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Rational Inequalities



Key Terms
Definition
Critical Points
The Sign Chart Method for Rational Inequalities
Endpoint Inclusion Rules
A Common Error: Cross-Multiplication
Getting the Inequality Into Standard Form
Worked Examples
Connection to Rational Equations
Summary of Rational Inequality Forms
Common Pitfalls
Rational Inequalities FAQ



Two Kinds of Critical Points and the Danger of Cross-Multiplication

A rational inequality asks where a fraction involving the variable is positive, negative, or zero. The sign chart method applies exactly as it does for polynomials, but with an added layer: the denominator contributes critical points of its own. Where the numerator is zero, the expression equals zero. Where the denominator is zero, the expression does not exist. Both types partition the number line into intervals, both affect the sign analysis, but they are treated differently when deciding which boundary points belong to the solution. The temptation to clear the denominator by cross-multiplying must be resisted — the denominator changes sign across the number line, making a single multiplication invalid.

Key Terms

Core Concepts

Rational Inequality— variable in denominator; sign chart with two kinds of critical points
Critical Point— numerator zeros (expression =0= 0) vs denominator zeros (undefined)
Sign Analysis— applied to the quotient via factor-by-factor sign tracking

Endpoint Rules

Inequality— numerator zeros may be included for ≤\leq/≥\geq; denominator zeros never
Interval Notation— denominator zeros always get parentheses

See All Algebra Definitions →


Definition

For a fraction equation, only the numerator decides the answer.
A fraction is zero exactly when its top is zero, so x−2x+1=0\frac{x - 2}{x + 1} = 0 means x−2=0x - 2 = 0, and x=2x = 2. The denominator only rules out the value that would break the fraction, here x=−1x = -1. It is reasonable to expect the same of x−2x+1>0\frac{x - 2}{x + 1} > 0: make the top positive, x−2>0x - 2 > 0, and the answer is x>2x > 2.
The guess: numerator positive, so x > 2−5−4−3−2−1012345the guess: x − 2 > 0(2, ∞)includedexcludedTreated like an equation: only the numerator decides
The guess: every xx greater than 22, read off the numerator alone.
Checking values shows the guess is wrong.
Put actual numbers into the fraction and compare with what the guess predicts.
x(x − 2)/(x + 1)really > 0?the guess says
−42yesno ✗
−32.5yesno ✗
−1.57yesno ✗
−1undefinednono
0−2nono
20nono
30.25yesyes
Right of 22 the guess holds. But at x=−4x = -4, −3-3 and −1.5-1.5 the fraction is positive and the guess says no. A whole stretch of answers is missing.
(x - 2)/(x + 1) is positive left of -1 and right of 2xy−6−5−4−3−2−112345−6−5−4−3−2−112345678y = (x − 2)/(x + 1)positive here toox > 2: positiveAbove the axis on two pieces; the guess found only one
The curve is above the axis on two separate pieces: right of 22, which the guess found, and left of the wall at x=−1x = -1, which it missed. Those are the three "✗" rows of the table.
The miss was predictable: left of −1 the denominator is negative.
A fraction's sign comes from both parts. A negative over a negative is positive, so wherever x−2x - 2 and x+1x + 1 are both negative, the fraction is positive even though the top is not. Recording the sign of each part on each piece of the line shows it in advance.
piece of the linex − 2x + 1(x − 2)/(x + 1)
left of −1−−+
between −1 and 2−+−
right of 2+++
The expression P(x)Q(x)\frac{P(x)}{Q(x)} is a quotient of two polynomials.
RULE
Its sign at any point where it is defined equals the sign of the product P(x)⋅Q(x)P(x) \cdot Q(x), because a fraction is positive when numerator and denominator share the same sign, and negative when they differ.
This observation is what makes the sign chart — originally a tool for products — applicable to quotients as well.
The definition
A rational inequality is an inequality involving at least one fraction in which the variable appears in the denominator, such as
Definition
x−2x+1>0\frac{x - 2}{x + 1} > 0
Its equation counterpart: rational equations.
A rational inequality is a statement about a fraction's sign, and its answer is a set.
It is not a number and not a function. It asks where a fraction is positive, negative or zero, and its answer collects those values — for x−2x+1>0\frac{x - 2}{x + 1} > 0, the two rays left of −1-1 and right of 22. The variable has to be in the denominator: x+35≤1\frac{x + 3}{5} \leq 1 is not rational — its denominator is a constant, making it a linear inequality in disguise.
The solution set of (x - 2)/(x + 1) > 0−5−4−3−2−1012345(x − 2)/(x + 1) > 0(−∞, −1) ∪ (2, ∞)includedexcludedTwo rays; −1 left out because the fraction does not exist there
The two "yes" stretches of the table. The open dot at 22 is where the fraction is 00; the open dot at −1-1 is where it does not exist.
The failure becomes the method: every zero of the denominator goes on the chart, and none of them is ever in the answer.
Wherever the denominator can turn negative, clearing it by multiplying through is unsafe, so the sign chart takes its place. Its boundaries are the zeros of the numerator, where the fraction is 00, and the zeros of the denominator, where it is undefined. The two behave differently at the end: with ≥\geq or ≤\leq a numerator zero joins the answer, but a denominator zero never does, because there is no value there to compare.
With >= the numerator zero joins; the denominator zero never does(x − 2)/(x + 1) > 0(−∞, −1) ∪ (2, ∞)−5−4−3−2−1012345(x − 2)/(x + 1) ≥ 0(−∞, −1) ∪ [2, ∞)includedexcludedWith ≥ the dot at 2 fills in; the dot at −1 stays open
Switching >> to ≥\geq fills the dot at 22, where the fraction equals 00. The dot at −1-1 stays open under every symbol.
The rest of the page turns this into a procedure: the two kinds of critical points, the sign chart step by step, the endpoint rules, why cross-multiplication fails, getting an inequality into standard form, worked examples, and how all of this relates to rational equations.

Critical Points

A rational inequality has two kinds of critical points, and confusing them is a common source of error.

Numerator zeros are values where P(x)=0P(x) = 0. At these points the rational expression equals zero: 0Q(x)=0\frac{0}{Q(x)} = 0 (provided Q(x)≠0Q(x) \neq 0). These are candidates for inclusion in the solution set when the inequality is non-strict (≤\leq or ≥\geq), because zero satisfies ≤0\leq 0 and ≥0\geq 0.

Denominator zeros are values where Q(x)=0Q(x) = 0.

RULE
At these points the expression is undefined — division by zero has no value. These are never part of the solution set, regardless of the inequality symbol.

Even for P(x)Q(x)≤0\frac{P(x)}{Q(x)} \leq 0, a denominator zero is excluded because the expression does not exist there.

Fact: Changing sign without passing zero
1x\frac{1}{x} is −1-1 at x=−1x = -1 and 11 at x=1x = 1, yet it is never 00: it jumps across the wall at x=0x = 0. A polynomial cannot do this, since its graph has no breaks, so it must pass through 00 to change sign.

Both types of critical points go on the sign chart. Both create interval boundaries. The expression can change sign at either type. The difference appears only at the endpoint-inclusion step: numerator zeros may be included, denominator zeros are always excluded.

For the inequality (x−3)(x+1)(x−5)(x+4)≥0\frac{(x - 3)(x + 1)}{(x - 5)(x + 4)} \geq 0, the numerator zeros are x=3x = 3 and x=−1x = -1, and the denominator zeros are x=5x = 5 and x=−4x = -4. All four values appear on the sign chart. The solution will include x=3x = 3 and x=−1x = -1 (where the expression is zero) but will exclude x=5x = 5 and x=−4x = -4 (where it is undefined).
Critical-point type Defined by Value of expression there Inclusion in solution
Numerator zero P(x) = 0,  Q(x) ≠ 0 equals 0 possible — depends on strictness
Denominator zero Q(x) = 0 undefined never
f(x)-505x-4-3-2-101231-2(x − 1) / (x + 2) < 0
Numerator zero at 1, pole at −2

Both critical points divide the axis, but they behave differently: at x = 1 the expression is zero and the curve crosses the axis, while at x = −2 the denominator vanishes and the curve runs off along the dashed asymptote. The solution, the interval between them, can include the first endpoint but never the second. Distinguish the two kinds on the inequality visual explorer.

Every rational sign chart is built from these two kinds of marks and nothing else.

The Sign Chart Method for Rational Inequalities

The sign chart for a rational inequality follows the same logic as for polynomial inequalities, with the numerator and denominator factors treated on equal footing during the sign-tracking phase.

Step 1: Factor the numerator and denominator completely. Every factor must be visible as a separate entity.

Step 2: Identify all critical points — zeros of the numerator and zeros of the denominator. Order them on the number line from smallest to largest.

Step 3: Create the sign chart. One row per factor (both numerator and denominator factors), with columns for each interval between consecutive critical points. Record the sign of each factor in each interval.

Step 4: Determine the sign of the rational expression in each interval. The sign is the product of all factor signs — or equivalently, count the number of negative factors: an even count means the expression is positive, odd means negative.

Step 5: Select the intervals matching the inequality.

Step 6: Handle endpoints. Numerator zeros are included for ≤\leq or ≥\geq, excluded for << or >>. Denominator zeros are excluded always. Mark included endpoints with square brackets and excluded ones with parentheses.

Example
For x−2x+1>0\frac{x - 2}{x + 1} > 0, the critical points are x=2x = 2 (numerator) and x=−1x = -1 (denominator). Three intervals arise. In (−∞,−1)(-\infty, -1): (x−2)(x - 2) is negative, (x+1)(x + 1) is negative — quotient is positive. In (−1,2)(-1, 2): (x−2)(x - 2) is negative, (x+1)(x + 1) is positive — quotient is negative. In (2,∞)(2, \infty): both positive — quotient is positive. The solution to >0> 0 is (−∞,−1)∪(2,∞)(-\infty, -1) \cup (2, \infty). Both critical points are excluded: x=−1x = -1 because it is a denominator zero, x=2x = 2 because the inequality is strict.
Step Action What it produces
1 factor the numerator and denominator completely every factor visible as a separate entity
2 identify all critical points (numerator zeros and denominator zeros) ordered list of boundaries on the number line
3 build the chart: one row per factor, one column per interval sign of each factor in each interval
4 combine factor signs into the sign of the quotient sign of P(x) ⁄ Q(x) in each interval
5 select intervals matching the inequality direction candidate solution set
6 apply endpoint rules: numerator zeros for ≤ ⁄ ≥; denominator zeros never final solution in interval notation
Every factor counts alike while the signs are tracked; only at the endpoints does it matter which ones are in the denominator.

Endpoint Inclusion Rules

The endpoint rules for rational inequalities are stricter than those for polynomial inequalities, and the reason is simple: a polynomial is defined everywhere, but a rational expression is not.

For a non-strict inequality like P(x)Q(x)≤0\frac{P(x)}{Q(x)} \leq 0, the expression must be less than or equal to zero. At a numerator zero where the denominator is nonzero, the expression equals zero, which satisfies ≤0\leq 0. The point is included. At a denominator zero, the expression has no value — it is neither less than zero nor equal to zero nor anything else. The point is excluded.

For a strict inequality like P(x)Q(x)<0\frac{P(x)}{Q(x)} < 0, all critical points are excluded. Numerator zeros are excluded because the expression equals zero there, and zero does not satisfy <0< 0. Denominator zeros are excluded because the expression is undefined.

This produces four cases in practice. Under ≤\leq or ≥\geq: numerator zeros included, denominator zeros excluded. Under << or >>: both excluded. There is no case where a denominator zero is included — the expression simply does not exist at that point.

In interval notation, included endpoints receive a square bracket [[ or ]], and excluded endpoints receive a parenthesis (( or )). A denominator zero always gets a parenthesis. A numerator zero gets a bracket for ≤\leq or ≥\geq and a parenthesis for << or >>.
Critical point Strict (<, >) Non-strict (≤, ≥)
Numerator zero  (P(x) = 0) ✗ ✓
Denominator zero  (Q(x) = 0) ✗ ✗

A Common Error: Cross-Multiplication

The most natural instinct when facing P(x)Q(x)>0\frac{P(x)}{Q(x)} > 0 is to multiply both sides by Q(x)Q(x) and work with P(x)>0P(x) > 0 or P(x)<0P(x) < 0. This instinct is wrong.

The problem is that Q(x)Q(x) changes sign. In some intervals Q(x)Q(x) is positive, and multiplying preserves the inequality direction. In other intervals Q(x)Q(x) is negative, and multiplying reverses it. A single multiplication cannot account for both cases simultaneously.

Consider x−3x+2>1\frac{x - 3}{x + 2} > 1. Multiplying both sides by (x+2)(x + 2) gives x−3>x+2x - 3 > x + 2 when x+2>0x + 2 > 0, which simplifies to −3>2-3 > 2 — false everywhere. But when x+2<0x + 2 < 0, the multiplication flips the inequality: x−3<x+2x - 3 < x + 2, giving −3<2-3 < 2 — true everywhere in that region. Tracking these cases is cumbersome and error-prone.

The correct approach avoids cross-multiplication entirely. Move everything to one side first: x−3x+2−1>0\frac{x - 3}{x + 2} - 1 > 0. Combine into a single fraction: x−3−(x+2)x+2>0\frac{x - 3 - (x + 2)}{x + 2} > 0, which simplifies to −5x+2>0\frac{-5}{x + 2} > 0. Now the sign chart has one critical point, x=−2x = -2 (denominator zero). The numerator −5-5 is always negative. In (−∞,−2)(-\infty, -2), the denominator is negative, so the fraction is positive. In (−2,∞)(-2, \infty), the denominator is positive, so the fraction is negative. The solution is (−∞,−2)(-\infty, -2).

The sign chart handled it cleanly. Cross-multiplication would have required splitting into cases and risked losing track of which direction applied where.
(x - 3)/(x + 2) > 1 holds exactly for x < -2xy−8−7−6−5−4−3−2−1123456−6−5−4−3−2−112345678(x − 3)/(x + 2)y = 1above 1 for x < −2multiplying by x + 2 misses it(x − 3)/(x + 2) > 1 holds exactly for x < −2
What cross-multiplying gets wrong

The section's (x − 3)/(x + 2) > 1. Multiplying by x + 2 as if it were positive gives −3 > 2, which suggests no solution. The graph disagrees: left of the wall at x = −2 the whole branch lies above y = 1, because there x + 2 is negative and the multiplication should have flipped the inequality. Rewriting as −5/(x + 2) > 0 gives the true answer, x < −2.

Moving everything to one side and using a sign chart avoids the case split entirely.
The correct route for the section's inequality, one move at a time. No move multiplies by x+2x + 2, so the direction of the inequality never has to be tracked.
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Rational inequalities

One side, one fraction, compared with zero

Instead of multiplying by x+2x + 2, whose sign changes, move everything to one side and combine it into a single fraction.
01
Fraction compared with 1Before
x−3x+2>1\dfrac{x - 3}{x + 2} > 1
↓Move everything to one side— subtract 1 from both sides
02
Compared with zero
x−3x+2−1>0\dfrac{x - 3}{x + 2} - 1 > 0
↓Combine into a single fraction— over the common denominator x + 2
03
Single fraction
x−3−(x+2)x+2>0\dfrac{x - 3 - (x + 2)}{x + 2} > 0
↓Simplify the numerator— x − 3 − x − 2 = −5
04
Ready for the sign chartAfter
−5x+2>0\dfrac{-5}{x + 2} > 0
One critical point, x=−2x = -2. The numerator −5-5 is always negative, so the fraction is positive exactly where x+2<0x + 2 < 0. The solution is (−∞,−2)(-\infty, -2).
Rewriting (x − 3)/(x + 2) > 1 for the sign chart·/algebra/inequalities/rationalLearn Math Class
Every move keeps the inequality exactly equivalent, and the last form has a single critical point, x=−2x = -2. The next section uses the same moves to bring any rational inequality into standard form.

Getting the Inequality Into Standard Form

The sign chart requires a single rational expression compared to zero. When the inequality is not already in this form, it must be rearranged before the method can be applied.

The inequality P(x)Q(x)>k\frac{P(x)}{Q(x)} > k is rewritten as P(x)Q(x)−k>0\frac{P(x)}{Q(x)} - k > 0. The two terms are then combined into a single fraction over a common denominator:

Getting the Inequality Into Standard Form
P(x)−k⋅Q(x)Q(x)>0\frac{P(x) - k \cdot Q(x)}{Q(x)} > 0

This is now a standard rational inequality with everything on one side and zero on the other. The numerator P(x)−k⋅Q(x)P(x) - k \cdot Q(x) and the denominator Q(x)Q(x) are factored, critical points are identified, and the sign chart proceeds as usual.

For the inequality 1x−3≤2x+1\frac{1}{x - 3} \leq \frac{2}{x + 1}, subtract the right side: 1x−3−2x+1≤0\frac{1}{x - 3} - \frac{2}{x + 1} \leq 0. The common denominator is (x−3)(x+1)(x - 3)(x + 1):

Getting the Inequality Into Standard Form
(x+1)−2(x−3)(x−3)(x+1)≤0\frac{(x + 1) - 2(x - 3)}{(x - 3)(x + 1)} \leq 0

Getting the Inequality Into Standard Form
x+1−2x+6(x−3)(x+1)≤0\frac{x + 1 - 2x + 6}{(x - 3)(x + 1)} \leq 0

Getting the Inequality Into Standard Form
−x+7(x−3)(x+1)≤0\frac{-x + 7}{(x - 3)(x + 1)} \leq 0

The critical points are x=7x = 7 (numerator zero), x=3x = 3 and x=−1x = -1 (denominator zeros). The sign chart takes over from here.

Skipping the combination step — trying to work with two separate fractions — leads to the same cross-multiplication trap. A single fraction compared to zero is the form the sign chart is built for.

Pitfall: Subtracting the constant from the numerator only
Writing x−3x+2−1\frac{x - 3}{x + 2} - 1 as x−4x+2\frac{x - 4}{x + 2} takes the 11 off the numerator alone, and x−4x+2>0\frac{x - 4}{x + 2} > 0 gives (−∞,−2)∪(4,∞)(-\infty, -2) \cup (4, \infty). With 1=x+2x+21 = \frac{x + 2}{x + 2}, the fraction is −5x+2\frac{-5}{x + 2} and the solution is (−∞,−2)(-\infty, -2).

The combination step is ordinary fraction subtraction. Before two terms are merged, each one is written over the common denominator, so in the general form the constant kk enters the numerator as k⋅Q(x)k \cdot Q(x), and in the example the fraction 2x+1\frac{2}{x + 1} enters it as 2(x−3)2(x - 3).

Only after that rewriting are the numerators combined and simplified. The result is a single numerator over a single denominator, ready to be factored for the sign chart.

Worked Examples

A simple rational inequality: solve x−2x+1>0\frac{x - 2}{x + 1} > 0. Critical points: x=2x = 2 (numerator), x=−1x = -1 (denominator). Three intervals. In (−∞,−1)(-\infty, -1): both factors negative, quotient positive. In (−1,2)(-1, 2): numerator negative, denominator positive, quotient negative. In (2,∞)(2, \infty): both positive, quotient positive. The solution to >0> 0 is (−∞,−1)∪(2,∞)(-\infty, -1) \cup (2, \infty).

An inequality requiring combination: solve 1x−3≤2x+1\frac{1}{x - 3} \leq \frac{2}{x + 1}. After combining into −x+7(x−3)(x+1)≤0\frac{-x + 7}{(x - 3)(x + 1)} \leq 0, the critical points are x=−1x = -1, x=3x = 3 (denominator zeros), and x=7x = 7 (numerator zero). Four intervals. End behavior: for large positive xx, the numerator −x+7-x + 7 is negative, (x−3)(x - 3) is positive, (x+1)(x + 1) is positive — product of signs is negative. Moving left using multiplicity (all factors have odd multiplicity, so sign flips at every critical point): (7,∞)(7, \infty) is negative, (3,7)(3, 7) is positive, (−1,3)(-1, 3) is negative, (−∞,−1)(-\infty, -1) is positive. The inequality ≤0\leq 0 selects negative intervals and includes the numerator zero: (−1,3)∪[7,∞)(-1, 3) \cup [7, \infty). The denominator zeros x=−1x = -1 and x=3x = 3 are excluded.

Pitfall: Bracketing a denominator zero
Under ≤0\leq 0, giving every critical point a bracket turns the answer into [−1,3]∪[7,∞)[-1, 3] \cup [7, \infty), but at x=−1x = -1 and x=3x = 3 the fraction −x+7(x−3)(x+1)\frac{-x + 7}{(x - 3)(x + 1)} is undefined. Only the numerator zero gets a bracket: (−1,3)∪[7,∞)(-1, 3) \cup [7, \infty).

In this example the inequality is non-strict, and all three critical points sit at the ends of the selected intervals. Which of them belong to the answer is a separate question from which intervals are selected, and it is settled point by point with the endpoint rules from the earlier section.

An inequality with a repeated factor: solve (x−1)2x+3≥0\frac{(x - 1)^2}{x + 3} \geq 0. Critical points: x=1x = 1 (numerator, multiplicity 22) and x=−3x = -3 (denominator). The factor (x−1)2(x - 1)^2 is non-negative everywhere, so it does not change the sign — only the denominator (x+3)(x + 3) controls the sign. In (−∞,−3)(-\infty, -3): (x+3)(x + 3) is negative, so the expression is non-positive. In (−3,∞)(-3, \infty): (x+3)(x + 3) is positive, so the expression is non-negative. The inequality ≥0\geq 0 selects (−3,∞)(-3, \infty), and since the expression equals zero at x=1x = 1, that point is included (it already lies within the selected interval). The denominator zero x=−3x = -3 is excluded. The solution is (−3,∞)(-3, \infty).

Pitfall: Alternating signs past a squared factor
Flipping the sign at every critical point of (x−1)2x+3≥0\frac{(x - 1)^2}{x + 3} \geq 0 makes (−∞,−3)(-\infty, -3) positive and gives (−∞,−3)∪[1,∞)(-\infty, -3) \cup [1, \infty). But x=−4x = -4 gives 25−1=−25\frac{25}{-1} = -25. The sign changes only at x=−3x = -3, and the answer is (−3,∞)(-3, \infty).

The second example could move from right to left flipping the sign at each critical point, because every factor there has odd multiplicity. Here the factor (x−1)(x - 1) is squared, so the multiplicity of each critical point is read before the chart is filled in.

Since the selected interval (−3,∞)(-3, \infty) already contains x=1x = 1, the endpoint rule for the numerator zero changes nothing in this example. Under the strict inequality >0> 0 it would matter: x=1x = 1 would be removed, leaving (−3,1)∪(1,∞)(-3, 1) \cup (1, \infty).

To solve your own rational inequality with each step shown, use the rational inequality solver.
(7 - x)/((x - 3)(x + 1)) <= 0 on (-1, 3) and [7, inf)xy−4−3−2−1123456789101112−3−2−1123−1, 3: excluded7: includedsolution: −1 < x < 3 or x ≥ 7Denominator zeros are walls; the numerator zero 7 joins ≤ 0
Two kinds of critical points on one graph

The section's (7 − x)/((x − 3)(x + 1)) ≤ 0. The denominator zeros −1 and 3 are walls where the graph breaks and the expression is undefined, so they are excluded even though the inequality is non-strict. The numerator zero 7 is where the graph touches 0, so it is included. The graph is on or below the axis on −1 < x < 3 and from 7 onward, where it stays just below the axis.

Numerator zeros follow the inequality symbol; denominator zeros are always excluded.

Connection to Rational Equations

Rational equations and rational inequalities begin with the same kind of expression — a quotient P(x)Q(x)\frac{P(x)}{Q(x)} — but ask different questions, and therefore require different tools.

RULE
A rational equation P(x)Q(x)=0\frac{P(x)}{Q(x)} = 0 asks where the expression equals zero. The answer depends only on the numerator: P(x)=0P(x) = 0 (with denominator zeros excluded).

The denominator determines the domain restrictions, but once those are enforced, only the numerator matters. Clearing the denominator by multiplying both sides is safe for equations because the target value is zero, and zero times anything is zero.

A rational inequality P(x)Q(x)>0\frac{P(x)}{Q(x)} > 0 asks where the expression is positive. This depends on both the numerator and the denominator — the sign of each matters in every interval. Clearing the denominator is not safe because the multiplication may need to flip the inequality in some intervals but not others.

The sign chart handles the inequality question by treating numerator and denominator factors on equal footing: all contribute critical points, all contribute sign information, and the overall sign is determined by the combination. The equation question is a degenerate case where only the numerator zeros matter and the sign chart is unnecessary.

An equation needs only the numerator's zeros; an inequality needs the signs of numerator and denominator alike.

Summary of Rational Inequality Forms

All rational inequalities reduce to a single fraction P(x)Q(x)\frac{P(x)}{Q(x)} compared to 00 once the appropriate preparation step is applied. The table below collects the most common starting forms, an example of each, the preparation step that brings it to standard form, and how the sign chart proceeds from there.
Starting form Example Preparation step Then
Already standard:  P(x) ⁄ Q(x) compared to 0 (x − 2) ⁄ (x + 1) > 0 none factor numerator and denominator; sign chart
Compared to a constant:  P(x) ⁄ Q(x) compared to k  (k ≠ 0) 1 ⁄ (x − 3) ≤ 1 subtract k; combine into a single fraction over Q(x) factor; sign chart
Two rational expressions:  P₁ ⁄ Q₁ compared to P₂ ⁄ Q₂ 1 ⁄ (x − 3) ≤ 2 ⁄ (x + 1) move everything to one side; combine over a common denominator factor; sign chart

Common Pitfalls

Cross-multiplication has a section of its own above. Three more mistakes account for many wrong answers to rational inequalities; each one is marked where it happens in the sections above, and here each is worked through in full, with the wrong and the right working side by side.
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1Subtracting the constant from the numerator only
An inequality compared with a constant is moved into standard form, but the constant is subtracted from the numerator without first being written over the denominator.
Start
x−3x+2>1\frac{x - 3}{x + 2} > 1
Wrong turn
Error here
  ⟹  x−3−1x+2>0\implies \frac{x - 3 - 1}{x + 2} > 0
  ⟹  x−4x+2>0\implies \frac{x - 4}{x + 2} > 0
The critical points x=−2x = -2 and x=4x = 4 give the solution (−∞,−2)∪(4,∞)(-\infty, -2) \cup (4, \infty).
Right path
Write 11 as x+2x+2\frac{x + 2}{x + 2} before subtracting.
x−3−(x+2)x+2>0\frac{x - 3 - (x + 2)}{x + 2} > 0
  ⟹  −5x+2>0\implies \frac{-5}{x + 2} > 0
The numerator is always negative, so the fraction is positive only where x+2<0x + 2 < 0. The solution is (−∞,−2)(-\infty, -2).
Why it happensThe constant has no visible denominator, so it looks as if it can be taken straight off the numerator, and the common-denominator step is skipped.
How to checkTest a value from the wrong answer that lies outside the right one. x=5x = 5 gives 5−35+2=27\frac{5 - 3}{5 + 2} = \frac{2}{7}, which is not greater than 11. A value from the right answer, x=−3x = -3, gives −6−1=6>1\frac{-6}{-1} = 6 > 1.
2Bracketing a denominator zero
Under a non-strict inequality, every critical point at the end of a selected interval is given a square bracket, including the zeros of the denominator.
Wrong turn
Error here
For −x+7(x−3)(x+1)≤0\frac{-x + 7}{(x - 3)(x + 1)} \leq 0, the negative intervals are (−1,3)(-1, 3) and (7,∞)(7, \infty), and with ≤\leq all their endpoints are included: [−1,3]∪[7,∞)[-1, 3] \cup [7, \infty).
Right path
x=7x = 7 is a numerator zero, where the expression equals 00, so it is included. x=−1x = -1 and x=3x = 3 are denominator zeros, where the expression is undefined, so they are excluded under every symbol. The solution is (−1,3)∪[7,∞)(-1, 3) \cup [7, \infty).
Why it happensThe habit of including every endpoint under ≤\leq and ≥\geq comes from polynomial inequalities, where every critical point is a zero of the expression. A rational inequality has a second kind of critical point that this habit does not cover.
How to checkSubstitute each bracketed endpoint into the expression. At x=3x = 3 the denominator is (3−3)(3+1)=0(3 - 3)(3 + 1) = 0, so the expression has no value to compare with 00. At x=7x = 7 it is 032=0\frac{0}{32} = 0, which satisfies ≤0\leq 0.
3Alternating signs past a squared factor
The sign chart is filled in by flipping the sign at every critical point, even where the factor that vanishes there has even multiplicity.
Wrong turn
Error here
For (x−1)2x+3≥0\frac{(x - 1)^2}{x + 3} \geq 0, the sign is positive on (1,∞)(1, \infty), then flipped to negative on (−3,1)(-3, 1) and to positive on (−∞,−3)(-\infty, -3). With the numerator zero included, the answer is (−∞,−3)∪[1,∞)(-\infty, -3) \cup [1, \infty).
Right path
(x−1)2(x - 1)^2 is never negative, so the sign does not change at x=1x = 1. Only x+3x + 3 changes sign, at x=−3x = -3: the expression is negative on (−∞,−3)(-\infty, -3) and non-negative on (−3,∞)(-3, \infty). The solution is (−3,∞)(-3, \infty).
Why it happensThe right-to-left flipping shortcut works whenever every factor has odd multiplicity, as in the second worked example, and it gets applied without checking the exponents.
How to checkTest one value in each interval. x=−4x = -4 gives 25−1=−25\frac{25}{-1} = -25, so (−∞,−3)(-\infty, -3) fails; x=0x = 0 gives 13>0\frac{1}{3} > 0, so (−3,1)(-3, 1) belongs to the answer.
Common pitfalls in rational inequalities·/algebra/inequalities/rationalLearn Math Class
Each of the three slips goes wrong at the critical points: one produces the wrong numerator and so the wrong points, one includes points where the expression is undefined, and one changes the sign at a point where the expression only touches zero. Testing one value inside every interval, and checking each endpoint in the original expression, catches them all.

Rational Inequalities FAQ

What goes wrong if you leave fractions on both sides?

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The sign chart has nothing to read. It needs one rational expression compared to zero, so 1x−3≤2x+1\frac{1}{x-3} \leq \frac{2}{x+1} must first become −x+7(x−3)(x+1)≤0\frac{-x+7}{(x-3)(x+1)} \leq 0 by subtracting and combining over a common denominator. Working across two separate fractions leads straight back into the cross-multiplication trap.Read more →

Is P(x)Q(x)>0\frac{P(x)}{Q(x)} > 0 the same as P(x)Q(x)>0P(x)Q(x) > 0?

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Yes, for a strict inequality. Wherever the fraction is defined it has the same sign as the product, because both are positive exactly when numerator and denominator share a sign. With ≥\geq or ≤\leq the swap needs care: the product also equals 00 at the denominator zeros, so those points have to be removed again, since the fraction does not exist there.Read more →

Is x+35≤1\frac{x + 3}{5} \leq 1 a rational inequality?

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No. A rational inequality needs the variable in a denominator, and here the denominator is the constant 55. Multiplying both sides by 55, a positive number, keeps the direction and gives x+3≤5x + 3 \leq 5, so x≤2x \leq 2: a linear inequality in disguise. Once xx sits in a denominator, that denominator can change sign, and a sign chart takes over.Read more →

Can an expression change sign without passing through zero?

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Yes, a fraction can, at a zero of its denominator. 1x\frac{1}{x} is −1-1 at x=−1x = -1 and 11 at x=1x = 1, yet it is never 00: it jumps across the wall at x=0x = 0. A polynomial cannot do this, because its graph has no breaks. That is why denominator zeros go on the sign chart alongside numerator zeros.Read more →