Rational Equations

Key Terms

Definition

Domain Restrictions

Solving by Clearing Denominators

Extraneous Solutions

Equations with Monomial Denominators

Equations with Polynomial Denominators

Proportions and Cross-Multiplication

Equations Reducible to Rational Form

Summary of Rational Equation Subtypes

Common Pitfalls

Rational Equations FAQ

Equations with Variables in the Denominator

When the unknown appears below a fraction bar, the equation acquires a feature that no polynomial equation possesses: certain values of the variable are forbidden outright, excluded from the domain before any solving begins. A rational equation is defined by this presence of the variable in at least one denominator. The standard solving technique — clearing all denominators by multiplying through — converts the equation into a polynomial one, but the conversion is not always reversible. Values that zero out a denominator may survive the algebra and masquerade as solutions, making verification against the original domain restrictions an unavoidable final step.

Key Terms

Equation Concepts

Domain Restriction— values that zero out a denominator must be excluded before solving

Extraneous Solution— clearing denominators may introduce values that violate domain restrictions

Equivalent Equations— multiplying by the LCD is not always reversible

Supporting

Expression— the rational expressions forming each side of the equation

Solution Set— must be intersected with the domain

See All Algebra Definitions →

Definition

One pipe fills a tank in 3 hours, another in 6 hours. Working together, how long do they take?

In one hour the first pipe fills

\frac{1}{3}

of the tank and the second

\frac{1}{6}

. If together they need

t

hours, they fill

\frac{1}{t}

of the tank per hour, so

\frac{1}{3} + \frac{1}{6} = \frac{1}{t}

. The unknown has landed below a fraction bar.

Trying values of t, only t = 2 makes the two sides agree — and t = 0 cannot even be tried.

try t (hours)	both pipes per hour: 1/3 + 1/6	needed per hour: 1/t	equal?
1	1/2	1	no
2	1/2	1/2	yes
4	1/2	1/4	no
0	1/2	1/0 — undefined	not allowed

Together the pipes fill the tank in

2

hours. Putting

t = 0

into

\frac{1}{t}

would mean dividing by zero, so that value is ruled out before any arithmetic is done.

With the unknown in a denominator, some values are forbidden from the start.

The distinction matters because a variable denominator can equal zero, and division by zero is undefined. A constant denominator like

3

is always nonzero and creates no restrictions. The moment

x

appears below the fraction bar, the equation inherits domain constraints that must be identified and enforced throughout the solving process.

The definition

A rational equation is an equation in which the variable appears in the denominator of at least one fraction.

A rational equation is any equation containing at least one fraction in which the variable appears in the denominator. The equation

Definition

\frac{3}{x - 1} + \frac{2}{x + 4} = 1

is rational because the variable

x

sits inside two denominators. By contrast, the equation

\frac{x + 1}{3} = 2

is not rational — its denominator is a constant, and the equation is simply a linear equation with a fractional coefficient.

\frac{1}{3} + \frac{1}{6} = \frac{1}{t}

is a rational equation, and its solution is

t = 2

Rational equations are equations between rational expressions — quotients of polynomials.

Rational equations arise naturally from rates, proportions, and any setting where a quantity is expressed as a ratio of two variable expressions. Algebraically, they are equations involving rational expressions — quotients

\frac{P(x)}{Q(x)}

where

P(x)

and

Q(x)

are polynomials.

Values that make a denominator zero must be excluded before solving and checked again afterwards.

Multiplying through by the denominators can produce answers that make one of them zero; such answers are rejected, as extraneous solutions explains. The values to exclude are found first, in domain restrictions.

The rest of the page covers domain restrictions, clearing denominators, extraneous solutions, monomial and polynomial denominators, proportions and cross-multiplication, and equations reducible to rational form.

Domain Restrictions

Before manipulating a rational equation in any way, the values that make any denominator zero must be identified and excluded from the domain.

RULE

These excluded values are never solutions, regardless of what subsequent algebra might suggest.

Consider the equation

\frac{x}{x - 3} = \frac{9}{x - 3} + 2

. The denominator

x - 3

equals zero when

x = 3

, so

x = 3

is excluded from the domain. This restriction is declared at the outset and carries through every step of the solution.

When multiple fractions appear, each denominator contributes its own restrictions. The equation

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

excludes both

x = 0

and

x = 2

. Factoring is sometimes necessary to identify all restrictions: the denominator

x^2 - 4

factors as

(x - 2)(x + 2)

, excluding both

x = 2

and

x = -2

.

The domain is the set of all real numbers minus the excluded values. Only candidates that survive within this domain qualify as solutions. Stating the domain explicitly at the start of the solution is not a formality — it is the safeguard against accepting extraneous results.

The excluded value is a wall, not a solution

The section's x/(x − 3) = 9/(x − 3) + 2, each side drawn as its own graph. Both break at the dashed line x = 3, where the shared denominator is zero: no point of either graph exists there, so 3 is declared out of the domain before solving. The graphs meet once, at x = −3, where both sides equal 1/2.

Stating the excluded values first is what later catches an extraneous candidate.

The card draws the line that is declared before any algebra starts: which values can never be solutions.

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Rational equations

Domain restrictions

Values that make any denominator zero are excluded before the equation is touched.

Domain

all real numbers minus the excluded values

no denominator in the original equation equals zero

✓Candidates for solutions

Any value that keeps every denominator nonzero

Only candidates that survive within the domain qualify as solutions.

✗Never solutions

x = 3

\dfrac{x}{x - 3} = \dfrac{9}{x - 3} + 2

The denominator

x - 3

is zero there, whatever the later algebra suggests.

x = 0

and

x = 2

\dfrac{1}{x} + \dfrac{1}{x - 2} = \dfrac{5}{x(x - 2)}

Each denominator contributes its own restriction.

Factor first

x^2 - 4

A denominator may hide restrictions:

x^2 - 4 = (x - 2)(x + 2)

excludes both

x = 2

and

x = -2

Domain restrictions for rational equations·/algebra/equations/rationalLearn Math Class

Stating the domain at the outset is the safeguard against accepting extraneous results.

Solving by Clearing Denominators

The standard method for solving a rational equation is to multiply every term on both sides by the least common denominator of all fractions. This eliminates every fraction in a single step, converting the rational equation into a polynomial equation that can be solved with familiar techniques.

For the equation

\frac{3}{x - 1} + \frac{2}{x + 4} = 1

, the LCD is

(x - 1)(x + 4)

. Multiplying every term by this product:

Solving by Clearing Denominators

3(x + 4) + 2(x - 1) = (x - 1)(x + 4)

Expanding both sides:

Solving by Clearing Denominators

3x + 12 + 2x - 2 = x^2 + 3x - 4

Solving by Clearing Denominators

5x + 10 = x^2 + 3x - 4

Rearranging into standard form:

Solving by Clearing Denominators

x^2 - 2x - 14 = 0

The quadratic formula gives

x = 1 \pm \sqrt{15}

. Neither value equals

1

-4

, so both lie within the domain and both are genuine solutions.

Pitfall: Multiplying only the fractions

Multiplying only the fractions by

(x - 1)(x + 4)

and leaving the

1

as it is gives

5x + 10 = 1

, so

x = -\frac{9}{5}

, which does not satisfy the equation. With the

1

multiplied too, the solutions are

x = 1 \pm \sqrt{15}

In this example the right side is the plain number

1

, a term with no visible denominator. When the whole equation is multiplied by the LCD, that term is multiplied as well, and the product

1 \cdot (x - 1)(x + 4)

is what supplies the

x^2

on the right of the cleared equation.

The LCD must include every distinct factor that appears in any denominator, each raised to the highest power at which it occurs. Identifying the LCD accurately is essential — multiplying by an expression that leaves out a factor of some denominator leaves those fractions uncleared, while using an unnecessarily large one introduces extra factors that complicate the resulting polynomial without benefit.

To solve your own rational equation with each step shown, use the rational equation solver.

Step	Action	Watch for
1	Factor every denominator completely	hidden restrictions inside polynomials
2	Identify domain restrictions	every denominator contributes
3	Determine the LCD	each distinct factor at highest power
4	Multiply every term on both sides by the LCD	clears all fractions in one step
5	Solve the resulting polynomial equation	use standard techniques for that degree
6	Verify each candidate against the original domain	reject any extraneous solutions

Multiplying every term by the LCD turns a rational equation into a polynomial one in a single step.

Extraneous Solutions

RULE

Multiplying both sides of an equation by an expression containing the variable is not a reversible operation. When that expression equals zero for some value of

x

, the multiplication maps a false statement to a true one, potentially creating solutions that do not satisfy the original equation.

These are extraneous solutions.

Return to the equation

\frac{x}{x - 3} = \frac{9}{x - 3} + 2

. The domain excludes

x = 3

. Multiply both sides by

(x - 3)

Extraneous Solutions

x = 9 + 2(x - 3)

Extraneous Solutions

x = 9 + 2x - 6

Extraneous Solutions

x = 2x + 3

Extraneous Solutions

-x = 3

Extraneous Solutions

x = -3

Checking:

\frac{-3}{-3 - 3} = \frac{-3}{-6} = \frac{1}{2}

and

\frac{9}{-3 - 3} + 2 = \frac{9}{-6} + 2 = -\frac{3}{2} + 2 = \frac{1}{2}

. The value

x = -3

checks out.

Example

Now consider

\frac{x}{x - 3} - \frac{3}{x - 3} = 1

. Multiply by

(x - 3)

x - 3 = x - 3

, which simplifies to

0 = 0

. This is an identity — yet

x = 3

is not a valid solution because it was excluded from the domain. Every real number except

3

satisfies the original equation.

Without the domain check,

x = 3

would appear to be among the solutions.

Pitfall: Skipping the domain check

Taking

0 = 0

from

\frac{x}{x - 3} - \frac{3}{x - 3} = 1

to mean every real number keeps

x = 3

, where both fractions are undefined. The solution set is every real number except

3

The rule is absolute: every candidate obtained after clearing denominators must be substituted back into the original equation to confirm it does not violate any domain restriction.

For a single candidate the check is quick. It is first compared with the excluded values listed at the start; if it is not one of them, it is substituted into both sides of the original equation, as was done for

x = -3

above, to confirm that the two sides agree.

An identity with one point missing

The section's x/(x − 3) − 3/(x − 3) = 1. The left side simplifies to (x − 3)/(x − 3), which is 1 wherever it is defined, so its graph lies exactly on the right side y = 1, except for the open circle at x = 3, where it is undefined. Clearing denominators reports 0 = 0; the graph shows the solution set is every real number except 3.

An identity after clearing still inherits every restriction of the original equation.

Equations with Monomial Denominators

The simplest rational equations have denominators that are single terms — monomials like

x

x^2

, or

3x

. The LCD is easy to identify, and the resulting polynomial equation is typically low in degree.

Consider

\frac{5}{x} + 3 = \frac{7}{x}

. The only denominator is

x

, so the domain excludes

x = 0

. Multiplying every term by

x

Equations with Monomial Denominators

5 + 3x = 7

Equations with Monomial Denominators

3x = 2

Equations with Monomial Denominators

x = \frac{2}{3}

Since

\frac{2}{3} \neq 0

, the solution is valid.

When different powers of

x

appear in the denominators, the LCD uses the highest power. The equation

\frac{1}{x} + \frac{1}{x^2} = \frac{3}{4}

has LCD

4x^2

. Multiplying through:

Equations with Monomial Denominators

4x + 4 = 3x^2

Equations with Monomial Denominators

3x^2 - 4x - 4 = 0

The quadratic formula yields

x = \frac{4 \pm \sqrt{16 + 48}}{6} = \frac{4 \pm 8}{6}

, giving

x = 2

x = -\frac{2}{3}

. Neither is zero, so both are valid solutions.

Monomial-denominator equations serve as the natural entry point for the technique of clearing denominators because the LCD is always a single power of the variable, making the multiplication step transparent.

With single-term denominators the LCD is one monomial, and

x = 0

is the only value to rule out.

Equations with Polynomial Denominators

When denominators are polynomials of degree two or higher, they must be factored completely before the LCD can be determined. Factoring reveals both the domain restrictions and the structure of the LCD.

Consider the equation

\frac{2}{x^2 - 1} = \frac{1}{x - 1}

. The left denominator factors as

(x - 1)(x + 1)

, so the domain excludes

x = 1

and

x = -1

. The LCD is

(x - 1)(x + 1)

. Multiplying:

Equations with Polynomial Denominators

2 = x + 1

Equations with Polynomial Denominators

x = 1

But

x = 1

is excluded from the domain. The candidate is extraneous, and the equation has no solution.

More complex denominators require careful bookkeeping. For the equation

Equations with Polynomial Denominators

\frac{3}{x^2 + 5x + 6} + \frac{1}{x + 2} = \frac{2}{x + 3}

factor the quadratic:

x^2 + 5x + 6 = (x + 2)(x + 3)

. The domain excludes

x = -2

and

x = -3

. The LCD is

(x + 2)(x + 3)

. Multiplying every term:

Equations with Polynomial Denominators

3 + (x + 3) = 2(x + 2)

Equations with Polynomial Denominators

3 + x + 3 = 2x + 4

Equations with Polynomial Denominators

6 + x = 2x + 4

Equations with Polynomial Denominators

x = 2

Since

2 \neq -2

and

2 \neq -3

, the solution

x = 2

is valid.

The factoring step is not optional. Attempting to clear unfactored denominators leads to an LCD that is larger than necessary, inflating the degree of the resulting polynomial and creating more opportunities for error.

A candidate that lands on an excluded value

The section's 2/(x² − 1) = 1/(x − 1). Clearing denominators gives 2 = x + 1, so x = 1, but x = 1 is one of the excluded values (dashed). The graphs confirm it: the two sides never meet anywhere in the domain, and the only place the algebra points to is the vertical line where both are undefined. The equation has no solution.

Factoring the denominators first is what reveals x = 1 as excluded.

Proportions and Cross-Multiplication

A proportion is a rational equation in which a single fraction on the left equals a single fraction on the right:

Proportions and Cross-Multiplication

\frac{A}{B} = \frac{C}{D}

RULE

Cross-multiplication converts this into the polynomial equation

AD = BC

, provided

B \neq 0

and

D \neq 0

This is a special case of clearing denominators — the LCD is

BD

, and multiplying both sides by

BD

produces

AD = BC

directly.

Example

For example,

\frac{x + 1}{x - 2} = \frac{3}{4}

becomes

4(x + 1) = 3(x - 2)

, which simplifies to

4x + 4 = 3x - 6

, giving

x = -10

. Since

-10 \neq 2

, the solution is valid.

Cross-multiplication is fast but limited to equations already in the form of two equal fractions. When more than two fractions appear, or when fractions are added or subtracted, the general LCD method is required. Even in the proportional case, domain restrictions must be stated and checked: if the variable appears in

B

D

, any value that zeros out either denominator is excluded.

A common structural clue that an equation is a proportion in disguise: if every term can be combined so that exactly one fraction appears on each side, cross-multiplication applies.

Cross-multiplication is clearing denominators for two fractions, so the excluded values must still be checked.

Equations Reducible to Rational Form

Several equation types that do not initially look like rational equations can be rewritten as such, inheriting the full domain-restriction and clearing-denominator framework.

Negative exponents are the most direct case. The equation

x^{-1} + x^{-2} = 6

is equivalent to

\frac{1}{x} + \frac{1}{x^2} = 6

, which is a standard rational equation with monomial denominators. Rewriting negative exponents as fractions is not a choice of style — it is a recognition of what the equation actually is.

Pitfall: Reading a negative exponent as a negative number

Reading

x^{-1} + x^{-2} = 6

-x - x^2 = 6

gives

x^2 + x + 6 = 0

, which has no real roots. As fractions,

\frac{1}{x} + \frac{1}{x^2} = 6

has the solutions

x = \frac{1}{2}

and

x = -\frac{1}{3}

Rewritten this way, the equation has one excluded value,

x = 0

, and its LCD is

x^2

. From here it is solved exactly like the monomial-denominator equations above: clear the fractions, solve the resulting quadratic, and check each candidate against

x \neq 0

.

Certain symmetric expressions become rational under substitution. The equation

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

can be treated directly as a rational equation by clearing

x

from the denominator:

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

, then

2x^2 - 5x + 2 = 0

. Alternatively, the substitution

u = x + \frac{1}{x}

captures a structure that appears in many competition and textbook problems.

Fact: Never strictly between

-2

and

2

For real

x \neq 0

, the sum

x + \frac{1}{x}

is at least

2

or at most

-2

, because

x^2 - kx + 1 = 0

has discriminant

k^2 - 4

. So an equation like

x + \frac{1}{x} = 1

has no real solution at all.

Here the quadratic

2x^2 - 5x + 2 = 0

factors as

(2x - 1)(x - 2) = 0

, so

x = 2

x = \frac{1}{2}

. Neither is

0

, so both are solutions, and each one is the reciprocal of the other:

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

. For a general right side

k

, clearing the denominator in

x + \frac{1}{x} = k

gives

x^2 - kx + 1 = 0

, and whether that quadratic has real roots depends on the size of

k

.

Equations involving rates, reciprocals of linear expressions, or differences of fractions often reduce to rational equations after algebraic manipulation. In every case, the same discipline applies: identify domain restrictions from any variable denominator, clear denominators systematically, solve the resulting polynomial, and verify all candidates against the original equation.

Original form	Rewrite as	Notes
Negative exponents	x⁻¹ + x⁻² = 6 → 1/x + 1/x² = 6	direct conversion to fraction notation
Symmetric: x + 1/x = k	clear x: x² − kx + 1 = 0	or substitute u = x + 1/x
Reciprocals of linear expressions	standard rational form	find LCD across all denominators

Negative exponents and reciprocals hide rational equations; once rewritten, they are solved and checked like any other.

Summary of Rational Equation Subtypes

The four configurations covered above all use the same underlying technique — identify domain, find LCD, clear denominators, solve, verify — but differ in how the LCD is identified and what specific risks they introduce. The table below compares them at a glance.

Subtype	Denominator structure	LCD	Watch for
Monomial	single term (x, x², 3x)	highest power of the variable	x = 0 excluded
Polynomial	factorable polynomial	product of distinct factors at highest power	factor first to expose all restrictions
Proportion (A/B = C/D)	exactly one fraction on each side	BD (cross-multiplication)	both B and D must be nonzero
Reducible	not initially rational (neg exponents, x + 1/x, …)	determined after rewrite	identify the hidden form first

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Before you read

Four kinds of rational equation

Rational equations differ in what sits in the denominators. The kind decides how to clear them.

Rational equations4 subtypes

Monomial denominators

LCDthe highest power of

x

$x \neq 0$

Polynomial denominators

Firstfactor every denominator

LCDproduct of the distinct factors

Proportion

A/B = C/D

Methodcross-multiply to

AD = BC

$B, D \neq 0$

Reducible

Looks likenegative exponents or

x + 1/x

Firstrewrite as a standard rational equation

Takeaway

Every kind ends with a check

Whatever the method, a candidate that makes a denominator zero is excluded.

Four kinds of rational equation·/algebra/equations/rationalLearn Math Class

Common Pitfalls

Three mistakes account for most wrong answers to rational equations. Each one is marked where it happens in the sections above; here each is worked through in full, with the wrong and the right working side by side.

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1Multiplying only the fractions

The equation contains a term with no denominator, and only the fractions are multiplied by the LCD. The term without a denominator is copied into the cleared equation unchanged.

Start

\frac{3}{x - 1} + \frac{2}{x + 4} = 1

Wrong turn

Error here

\implies 3(x + 4) + 2(x - 1) = 1

5x + 10 = 1

\implies x = -\frac{9}{5}

Right path

The

1

is multiplied by the LCD as well.

3(x + 4) + 2(x - 1) = (x - 1)(x + 4)

x^2 - 2x - 14 = 0

\implies x = 1 \pm \sqrt{15}

Neither value is

1

-4

, so both are solutions.

Why it happensThe LCD is found by looking at the denominators, so attention stays on the fractions, and a whole number looks as if it needs nothing.

How to checkSubstitute the result into the original equation.

x = -\frac{9}{5}

gives

\frac{3}{-14/5} + \frac{2}{11/5} = -\frac{15}{14} + \frac{10}{11} = -\frac{25}{154}

, not

1

2Skipping the domain check

The cleared equation is solved, and its solutions are reported as the answer without being compared with the excluded values.

Start

\frac{x}{x - 3} - \frac{3}{x - 3} = 1

\implies x - 3 = x - 3

\implies 0 = 0

Wrong turn

Error here

So every real number is reported as a solution.

Right path

The domain excludes

x = 3

, so the solution set is every real number except

3

. In the same way,

\frac{2}{x^2 - 1} = \frac{1}{x - 1}

clears to

2 = x + 1

, so

x = 1

; but

x = 1

is excluded, and that equation has no solution.

Why it happensMultiplying by the LCD removes the denominators, and with them the only visible sign of the restrictions. The cleared equation looks like an ordinary polynomial equation, which has no excluded values.

How to checkSubstitute each candidate into the original equation, not the cleared one. At

x = 3

the term

\frac{3}{3 - 3}

is a division by zero, and at

x = 1

so is

\frac{1}{1 - 1}

3Reading a negative exponent as a negative number

In an equation written with negative exponents,

x^{-1}

and

x^{-2}

are read as

-x

and

-x^2

, so the equation is never rewritten as a rational one.

Start

x^{-1} + x^{-2} = 6

Wrong turn

Error here

\implies -x - x^2 = 6

\implies x^2 + x + 6 = 0

The discriminant is

1 - 24 = -23 < 0

, so no real solution is reported.

Right path

x^{-1} = \frac{1}{x}

and

x^{-2} = \frac{1}{x^2}

, with

x \neq 0

. Multiplying by the LCD

x^2

\frac{1}{x} + \frac{1}{x^2} = 6

\implies x + 1 = 6x^2

\implies 6x^2 - x - 1 = 0

(3x + 1)(2x - 1) = 0

x = \frac{1}{2}

x = -\frac{1}{3}

. Neither is zero, so both are solutions.

Why it happensThe minus sign in the exponent looks like the minus sign of a negative number, but it stands for a reciprocal, not a change of sign.

How to checkSubstitute the results.

x = \frac{1}{2}

gives

2 + 4 = 6

, and

x = -\frac{1}{3}

gives

-3 + 9 = 6

Common pitfalls in rational equations·/algebra/equations/rationalLearn Math Class

All three slips happen around the step that clears the denominators: a term is left out of the multiplication, values excluded by the original denominators slip back in, or a negative exponent is never recognised as a fraction to be cleared. Substituting into the original equation, not the cleared one, settles the first two. For the third, rewriting

x^{-1}

\frac{1}{x}

before anything else prevents the slip.

Rational Equations FAQ

Can a rational equation have infinitely many solutions?

Yes, but usually with a hole in them. Clearing denominators in

\frac{x}{x - 3} - \frac{3}{x - 3} = 1

leaves

x - 3 = x - 3

, an identity, so every real number appears to work. The excluded value survives that check though:

x = 3

zeroes a denominator, so the solution set is every real number except

3

.Read more →

Does it have to be the least common denominator, or will any common denominator do?

Any common multiple clears the fractions, but the least one keeps the resulting polynomial smallest. Choose something that is not a multiple of every denominator and some fractions survive the multiplication; choose something needlessly large and you carry extra factors into a messier polynomial for no gain. The LCD takes every distinct factor at the highest power it appears.Read more →

How do you spot that an equation is really a proportion?

Combine terms and count the fractions: if exactly one fraction sits on each side, cross-multiplication applies and

\frac{A}{B} = \frac{C}{D}

becomes

AD = BC

. With three or more fractions, or with fractions being added on one side, fall back to the general LCD method. Either way the domain restrictions still have to be checked.Read more →

Why do you add rates, not times, when two pipes fill a tank together?

Because the work done per hour adds up, while the hours do not. A pipe that fills a tank in

3

hours does

\frac{1}{3}

of it each hour, and one that takes

6

hours does

\frac{1}{6}

. Together they do

\frac{1}{2}

per hour, so

\frac{1}{3} + \frac{1}{6} = \frac{1}{t}

gives

t = 2

hours. Adding the times would give

9

hours, longer than either pipe alone.Read more →

Why does $\frac{2}{x^2 - 1} = \frac{1}{x - 1}$ have no solution?

Because the only candidate is an excluded value. Factoring gives

x^2 - 1 = (x - 1)(x + 1)

, so

x = 1

and

x = -1

are ruled out first; clearing denominators then gives

2 = x + 1

, so

x = 1

. That value makes both denominators zero, so it is extraneous, and nothing else is left. The graphs agree: the two sides never meet anywhere in the domain.Read more →

Can $x + \frac{1}{x}$ equal $1$ ?

No, not for any real

x

. Clearing the denominator in

x + \frac{1}{x} = k

gives

x^2 - kx + 1 = 0

, whose discriminant is

k^2 - 4

, so real solutions need

k \geq 2

k \leq -2

; for

k = 1

it is

-3

. A positive

x

makes the sum at least

2

, reached only at

x = 1

, and a negative

x

makes it at most

-2

.Read more →

Rational Equations

Equations with Variables in the Denominator

Key Terms

Equation Concepts

Supporting

Definition

Domain Restrictions

Domain restrictions

Solving by Clearing Denominators

Extraneous Solutions

Equations with Monomial Denominators

Equations with Polynomial Denominators

Proportions and Cross-Multiplication

Equations Reducible to Rational Form

Summary of Rational Equation Subtypes

Common Pitfalls

Rational Equations FAQ

Can a rational equation have infinitely many solutions?

Does it have to be the least common denominator, or will any common denominator do?

How do you spot that an equation is really a proportion?

Why do you add rates, not times, when two pipes fill a tank together?

Why does 2x2−1=1x−1\frac{2}{x^2 - 1} = \frac{1}{x - 1}x2−12​=x−11​ have no solution?

Can x+1xx + \frac{1}{x}x+x1​ equal 111?

Why does $\frac{2}{x^2 - 1} = \frac{1}{x - 1}$ have no solution?

Can $x + \frac{1}{x}$ equal $1$ ?