at all (the $\frac{1}{2}$ above is easy to remember, but a lone $+\,1$ is easy to forget). And factor the denominators before you choose the LCD, or you will not recognize that $x^2 - 4$ is already made of $x - 2$ and $x + 2$ .

Worked example 1 Solve $\dfrac{x}{x+2} + \dfrac{1}{2} = \dfrac{3}{x+2}$

The denominators are $x + 2$ and $2$ , so $x = -2$ is excluded, and the LCD is $2(x+2)$ .

Multiply each of the three terms by $2(x+2)$ and cancel:

2(x+2) \cdot \frac{x}{x+2} = 2x, \qquad 2(x+2) \cdot \frac{1}{2} = x + 2, \qquad 2(x+2) \cdot \frac{3}{x+2} = 6.

The equation that survives has no fractions in it at all:

2x + (x + 2) = 6 \qquad\Longrightarrow\qquad 3x = 4 \qquad\Longrightarrow\qquad x = \frac{4}{3}.

The only excluded value was $-2$ , and $\frac{4}{3}$ is not it, so this candidate is a genuine solution. Substituting back confirms the arithmetic: $x + 2 = \frac{10}{3}$ , so the left side is $\frac{4/3}{10/3} + \frac{1}{2} = \frac{2}{5} + \frac{1}{2} = \frac{9}{10}$ , and the right side is $\frac{3}{10/3} = \frac{9}{10}$ .

The one thing that can go wrong

In the first lesson of this course you proved that multiplying both sides of an equation by an expression that can equal zero is not reversible: at a zero of the multiplier, both sides collapse to $0 = 0$ , which is true no matter what stood there a moment ago. A value can then satisfy the new equation for a reason that has nothing to do with the old one. That value is extraneous, meaning it solves the new equation but not the original. You met the word there, and again in absolute value equations, where a different step manufactures the same kind of impostor.

A rational equation can pin that danger down exactly, because the variable sits in a denominator. Start from the equation the first lesson opened with:

\frac{x}{x-2} = \frac{2}{x-2}.

Reason it out with no machinery at all. Suppose some number $r$ solves it. Then $r \neq 2$ , or neither side would exist. The two fractions have the same nonzero denominator, so they are equal exactly when their numerators are equal, which forces $r = 2$ . That contradicts $r \neq 2$ . So no number whatsoever solves this equation, and its solution set is empty.

Now run the method. Multiply both sides by $x - 2$ and you get $x = 2$ , the one value the equation excludes. Every line of that algebra is correct, and the step still imported a solution: $x = 2$ solves the cleared equation even though the original equation cannot even be evaluated there, since its denominator $x - 2$ is zero at exactly that point.

Look at exactly where the impostor got in. When you multiply $\frac{x}{x-2}$ by $x - 2$ and cancel, you write down $x$ . But $\frac{x}{x-2}\cdot(x-2)$ and $x$ are not the same object: they agree at every value except $x = 2$ , where the first is undefined and the second is $2$ . Canceling plugged the hole, and the polynomial equation you end up with is defined right where the original was not, so it is free to have a root there.

Multiplying by x - 2 carries every solution of the original forward, but it cannot be undone at x = 2. That is exactly where the cleared equation grows a root the original never had.

That single example shows what goes wrong, and the same reasoning explains why it can only go wrong there. The LCD is built from the same factors as the denominators, so it is zero at exactly the excluded values and nowhere else. That means for every allowed value of $x$ , the LCD is some ordinary nonzero number, and multiplying or dividing both sides of a true equation by a nonzero number never changes whether it is true. So on the allowed values, the original equation and the cleared equation agree completely: clearing the denominators cannot lose a solution there, and it cannot invent one there either. The two equations can only disagree at an excluded value, exactly as they just did at $x = 2$ .

So the roots of the cleared equation are candidates, and a candidate fails exactly when it is excluded from the original domain. There is no third possibility, which is why the check is a complete test and not a nervous habit: you never have to substitute anything back in to know a genuine root is genuine.

That promise belongs to this method and does not transfer automatically. To solve $\lvert x - 1 \rvert = 2x - 4$ you test each candidate against the condition $2x - 4 \ge 0$ , not against any domain, because absolute value is defined everywhere and nothing is excluded to begin with. Same word, extraneous, but an entirely different test, because a different step did the damage. Radical equations will earn their own version of this promise later, for a different reason again.

There is a mirror mistake worth a warning here too: dividing both sides by an expression that holds the variable can lose a solution instead of adding one.

The method, and what checking means

Factor every denominator and list the excluded values. This is the domain, and the original equation fixes it. Nothing you do afterwards can change it.
Multiply every term on both sides by the LCD and cancel. A polynomial equation is left.
Solve the polynomial equation. Its roots are candidates, not answers. If the variable cancels away entirely, read whatever statement is left standing. A false one means there are no candidates at all, and a true one means every value is a candidate.
Reject every excluded candidate. What survives is the solution set, and it is allowed to be empty.

Step 4 is where the check lives, and it pays to know exactly what it must do. As the reasoning above showed, membership in the domain is the whole test, so you never have to substitute a candidate back into the original to know it is genuine. Substituting back is still worth doing whenever the arithmetic is kind. The reason is not that the theory demands it, but that it catches slips in steps 2 and 3, which no domain check can see.

In the first lesson you sorted equations by their solution set into conditional, identity, and contradiction. A rational equation can be any of the three, with the excluded values punched out of the answer, and step 3 is where you find out which one you have. Usually it is conditional, and you get one or more genuine solutions, however many the cleared polynomial equation turns out to have. There is more than one way to end with none. Every candidate can be excluded, as in $\frac{x}{x-2} = \frac{2}{x-2}$ . The variable can cancel and leave a false statement, as $\frac{4}{x-3} = \frac{x}{x-3} - 1$ collapses to $4 = 3$ . Or the cleared polynomial equation can simply have no real roots at all, the same way a plain quadratic sometimes does, with no cancellation and nothing excluded in sight. Every one of those reads “no solution”, and each is an answer rather than evidence of a mistake.

The third outcome is the one that catches people out. The variable can cancel and leave a statement that is true for every value, which is what happens to

\frac{3x}{x-2} = 3 + \frac{6}{x-2} \qquad\Longrightarrow\qquad 3x = 3(x-2) + 6 \qquad\Longrightarrow\qquad 3x = 3x.

Every value is now a candidate, so every value in the domain is a solution, and the answer is the domain itself: every real number except $2$ . The answer is not “every real number”, however tempting that sounds. An identity holds wherever both sides exist, and at $x = 2$ they do not exist, so the original equation still has nothing to say there. That matches the reasoning above exactly: the original and the cleared equation agree only on the allowed values, so $x = 2$ was never part of either one’s solution set, no matter how the cleared equation reads.

Check your understanding

Solving $\dfrac{2x}{x+1} = 2 - \dfrac{2}{x+1}$ leads to $2x = 2(x+1) - 2$ , which simplifies to $2x = 2x$ . What is the solution set?

A Every real number B Every real number except

x = -1

C No solution D

x = -1

When the algebra hands you an excluded value

Worked example 2 Solve $\dfrac{1}{x-2} + \dfrac{x}{x+2} = \dfrac{8}{x^2-4}$

Factor first: $x^2 - 4 = (x-2)(x+2)$ . The excluded values are $x = 2$ and $x = -2$ , and the LCD is $(x-2)(x+2)$ .

Multiply all three terms by the LCD and cancel:

\begin{aligned} (x-2)(x+2) \cdot \frac{1}{x-2} &= x + 2, \\ (x-2)(x+2) \cdot \frac{x}{x+2} &= x(x-2), \\ (x-2)(x+2) \cdot \frac{8}{(x-2)(x+2)} &= 8. \end{aligned}

So the cleared equation is $(x+2) + x(x-2) = 8$ . Expand and collect:

\begin{aligned} x^2 - x + 2 = 8 \qquad&\Longrightarrow\qquad x^2 - x - 6 = 0 \\ &\Longrightarrow\qquad (x-3)(x+2) = 0. \end{aligned}

The candidates are $x = 3$ and $x = -2$ . Now check them against the list of excluded values, not against the cleared equation, which they both satisfy by construction. The value $-2$ is excluded, so it is extraneous and gets thrown out no matter how correctly it was derived. The value $3$ is in the domain, so it is genuine:

\frac{1}{1} + \frac{3}{5} = \frac{8}{5}, \qquad \frac{8}{3^2 - 4} = \frac{8}{5}. \qquad \checkmark

The solution is $x = 3$ , and the equation has exactly one.

Every solution of the original equation is a root of the cleared equation, so the solutions sit inside the candidates. The gap between the two boxes holds the excluded candidates, and each one of them is an extraneous root. (The other excluded value, 2, is not a root of the cleared equation, so it never becomes a candidate at all.)

Check your understanding

Solve $\dfrac{x^2}{x-3} = \dfrac{9}{x-3}$ .

x = 3

and

x = -3

x = 3

x = -3

D no solution

Equations that model a job or a trip

Rational equations are worth solving because rates produce them on their own. If a job takes $t$ hours, then in one hour you finish $\frac{1}{t}$ of it. Rates add: a worker finishing $\frac{1}{a}$ of the job per hour and a worker finishing $\frac{1}{b}$ of it per hour together finish $\frac{1}{a} + \frac{1}{b}$ per hour. So the time $t$ they take together satisfies

\frac{1}{a} + \frac{1}{b} = \frac{1}{t}.

Travel is the same story with distance in place of work: time is $\frac{\text{distance}}{\text{speed}}$ , so an unknown speed lands in a denominator immediately.

Worked example 3 A large pump drains a tank in $6$ hours. Running alongside a smaller pump, it drains in $4$ hours. How long would the smaller pump take alone?

Let $t$ be the hours the smaller pump needs on its own. In one hour the large pump drains $\frac{1}{6}$ of the tank and the smaller drains $\frac{1}{t}$ of it, and together they drain $\frac{1}{4}$ of it:

\frac{1}{6} + \frac{1}{t} = \frac{1}{4}.

The only excluded value is $t = 0$ , and the LCD is $12t$ . Multiply every term by it:

2t + 12 = 3t \qquad\Longrightarrow\qquad t = 12.

The candidate $12$ is not excluded, so it is a genuine solution: $\frac{1}{6} + \frac{1}{12} = \frac{2}{12} + \frac{1}{12} = \frac{1}{4}$ .

The smaller pump would take $12$ hours alone. Check that the rate makes sense: alone it drains $\frac{1}{12}$ of the tank per hour, less than the large pump’s $\frac{1}{6}$ , and together the two rates add to exactly $\frac{1}{4}$ , which is what the problem promised.

Worked example 4 A boat covers $12$ km up a river and $12$ km back down. The current runs at $2$ km/h and the round trip takes $8$ hours. How fast is the boat in still water?

Let $v$ be the boat’s speed in still water, in km/h. Going upstream the current subtracts, so the boat makes $v - 2$ ; coming back it adds, so the boat makes $v + 2$ . For the boat to travel upstream at all, $v - 2$ must be positive, so the model already requires $v > 2$ . Each leg takes distance divided by speed, and the two legs add to $8$ hours:

\frac{12}{v-2} + \frac{12}{v+2} = 8.

The excluded values are $v = 2$ and $v = -2$ , and the LCD is $(v-2)(v+2)$ . Multiplying every term by it gives $12(v+2) + 12(v-2) = 8(v^2 - 4)$ , and the left side collapses:

\begin{aligned} 24v = 8v^2 - 32 \qquad&\Longrightarrow\qquad v^2 - 3v - 4 = 0 \\ &\Longrightarrow\qquad (v-4)(v+1) = 0. \end{aligned}

The candidates are $v = 4$ and $v = -1$ . Neither is excluded, so both are genuine solutions of the equation, and indeed $\frac{12}{-3} + \frac{12}{1} = -4 + 12 = 8$ checks out for $v = -1$ .

But $v = -1$ fails the restriction $v > 2$ stated at the start: it is not the domain that rules it out, since the equation is perfectly happy to evaluate at $v = -1$ , it is the boat. So $v = 4$ km/h, which the story confirms, since the boat then makes $2$ km/h upstream ( $6$ hours) and $6$ km/h downstream ( $2$ hours).

Look carefully at what just happened to $v = -1$ , because it is a second kind of rejection and it is easy to confuse with the first. That root is not extraneous. It lies in the domain and it satisfies the equation exactly. A rational equation used as a model carries two independent filters. The domain rejects values the equation cannot even evaluate, and it does so for every rational equation. The context rejects values the equation evaluates perfectly well but that cannot describe a boat, an hour, or a length. Run both filters, and say out loud which one you are running, because “reject the negative root” is a statement about rivers, not about mathematics.

Check your understanding

Solving a rational equation that models a rectangle's width gives the candidates $w = -3$ and $w = 5$ , and neither value is excluded from the equation. Which is the correct classification?

A Both are extraneous, since only one can be the answer B

w = -3

is extraneous, and

w = 5

is the solution C

w = -3

is rejected by context, not by the domain, and

w = 5

is the solution D Both are genuine solutions, since a width can be described by either sign

Check your understanding

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

x = 1

x = -1

C no solution D

x = 3

Common mistakes

Solving and stopping, or checking against the wrong equation. The candidates are not the answer until each is tested against the excluded values from the original denominators. Every candidate already satisfies the cleared equation, since that is where it came from, so testing it there proves nothing.
Finding the domain last. Clearing the denominators destroys the evidence: once you have multiplied through, $x^2 - 4$ is gone and nothing in the polynomial equation remembers that $-2$ was excluded. List the excluded values before you clear, not after.
Multiplying only the fractions. Every term on both sides gets multiplied by the LCD, including a lone constant or a lone $x$ . In $\frac{3}{x-1} + 2 = \frac{x}{x-1}$ the middle term becomes $2(x-1)$ , not $2$ .
Answering “every real number” to an identity without checking the domain first. When the variable cancels into something like $3x = 3x$ , every value in the domain is a solution. The answer is the domain the equation started with, which is every real number only when nothing was ever excluded from it. The excluded values are excluded no matter what the cleared equation says about them.
Dividing both sides by an expression with the variable in it. That is the mirror mistake, and it loses solutions instead of adding one. Factor and set the product to zero instead.
Treating “no solution” as failure. An equation whose only candidate is excluded genuinely has no solution, and saying so is the correct answer, not an admission of defeat.

Key takeaways

A rational equation carries a domain. Factor every denominator, set each factor to zero, and list the excluded values before doing any algebra.
To solve it, multiply every term on both sides by the LCD and solve the polynomial equation left behind. That step is not reversible, but because the LCD is zero at exactly the excluded values, it can only ever add one of those back as a fake root, nothing else.
The roots of the cleared equation are candidates. A candidate is a genuine solution exactly when it lies in the original domain; an excluded candidate is an extraneous root and is rejected, however correctly it was derived. The domain check is therefore a complete test: no substitution is needed to certify an answer.
Three outcomes are possible: the usual conditional case, with one or more solutions; no solution, because every candidate is excluded, because the variable cancels into a false statement, or because the cleared polynomial equation simply has no real roots at all; or an identity, where the variable cancels into a true statement and the answer is the whole domain, which is only “every real number” when nothing was excluded from the domain to begin with.
In a word problem there is a second, separate filter: a root can be genuine and still impossible, such as a negative time or speed. Reject it for that reason, and name the reason.

Practice

Multiple Choice Questions (MCQ)

Progressively harder sets of questions. Each opens on its own page.

Level 1 Foundational 12 questions Start → Level 2 Intermediate 12 questions Start → Level 3 Challenge 12 questions Start →

Core practice

Practice problems at the level of the course, to be worked out on paper. Hints one at a time, then the answer or the full worked solution, with your progress kept in this browser.

More practice (optional)

Extra sets, as hard as the Challenge set. Each one opens on its own page.

More resources (optional)

Other explanations of this lesson, if you want a second take.

Rational expressions, equations, and functions (Khan Academy)
Solve rational equations (OpenStax)
Solve applications with rational equations (OpenStax)

Go deeper (optional)

You can skip this and keep going. Read it if you want to know more.

Why clearing denominators loses nothing and adds only excluded values

Clearing the denominators loses nothing and adds only excluded values#

Write the equation as $A = B$ , where $A$ and $B$ are its two sides, and let $D$ be its domain: the real numbers that make none of the denominators zero. Let $m$ be the LCD. Because $m$ is built out of the factors of those denominators, $m$ is zero at exactly the excluded values, so $m(r) \neq 0$ for every $r$ in $D$ .

Multiplying through by $m$ and canceling produces a polynomial equation $P = Q$ . Canceling a factor is legitimate wherever that factor is nonzero, so for every $r$ in $D$ we have $P(r) = m(r)A(r)$ and $Q(r) = m(r)B(r)$ . Fix such an $r$ . If $A(r) = B(r)$ , multiplying both sides by the number $m(r)$ gives $P(r) = Q(r)$ . Conversely, if $P(r) = Q(r)$ , then $m(r)A(r) = m(r)B(r)$ , and dividing by the nonzero number $m(r)$ gives back $A(r) = B(r)$ . On the domain, then, the two equations have exactly the same solutions.

Two consequences follow, and together they are the entire method. First, every solution of the original lies in $D$ , so by that equivalence it satisfies $P = Q$ : clearing the denominators cannot lose a solution. Second, a root $r$ of $P = Q$ solves the original if and only if $r$ lies in $D$ . One direction is the equivalence just proved. The other is quicker still: if $r$ is not in $D$ , then some denominator of the original vanishes at $r$ . In that case one side of the original is not a number at $r$ , and an expression that does not exist cannot equal anything.

So the roots of $P = Q$ are candidates, the excluded ones are the frauds, and there is no third possibility.

How dividing by a variable expression loses a root

If multiplying by a possibly-zero expression can create a solution, dividing by one can destroy a solution, the mirror image of the whole lesson. From $x(x-2) = 3(x-2)$ you might cancel the common $(x-2)$ and report $x = 3$ , but $x = 2$ makes both sides zero, and canceling just threw it away. Move everything to one side and factor instead:

x(x-2) - 3(x-2) = 0 \qquad\Longrightarrow\qquad (x-2)(x-3) = 0,

which keeps both roots, $x = 2$ and $x = 3$ .

A bit of history (optional)

You work every line correctly, and still finish holding a number that does not solve the equation you started with. Where did it come from?

The name answers. Extraneous is the Latin extraneus, meaning from outside, foreign, not of the household. That same Latin word took a second road into English, through French. It arrived as strange, and as stranger. So an extraneous root really is a stranger, and the label is exact rather than decorative. The root was never a fact about your equation. It came in from outside, carried by the step in which you multiplied both sides by the LCD.

Naming it that way needs an idea the word cannot supply on its own: an expression may speak only on a stated set of inputs, and a value outside that set is not a wrong answer but no answer at all. So listing the excluded values first is not caution. It is what turns the final check into a complete test, because a candidate is genuine exactly when it was never a stranger.