The Central Idea
A rational equation contains one or more rational expressions, such as
\[ \frac{2}{x-1}=\frac{3}{x+2}. \]A rational inequality compares rational expressions:
\[ \frac{x+1}{x-3}>0. \]In both cases, denominators impose restrictions.
Before solving, identify every value that makes an original denominator zero.
Domain Restrictions Come First
Consider
\[ \frac{2}{x-1}=\frac{3}{x+2}. \]The denominators require
\[ x-1\neq0 \]and
\[ x+2\neq0. \]Therefore,
\[ \boxed{x\neq1,\qquad x\neq-2}. \]Any proposed solution equal to \(1\) or \(-2\) must be rejected.
Why Restrictions Matter
Algebraic operations may transform a rational equation into an equation that has solutions not permitted by the original expression.
These are called extraneous solutions.
The safest workflow is:
- record restrictions;
- solve the transformed equation;
- check every proposed solution against the original restrictions and, when useful, the original equation.
Clearing Denominators
Rational equations are often easiest to solve by multiplying both sides by a least common denominator, or LCD.
For
\[ \frac{2}{x-1}=\frac{3}{x+2}, \]the LCD is
\[ (x-1)(x+2). \]Multiply both sides by this expression.
Example: Solve by Clearing Denominators
Start with
\[ \frac{2}{x-1}=\frac{3}{x+2}. \]Restrictions:
\[ x\neq1,\qquad x\neq-2. \]Multiply both sides by \((x-1)(x+2)\):
\[ 2(x+2)=3(x-1). \]Expand:
\[ 2x+4=3x-3. \]Solve:
\[ x=7. \]Since \(7\) violates no restriction,
\[ \boxed{x=7}. \]Cross Multiplication Is a Special Case
When an equation has the form
\[ \frac{a}{b}=\frac{c}{d}, \]with \(b\neq0\) and \(d\neq0\), multiplying by \(bd\) gives
\[ ad=bc. \]This is commonly called cross multiplication.
It is not a separate law. It is simply the result of clearing denominators.
Example: Cross Multiplication
Solve
\[ \frac{x+1}{x-2}=\frac{4}{3}. \]Restriction:
\[ x\neq2. \]Cross multiply:
\[ 3(x+1)=4(x-2). \]Expand:
\[ 3x+3=4x-8. \]Therefore,
\[ x=11. \]Thus,
\[ \boxed{x=11}. \]Equations with Several Rational Terms
Consider
\[ \frac{1}{x}+\frac{2}{x+1}=3. \]Restrictions:
\[ x\neq0,\qquad x\neq-1. \]The LCD is
\[ x(x+1). \]Multiply every term by the LCD.
Example: Several Rational Terms
Starting with
\[ \frac{1}{x}+\frac{2}{x+1}=3, \]multiply through by \(x(x+1)\):
\[ x+1+2x=3x(x+1). \]Simplify:
\[ 3x+1=3x^2+3x. \]Therefore,
\[ 3x^2=1. \]So
\[ x=\pm\frac{1}{\sqrt3} = \pm\frac{\sqrt3}{3}. \]Neither value violates the restrictions. Hence,
\[ \boxed{x=\pm\frac{\sqrt3}{3}}. \]Every Term Must Be Multiplied by the LCD
When clearing denominators, the LCD must multiply every term on both sides.
For
\[ \frac{1}{x}+\frac{2}{x+1}=3, \]the correct step is
\[ x(x+1)\left(\frac1x\right) + x(x+1)\left(\frac2{x+1}\right) = x(x+1)(3). \]Omitting any term changes the equation.
Example: A Variable Appears in Only One Denominator
Solve
\[ \frac{x+3}{x-1}=2. \]Restriction:
\[ x\neq1. \]Multiply both sides by \(x-1\):
\[ x+3=2(x-1). \]Then
\[ x+3=2x-2, \]so
\[ \boxed{x=5}. \]An Equation Can Have No Solution
Consider
\[ \frac{x+1}{x-2}=1. \]Restriction:
\[ x\neq2. \]Multiply by \(x-2\):
\[ x+1=x-2. \]Subtract \(x\):
\[ 1=-2. \]This contradiction shows that the equation has no solution:
\[ \boxed{\varnothing}. \]An Equation Can Be True for Every Allowed Value
Consider
\[ \frac{x^2-1}{x-1}=x+1. \]Restriction:
\[ x\neq1. \]Factor the numerator:
\[ x^2-1=(x-1)(x+1). \]Therefore,
\[ \frac{(x-1)(x+1)}{x-1}=x+1 \]for every allowed \(x\).
The solution set is
\[ \boxed{\mathbb{R}\setminus\{1\}}. \]Extraneous Solutions
A proposed solution is extraneous if it solves a transformed equation but is not allowed in the original rational equation.
This frequently occurs when clearing denominators creates an equation that is defined at values where the original equation was not.
Restrictions identify such values immediately.
Example: An Extraneous Solution from a Restriction
Solve
\[ \frac{x}{x-2}=\frac{2}{x-2}. \]Restriction:
\[ x\neq2. \]Multiply both sides by \(x-2\):
\[ x=2. \]But \(x=2\) is excluded from the original equation.
Therefore,
\[ \boxed{\varnothing}. \]Checking in the Original Equation
Substitution into the original equation is a direct verification method.
Suppose a calculation produces
\[ x=7 \]for
\[ \frac{2}{x-1}=\frac{3}{x+2}. \]Substitute:
\[ \frac{2}{7-1}=\frac{2}{6}=\frac13, \] \[ \frac{3}{7+2}=\frac{3}{9}=\frac13. \]The two sides agree.
Equations with Factored Denominators
Solve
\[ \frac{1}{x-2} + \frac{1}{x+2} = \frac{x}{x^2-4}. \]Factor
\[ x^2-4=(x-2)(x+2). \]Restrictions:
\[ x\neq2,\qquad x\neq-2. \]The LCD is
\[ (x-2)(x+2). \]Complete the Factored-Denominator Example
Multiply through by the LCD:
\[ x+2+x-2=x. \]Simplify:
\[ 2x=x. \]Therefore,
\[ x=0. \]Since \(0\) is allowed,
\[ \boxed{x=0}. \]Quadratic Equations Can Result
Clearing denominators does not always produce a linear equation.
For example,
\[ \frac1x+\frac1{x-1}=1 \]has restrictions
\[ x\neq0,\qquad x\neq1. \]Multiplying by \(x(x-1)\) gives
\[ x-1+x=x(x-1). \]Example: Rational Equation Leading to a Quadratic
Continue:
\[ 2x-1=x^2-x. \]Move all terms to one side:
\[ x^2-3x+1=0. \]Apply the quadratic formula:
\[ x= \frac{3\pm\sqrt{9-4}}{2} = \frac{3\pm\sqrt5}{2}. \]Neither solution equals \(0\) or \(1\), so
\[ \boxed{x=\frac{3\pm\sqrt5}{2}}. \]Rational Inequalities Require a Different Strategy
For equations, clearing denominators is usually safe after restrictions are recorded.
For inequalities, multiplying by a variable expression is more dangerous because its sign may be unknown.
Multiplying an inequality by a negative quantity reverses the inequality sign.
Therefore rational inequalities are usually solved by sign analysis rather than by blindly clearing denominators.
Put the Inequality on One Side
A useful standard form is
\[ \boxed{ \frac{P(x)}{Q(x)}>0 } \]or one of the corresponding forms
\[ \frac{P(x)}{Q(x)}<0, \qquad \frac{P(x)}{Q(x)}\geq0, \qquad \frac{P(x)}{Q(x)}\leq0. \]Once the inequality has one rational expression compared with zero, its sign can be analyzed interval by interval.
Critical Values
The sign of
\[ \frac{P(x)}{Q(x)} \]can change only where the numerator or denominator is zero.
Therefore the important values are:
- zeros of \(P(x)\), where the rational expression may equal zero;
- zeros of \(Q(x)\), where the rational expression is undefined.
These values divide the number line into intervals for sign testing.
Example: A Basic Rational Inequality
Solve
\[ \frac{x+1}{x-3}>0. \]The numerator is zero at
\[ x=-1. \]The denominator is zero at
\[ x=3. \]Thus the critical values are
\[ -1,\qquad3. \]Build the Test Intervals
The critical values divide the real line into
\[ (-\infty,-1), \] \[ (-1,3), \]and
\[ (3,\infty). \]On each interval, the signs of the factors remain constant.
Test the Signs
For
\[ \frac{x+1}{x-3}>0, \]choose one test value from each interval.
If \(x=-2\),
\[ \frac{-1}{-5}>0. \]If \(x=0\),
\[ \frac{1}{-3}<0. \]If \(x=4\),
\[ \frac{5}{1}>0. \]Therefore the expression is positive on
\[ (-\infty,-1) \]and
\[ (3,\infty). \]Complete the Basic Inequality
Because the inequality is strict,
\[ \frac{x+1}{x-3}>0, \]the zero at \(x=-1\) is not included.
The denominator zero \(x=3\) can never be included.
Therefore,
\[ \boxed{ (-\infty,-1)\cup(3,\infty) }. \]When a Numerator Zero Is Included
Solve
\[ \frac{x+1}{x-3}\geq0. \]The sign pattern is the same as before.
But now equality is allowed.
Since the expression equals zero at \(x=-1\), that value is included.
The denominator zero \(x=3\) remains excluded.
Thus,
\[ \boxed{ (-\infty,-1]\cup(3,\infty) }. \]Denominator Zeros Are Never Included
Even if an inequality uses
\[ \geq \]or
\[ \leq, \]a denominator zero cannot be included because the rational expression is undefined there.
For example, in
\[ \frac{x-2}{x+4}\leq0, \]\(x=-4\) is always excluded.
Factor Before Sign Analysis
Consider
\[ \frac{x^2-4}{x-3}\leq0. \]Factor the numerator:
\[ x^2-4=(x-2)(x+2). \]The critical values are
\[ x=-2,\qquad x=2,\qquad x=3. \]Factoring reveals every point at which the sign may change.
Example: Three Critical Values
Solve
\[ \frac{(x-2)(x+2)}{x-3}\leq0. \]The intervals are
\[ (-\infty,-2), \quad (-2,2), \quad (2,3), \quad (3,\infty). \]Testing one point in each interval gives the sign pattern
\[ -,\quad+,\quad-,\quad+. \]We need the negative intervals together with numerator zeros.
Complete the Three-Critical-Value Example
The inequality is
\[ \frac{(x-2)(x+2)}{x-3}\leq0. \]The negative intervals are
\[ (-\infty,-2) \]and
\[ (2,3). \]The numerator zeros \(x=-2\) and \(x=2\) are included because equality is allowed.
The denominator zero \(x=3\) is excluded.
Therefore,
\[ \boxed{ (-\infty,-2]\cup[2,3) }. \]Repeated Factors Affect Sign Changes
Consider the factor
\[ (x-1)^2. \]Because this factor has even multiplicity, it is nonnegative and does not change sign as \(x\) passes through \(1\).
By contrast,
\[ x-1 \]has odd multiplicity and changes sign at \(x=1\).
Multiplicity therefore helps predict sign behavior.
Example: Repeated Numerator Factor
Solve
\[ \frac{(x-2)^2}{x+1}>0. \]Critical values:
\[ x=-1,\qquad x=2. \]The numerator is positive except at \(x=2\), where it is zero.
Therefore the sign is determined by \(x+1\), except that \(x=2\) must be excluded because the inequality is strict.
Hence,
\[ \boxed{ (-1,2)\cup(2,\infty) }. \]Example: Repeated Denominator Factor
Solve
\[ \frac{x-3}{(x+2)^2}<0. \]The denominator is positive for every \(x\neq-2\).
Therefore the sign is controlled by the numerator \(x-3\).
We need
\[ x-3<0, \]so
\[ x<3. \]Exclude \(x=-2\):
\[ \boxed{ (-\infty,-2)\cup(-2,3) }. \]Move Everything to One Side Before Testing
Suppose we want to solve
\[ \frac{1}{x}>2. \]It is dangerous to multiply immediately by \(x\) because the sign of \(x\) is unknown.
Instead, move everything to one side:
\[ \frac1x-2>0. \]Combine:
\[ \frac{1-2x}{x}>0. \]Now use critical-value sign analysis.
Example: Inequality with a Rational Expression on One Side
Solve
\[ \frac1x>2. \]Rewrite:
\[ \frac{1-2x}{x}>0. \]Critical values:
\[ x=0,\qquad x=\frac12. \]Test intervals:
\[ (-\infty,0), \quad \left(0,\frac12\right), \quad \left(\frac12,\infty\right). \]The signs are
\[ -,\quad+,\quad-. \]Therefore,
\[ \boxed{ \left(0,\frac12\right) }. \]Why Blind Multiplication Can Fail
Starting from
\[ \frac1x>2, \]multiplying both sides by \(x\) without knowing its sign is invalid.
If \(x>0\), the inequality direction stays the same.
If \(x<0\), the inequality direction reverses.
Sign analysis handles both possibilities systematically.
Rational Inequalities with Two Fractions
Consider
\[ \frac{1}{x-1} < \frac{2}{x+2}. \]Move everything to one side:
\[ \frac{1}{x-1} - \frac{2}{x+2} <0. \]Combine using the common denominator
\[ (x-1)(x+2). \]Example: Two-Fraction Inequality
Continue:
\[ \frac{x+2-2(x-1)} {(x-1)(x+2)} <0. \]Simplify the numerator:
\[ x+2-2x+2=4-x. \]Thus,
\[ \frac{4-x}{(x-1)(x+2)}<0. \]Critical values:
\[ x=-2,\qquad x=1,\qquad x=4. \]Complete the Two-Fraction Inequality
Test the intervals
\[ (-\infty,-2), \quad (-2,1), \quad (1,4), \quad (4,\infty). \]The signs of
\[ \frac{4-x}{(x-1)(x+2)} \]are
\[ +,\quad-,\quad+,\quad-. \]We need the negative intervals.
Because the inequality is strict, \(x=4\) is not included, and denominator zeros are never included.
\[ \boxed{ (-2,1)\cup(4,\infty) }. \]Canceled Factors Still Matter in Inequalities
Consider
\[ \frac{(x-2)(x+1)}{x-2}>0. \]The expression simplifies to
\[ x+1 \]only for
\[ x\neq2. \]Therefore the inequality becomes
\[ x+1>0, \qquad x\neq2. \]So
\[ x>-1 \]with \(x=2\) removed.
Hence,
\[ \boxed{ (-1,2)\cup(2,\infty) }. \]Holes Are Critical Even When the Sign Does Not Change
In
\[ \frac{(x-2)(x+1)}{x-2}, \]the factor \(x-2\) cancels algebraically but leaves an excluded input at \(x=2\).
The sign may be the same on both sides of \(2\), yet the solution interval must still be split because the function is undefined there.
Domain information survives cancellation.
Common Error: Ignoring Restrictions
Suppose
\[ \frac{x}{x-3} = \frac{3}{x-3}. \]Clearing denominators gives
\[ x=3. \]But \(x=3\) makes the original denominator zero.
Therefore the correct solution is
\[ \boxed{\varnothing}. \]Common Error: Cross Multiplying a Sum
Cross multiplication applies directly only to an equation with one fraction on each side:
\[ \frac{A}{B}=\frac{C}{D}. \]It does not apply directly to
\[ \frac1x+\frac1{x+1}=2. \]For equations with several rational terms, use an LCD and multiply every term.
Common Error: Multiplying an Inequality by an Unknown-Sign Denominator
From
\[ \frac{x+1}{x-2}>0, \]we cannot simply multiply by \(x-2\) and write
\[ x+1>0. \]The sign of \(x-2\) depends on \(x\). If it is negative, the inequality direction would reverse.
Use sign analysis instead.
Common Error: Including a Denominator Zero
Suppose the inequality is
\[ \frac{x-1}{x+2}\leq0. \]Even though the inequality allows equality, \(x=-2\) cannot be included.
The expression is undefined there.
Only zeros of the numerator can be included when equality is permitted.
Common Error: Forgetting to Combine Before Sign Analysis
For
\[ \frac1{x-1} - \frac2{x+2} <0, \]analyzing the signs of the two fractions separately does not directly determine the sign of their difference.
First combine them into a single rational expression:
\[ \frac{4-x}{(x-1)(x+2)}<0. \]Then analyze the sign of the product and quotient factors.
Common Error: Testing Only One Point
A rational inequality may change sign at every numerator or denominator zero.
Therefore a single test value cannot determine the entire solution set.
Test one value from every interval determined by the critical values.
A Reliable Procedure for Rational Equations
- Factor denominators when useful.
- Record every value excluded from the original domain.
- Find the least common denominator.
- Multiply every term by the LCD.
- Simplify the resulting equation.
- Solve using appropriate algebraic methods.
- Reject any solution that violates an original restriction.
- Substitute into the original equation when verification is useful.
A Reliable Procedure for Rational Inequalities
- Move all terms to one side so the other side is zero.
- Combine the expression into a single rational expression.
- Factor the numerator and denominator.
- Find every numerator zero and denominator zero.
- Place these critical values on the number line.
- Test the sign on each interval.
- Select the intervals satisfying the inequality.
- Include numerator zeros only when equality is allowed.
- Never include denominator zeros.
- Preserve any holes created by canceled factors.
Practice 1: Proportion
Solve:
\[ \frac{x+2}{x-1}=\frac{3}{2}. \]Practice 2: Clear the Denominator
Solve:
\[ \frac{x+4}{x-2}=3. \]Practice 3: Several Rational Terms
Solve:
\[ \frac1x+\frac1{x+2}=1. \]Practice 4: Extraneous Solution
Solve:
\[ \frac{x}{x-5}=\frac{5}{x-5}. \]Practice 5: Basic Rational Inequality
Solve:
\[ \frac{x-2}{x+1}>0. \]Practice 6: Inclusive Rational Inequality
Solve:
\[ \frac{x+3}{x-4}\leq0. \]Practice 7: Factored Numerator
Solve:
\[ \frac{x^2-9}{x-1}\geq0. \]Practice 8: Repeated Factor
Solve:
\[ \frac{(x-1)^2}{x+2}>0. \]Practice 9: Move to One Side
Solve:
\[ \frac1x<1. \]Practice 10: Identify the Error
A student solves
\[ \frac{x+1}{x-2}>0 \]by multiplying both sides by \(x-2\) and concluding
\[ x>-1. \]Explain the error and find the correct solution.
Practice Answers
Practice 1
Restriction:
\[ x\neq1. \]Cross multiply:
\[ 2(x+2)=3(x-1). \] \[ 2x+4=3x-3. \]Therefore,
\[ \boxed{x=7}. \]Practice 2
Restriction:
\[ x\neq2. \]Multiply by \(x-2\):
\[ x+4=3(x-2). \] \[ x+4=3x-6. \] \[ 10=2x. \]Thus,
\[ \boxed{x=5}. \]Practice 3
Restrictions:
\[ x\neq0,\qquad x\neq-2. \]Multiply by \(x(x+2)\):
\[ x+2+x=x(x+2). \] \[ 2x+2=x^2+2x. \] \[ x^2=2. \]Therefore,
\[ \boxed{x=\pm\sqrt2}. \]Practice 4
Restriction:
\[ x\neq5. \]Clearing denominators gives
\[ x=5. \]But \(5\) is excluded.
Therefore,
\[ \boxed{\varnothing}. \]Practice 5
Critical values:
\[ x=-1,\qquad x=2. \]The sign is positive on
\[ (-\infty,-1) \]and
\[ (2,\infty). \]Therefore,
\[ \boxed{ (-\infty,-1)\cup(2,\infty) }. \]Practice 6
Critical values:
\[ x=-3,\qquad x=4. \]The expression is negative between the critical values and zero at \(x=-3\).
Therefore,
\[ \boxed{[-3,4)}. \]Practice 7
Factor:
\[ x^2-9=(x-3)(x+3). \]Critical values:
\[ -3,\qquad1,\qquad3. \]Testing intervals gives the sign pattern
\[ -,\quad+,\quad-,\quad+. \]Equality is allowed at numerator zeros \(x=-3\) and \(x=3\), but not at the denominator zero \(x=1\).
Thus,
\[ \boxed{ [-3,1)\cup[3,\infty) }. \]Practice 8
Critical values:
\[ x=-2,\qquad x=1. \]The numerator is positive except at \(x=1\), where it equals zero.
The denominator is positive for \(x>-2\).
Because the inequality is strict, exclude \(x=1\).
\[ \boxed{ (-2,1)\cup(1,\infty) }. \]Practice 9
Move everything to one side:
\[ \frac1x-1<0. \] \[ \frac{1-x}{x}<0. \]Critical values:
\[ x=0,\qquad x=1. \]The expression is negative on
\[ (-\infty,0) \]and
\[ (1,\infty). \]Therefore,
\[ \boxed{ (-\infty,0)\cup(1,\infty) }. \]Practice 10
Multiplying by \(x-2\) is not valid without knowing whether \(x-2\) is positive or negative.
Instead analyze
\[ \frac{x+1}{x-2}>0. \]Critical values:
\[ x=-1,\qquad x=2. \]The expression is positive on
\[ (-\infty,-1) \]and
\[ (2,\infty). \]Therefore,
\[ \boxed{ (-\infty,-1)\cup(2,\infty) }. \]What to Remember
Rational equations and inequalities both require careful attention to denominators, but they are usually solved differently.
For rational equations:
- record domain restrictions first;
- use an LCD to clear denominators;
- solve the resulting equation;
- reject extraneous solutions.
For rational inequalities:
- move everything to one side;
- combine into one rational expression;
- factor numerator and denominator;
- use numerator and denominator zeros as critical values;
- test signs interval by interval;
- include numerator zeros only when equality is allowed;
- never include denominator zeros.
The governing principle is simple: algebraic transformations are valid only on values where the original rational expressions are defined.
Where We Go Next
Rational expressions naturally describe quantities that vary inversely. When one quantity grows while another decreases according to a reciprocal relationship, the algebra often takes the form
\[ y=\frac{k}{x}. \]The next lesson, Reciprocal Relationships and Inverse Variation, develops the structure, graphs, equations, and applications of reciprocal relationships.