Solving Rational Equations

Intuition

You can already solve polynomial equations, so an equation with the variable stuck in a denominator suggests its own plan. Get rid of the denominators and you are back on ground you know. Multiplying both sides by the least common denominator does exactly that, and the whole method fits on one line.

You also already know why that plan is dangerous. In the first lesson of this course you proved that a reversible step preserves the solution set. You also proved that multiplying by an expression which can equal zero is not reversible, so it can smuggle in an extraneous root. Absolute value equations then made you check every candidate for a related reason. What you do not have yet is a rule that says which candidates to suspect and how much checking you actually owe. For rational equations that rule exists and it is sharp: the only thing that can go wrong is the domain. That is what this lesson proves, and it is what turns the check from a nervous ritual into two seconds of work.

Learning goals

List the excluded values before any algebra
Multiply every term by the LCD to clear fractions
Treat the roots as candidates until the domain check
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.

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 recognise 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

The principle you need here is one you have already proved. In the first lesson of this course you saw that a reversible step produces an equivalent equation, one with exactly the same solution set. You also saw that multiplying both sides by an expression which can equal zero is not reversible: at a zero of the multiplier you cannot divide back. There both sides read $0 = 0$ regardless of what stood there, and a value can satisfy the new equation for a reason that has nothing to do with the old one. A value that solves the new equation but not the original is extraneous. You met the word there, and you met the discipline again in absolute value equations. In those equations, splitting $\lvert X \rvert = Y$ into $X = \pm Y$ quietly drops the condition $Y \ge 0$ and hands you candidates that have to be tested.

So the danger is not news. What is new is that for a rational equation it can be pinned down exactly, and earning that is what this section is for.

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. Nothing went wrong. Every line of that algebra is correct. The step imported a solution: at $x = 2$ the multiplier is zero, so both sides became $0$ , and $0 = 0$ tells you nothing about the fractions you started with.

It is worth seeing precisely where the intruder came in, because the usual account stops one step early. 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 $2$ . Cancelling plugged the hole. The polynomial equation you end up with is defined at the excluded values, and being defined there, it is free to have a root there.

That is the observation this whole method rests on, because the LCD is not just any multiplier. It is built out of the factors of the denominators, so it vanishes at exactly the excluded values and nowhere else. The zeros of the multiplier and the holes in the domain are the same set. Once those two sets coincide, the vague instruction “check your candidates” sharpens into a single question with a yes or no answer, and nothing else can hide.

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.

The argument is short enough to write out in full. It is also worth writing out, because it says precisely how much you have to check and precisely how much you do not.

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 cancelling produces a polynomial equation $P = Q$ . Cancelling 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.

That is the payoff, and it is a promise about this method and no other. A candidate fails exactly when it is excluded from the original domain. So the check is not a nervous habit but a complete test, and a cheap one: you never have to substitute anything anywhere.

Notice how method-specific that promise is by holding it against the case you already know. To solve $\lvert x - 1 \rvert = 2x - 4$ you test each candidate against the condition $2x - 4 \ge 0$ , and not against any domain. The reason is that no value of $x$ is excluded from that equation at all: absolute value is defined everywhere. Same word, extraneous, and an entirely different test, because a different step did the damage. Extraneous roots will turn up once more when you solve radical equations, for a third reason again. A guarantee proved for one method never transfers to a method that has not earned it, so what you carry forward from this lesson is the theorem, not the slogan.

The mirror image, which you also met in the first lesson, is worth restating here where the temptation is strongest. If multiplying by a possibly-zero expression can create a solution, dividing by one can destroy a solution. 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 you have just thrown it away. Never divide both sides by an expression holding the variable. Move everything to one side and factor:

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

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

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. By the proof above, 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 two genuine solutions. There are two separate ways to end with none: every candidate is excluded, as in $\frac{x}{x-2} = \frac{2}{x-2}$ . In the other, the polynomial equation has no roots at all because the variable cancels and leaves a false statement, as $\frac{4}{x-3} = \frac{x}{x-3} - 1$ collapses to $4 = 3$ . Both answers read “no solution”, and both are answers 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. The proof above says exactly this and no more: it matches the original with the cleared equation on $D$ , and it claims nothing whatsoever off $D$ .

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

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. That it comes out slower than the large pump is a good sign, since we called it the smaller one.

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 $v + 2$ . 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 by the proof above both are genuine solutions of the equation, and indeed $\frac{12}{-3} + \frac{12}{1} = -4 + 12 = 8$ checks out for $v = -1$ .

What rules $-1$ out is the boat, not the algebra: a speed in still water cannot be negative. 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

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. The candidates are not the answer. Until each one has been tested against the list of excluded values, the problem is not finished, and the whole point of this lesson is that the test can fail.
Checking against the cleared equation. Every candidate satisfies the cleared equation, since that is where it came from, so testing it there proves nothing at all. The check is against the original denominators.
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. When the variable cancels into something like $3x = 3x$ , every value in the domain is a solution, so the answer is the domain, not the whole number line. 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: cancelling $(x-2)$ from $x(x-2) = 3(x-2)$ discards the perfectly good root $x = 2$ . Factor 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. The fractions cancel and a polynomial equation is left.
That step is not reversible, so it can add a root, exactly as the first lesson warned. What is special here is that the LCD is zero at exactly the excluded values, which is what makes the next point true.
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, and no substitution is needed to certify an answer.
That completeness belongs to this method alone. Absolute value equations breed extraneous roots too, and there the test is $Y \ge 0$ rather than a domain.
Three outcomes are possible: the usual one or two solutions; no solution, either because every candidate is excluded or because the variable cancels into a false statement. The third is an identity, where the variable cancels into a true statement and the answer is the whole domain, which is never “every real number”.
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.

Solving Rational Equations

Learning goals

What a rational equation is

Clearing the denominators

The one thing that can go wrong

The method, and what checking means

When the algebra hands you an excluded value

Equations that model a job or a trip

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Free Response Questions (FRQ)