A complete guide to every inequality type you will encounter in precalculus and Algebra 2 — from simple linear inequalities through compound, absolute value, polynomial, and rational inequalities, with sign charts, interval notation, and real-world applications.
Stewart Precalculus §1.7
Before diving into solving techniques, you need to be fluent in the three ways mathematicians write solution sets: inequality notation, interval notation, and set-builder notation. Every textbook and exam uses all three interchangeably.
| Inequality | Set-Builder | Interval | Number Line |
|---|---|---|---|
| x > a | {x | x > a} | (a, ∞) | Open circle at a, shade right |
| x ≥ a | {x | x ≥ a} | [a, ∞) | Closed circle at a, shade right |
| x < b | {x | x < b} | (−∞, b) | Open circle at b, shade left |
| x ≤ b | {x | x ≤ b} | (−∞, b] | Closed circle at b, shade left |
| a < x < b | {x | a < x < b} | (a, b) | Open circles at a and b, shade between |
| a ≤ x ≤ b | {x | a ≤ x ≤ b} | [a, b] | Closed circles at both, shade between |
| a ≤ x < b | {x | a ≤ x < b} | [a, b) | Closed at a, open at b |
| x < a or x > b | {x | x < a or x > b} | (−∞,a)∪(b,∞) | Two rays outward |
| all real numbers | {x | x ∈ ℝ} | (−∞, ∞) | Entire number line shaded |
Parenthesis ( )
Endpoint is excluded. Used with strict inequalities < and >, and always with ±∞.
Square Bracket [ ]
Endpoint is included. Used with non-strict inequalities ≤ and ≥. Never with infinity.
Union ∪ vs. Intersection ∩
OR solutions use ∪. AND solutions use ∩ (intersection, overlap). Most AND answers simplify to a single interval.
Set-Builder vs. Interval Notation in Practice
Set-builder notation is the most expressive form: {x | x ≠ 0} easily describes the reals with zero removed, which requires two intervals: (−∞, 0) ∪ (0, ∞). For connected intervals, interval notation is preferred for its brevity. Precalculus textbooks (Stewart, Larson, Sullivan) use interval notation almost exclusively for solution sets.
When writing the solution to an inequality, always check: (1) does the boundary point satisfy the original inequality? If yes, use a bracket. If no, use a parenthesis. This one check catches most bracket errors.
The Foundation of All Inequality Work
A linear inequality in one variable has the form ax + b < c (or any of the four inequality symbols). The solution process mirrors solving a linear equation, with one critical difference: the sign-flip rule when multiplying or dividing by a negative.
Clear fractions (if any)
Multiply every term on both sides by the LCD. This eliminates denominators without changing the inequality direction — provided the LCD is positive (constants always are).
Distribute and combine like terms
Expand parentheses on both sides. Combine any like terms on each side separately before moving terms across the inequality.
Move variable terms to one side
Add or subtract variable terms to collect them on one side. Addition and subtraction never flip the inequality sign.
Move constants to the other side
Add or subtract constants. Again, no sign flip needed for addition or subtraction.
Divide by the coefficient — FLIP if negative
Divide both sides by the coefficient of x. If the coefficient is negative, REVERSE the inequality symbol. This is the only step that can change the direction.
Example 1 — Basic Linear Inequality
Solve: 3x − 7 < 8
3x − 7 < 8
3x < 15 (add 7 to both sides)
x < 5 (divide by 3, positive — no flip)
Solution:
x < 5 or (−∞, 5)
Open circle at 5, shade to the left on the number line.
Example 2 — Negative Coefficient (Sign Flip)
Solve: −4x + 3 ≥ 11
−4x + 3 ≥ 11
−4x ≥ 8 (subtract 3 from both sides)
x ≤ −2 (divide by −4 → FLIP ≥ to ≤)
Solution:
x ≤ −2 or (−∞, −2]
Closed circle at −2 (included), shade left.
Example 3 — Inequality with Fractions
Solve: (x/3) − 2 > (x/2) + 1
(x/3) − 2 > (x/2) + 1
6 · (x/3) − 6 · 2 > 6 · (x/2) + 6 · 1 (multiply by LCD=6)
2x − 12 > 3x + 6
−12 − 6 > 3x − 2x (rearrange)
−18 > x (equivalently, x < −18)
Solution:
x < −18 or (−∞, −18)
⚠ The Sign-Flip Rule — Never Forget
Multiply or divide both sides by a positive number: inequality direction stays the same. Multiply or divide by a negative number: inequality direction reverses. This also applies to compound inequalities — when you divide all three parts by a negative, flip both inequality symbols simultaneously.
AND (Intersection) and OR (Union)
A compound inequality combines two inequalities with the word AND or OR. The connector determines whether you intersect or union the solution sets.
AND Inequalities
Both conditions must be satisfied simultaneously. The solution is the intersection — only values in both solution sets.
x > −2 AND x < 5
Solution: (−2, 5)
Values between −2 and 5
Often written as a three-part inequality: −2 < x < 5. AND inequalities always produce a bounded or empty interval.
OR Inequalities
At least one condition must be satisfied. The solution is the union — all values in either solution set.
x < −3 OR x ≥ 4
Solution: (−∞, −3) ∪ [4, ∞)
Two separate rays
OR inequalities produce a union, often two disjoint intervals. If the intervals overlap, simplify to one interval.
When an AND compound inequality is written as a sandwich like −3 < 2x + 1 ≤ 7, apply every operation to all three parts simultaneously.
−3 < 2x + 1 ≤ 7
−1 −1 −1 (subtract 1 from all parts)
−4 < 2x ≤ 6
÷2 ÷2 ÷2 (divide all parts by 2)
−2 < x ≤ 3
Solution: (−2, 3]
Open circle at −2 (strict <), closed circle at 3 (non-strict ≤).
Solve each inequality separately, then take the union.
Example: 2x − 3 < 1 OR 3x + 2 ≥ 14
Left part:
2x − 3 < 1
2x < 4
x < 2 → (−∞, 2)
Right part:
3x + 2 ≥ 14
3x ≥ 12
x ≥ 4 → [4, ∞)
Union (OR):
(−∞, 2) ∪ [4, ∞)
Empty Set
If the AND intersection produces no overlap (e.g., x < 2 AND x > 5), the solution is the empty set, written as ∅ or {⌉. No value satisfies both conditions.
All Real Numbers
If an OR union covers everything (e.g., x < 5 OR x > 3), the solution is all reals: (−∞, ∞). Every real number satisfies at least one of the conditions.
Two Cases — Inside or Outside
The absolute value |x| measures the distance from x to zero on the number line. An absolute value inequality asks: for which values of x is that distance less than (or greater than) some target a? This geometric interpretation makes both cases intuitive.
|x| < a (Less Than)
Distance from x to 0 is less than a.
This is an AND compound inequality. The solution is a bounded interval — x lies between −a and a.
Memory: Less than = AND = between
|x| > a (Greater Than)
Distance from x to 0 is greater than a.
This is an OR compound inequality. The solution is an unbounded union — x lies outside [−a, a].
Memory: Greater than = OR = outside
| Form | Equivalent Condition | Interval |
|---|---|---|
| |E| < a (a > 0) | −a < E < a | (−a, a) |
| |E| ≤ a (a > 0) | −a ≤ E ≤ a | [−a, a] |
| |E| > a (a > 0) | E < −a OR E > a | (−∞,−a)∪(a,∞) |
| |E| ≥ a (a > 0) | E ≤ −a OR E ≥ a | (−∞,−a]∪[a,∞) |
| |E| < 0 | No solution | ∅ |
| |E| ≥ 0 | Always true | (−∞, ∞) |
Example A — |2x − 1| < 5
|2x − 1| < 5
−5 < 2x − 1 < 5 (less-than → AND)
−4 < 2x < 6 (add 1 to all parts)
−2 < x < 3 (divide by 2)
Solution: (−2, 3)
Example B — |3x + 2| ≥ 7
|3x + 2| ≥ 7
3x + 2 ≤ −7 OR 3x + 2 ≥ 7 (≥ → OR)
3x ≤ −9 OR 3x ≥ 5
x ≤ −3 OR x ≥ 5/3
Solution: (−∞, −3] ∪ [5/3, ∞)
Example C — |5 − 2x| ≤ 3 (watch the sign)
|5 − 2x| ≤ 3
−3 ≤ 5 − 2x ≤ 3
−8 ≤ −2x ≤ −2 (subtract 5 from all parts)
4 ≥ x ≥ 1 (divide by −2, FLIP both signs)
1 ≤ x ≤ 4 (rewrite left to right)
Solution: [1, 4]
|x − c| represents the distance between x and c on the number line. Therefore:
This interpretation is essential in calculus (epsilon-delta definitions) and in understanding error tolerances in applied problems.
Parabolas and Their Signs
A quadratic inequality involves a degree-2 polynomial: ax² + bx + c < 0 (or any symbol). The solution depends on where the parabola lies above or below the x-axis.
Move everything to one side
Rewrite the inequality so one side is 0. For example, x² − 3x > 4 becomes x² − 3x − 4 > 0.
Factor the quadratic
Factor completely. If the quadratic is not factorable over the rationals, use the quadratic formula to find the roots.
Find the zeros
Set each factor equal to zero to find the critical numbers — the x-values where the quadratic equals zero.
Plot critical numbers on a number line
The zeros divide the number line into intervals. A quadratic with two real roots creates three intervals.
Test a value in each interval
Choose a convenient test value in each interval and evaluate the sign of the quadratic. The sign is constant throughout each interval.
Select the correct intervals
Pick intervals where the inequality is satisfied. Include endpoints for ≤ or ≥; exclude them for strict inequalities.
Example 1 — x² − 5x + 6 < 0
x² − 5x + 6 < 0
(x − 2)(x − 3) < 0 (factor)
Zeros: x = 2 and x = 3
Sign chart for (x − 2)(x − 3):
| Interval | Test x | (x−2) | (x−3) | Product |
|---|---|---|---|---|
| (−∞, 2) | x=0 | − | − | + (above x-axis) |
| (2, 3) | x=2.5 | + | − | − (below x-axis) ✓ |
| (3, ∞) | x=4 | + | + | + (above x-axis) |
Solution: (2, 3)
Strict inequality → open circles at both endpoints.
Example 2 — x² − 3x − 4 ≥ 0
x² − 3x − 4 ≥ 0
(x − 4)(x + 1) ≥ 0 (factor)
Zeros: x = 4 and x = −1
Sign chart (product positive or zero):
| Interval | Test x | (x−4) | (x+1) | Product |
|---|---|---|---|---|
| (−∞, −1) | x=−2 | − | − | + ✓ |
| (−1, 4) | x=0 | − | + | − |
| (4, ∞) | x=5 | + | + | + ✓ |
Include x = −1 and x = 4 because the inequality is ≥ 0.
Solution: (−∞, −1] ∪ [4, ∞)
Example 3 — Discriminant Analysis: x² + x + 2 > 0
Attempt to factor or use the discriminant: b² − 4ac = 1 − 8 = −7 < 0.
Since the discriminant is negative, the quadratic has no real roots — the parabola never crosses the x-axis. Because the leading coefficient a = 1 > 0, the parabola opens upward and is entirely above the x-axis.
Solution: (−∞, ∞) — all real numbers
The quadratic is always positive; every x satisfies the inequality.
For a quadratic that is hard to factor, complete the square to find the vertex. The vertex form a(x − h)² + k reveals the minimum (a > 0) or maximum (a < 0) value of the quadratic.
Example: x² − 4x − 1 > 0
x² − 4x − 1 = (x² − 4x + 4) − 4 − 1 = (x − 2)² − 5
Zeros at (x − 2)² = 5, so x = 2 ± √5
Solution: (−∞, 2−√5) ∪ (2+√5, ∞)
Sign Chart / Test-Interval Method
Any polynomial inequality of degree 3 or higher is solved using the sign chart method. The key insight: a polynomial is continuous, so it can only change sign at its real zeros. Between consecutive zeros, the sign is constant — you only need to check one point per interval.
Standard form: one side equals zero
Rewrite so the inequality has zero on the right. Do NOT expand unnecessarily — keep the factored form.
Factor completely
Find all linear and irreducible quadratic factors. Use the rational root theorem, synthetic division, and grouping as needed.
Find all real zeros
Set each factor equal to zero. Complex or repeated roots affect multiplicity but not basic sign analysis.
Mark zeros on a number line
Place all real critical numbers on a number line in order. n zeros create n+1 test intervals.
Determine the sign in each interval
Pick any convenient test value in each interval. Evaluate each factor's sign, then combine (count the number of negative factors — odd count gives overall negative sign).
Build the solution
Select intervals where the polynomial has the required sign. Endpoints are included only when the inequality uses ≤ or ≥ and the point is a zero (not a hole or asymptote).
Example — (x + 1)(x − 2)(x − 4) ≤ 0
Already factored. Zeros: x = −1, x = 2, x = 4. Four test intervals.
| Interval | Test x | (x+1) | (x−2) | (x−4) | Product |
|---|---|---|---|---|---|
| (−∞, −1) | −2 | − | − | − | − ✓ |
| (−1, 2) | 0 | + | − | − | + |
| (2, 4) | 3 | + | + | − | − ✓ |
| (4, ∞) | 5 | + | + | + | + |
Include zeros x = −1, x = 2, x = 4 because inequality is ≤ (non-strict).
Solution: (−∞, −1] ∪ [2, 4]
A repeated zero affects whether the polynomial changes sign at that point:
Odd Multiplicity
The polynomial crosses the x-axis — the sign changes at this zero. A simple zero (x − a) has multiplicity 1 (odd): sign changes.
Even Multiplicity
The polynomial touches the x-axis and bounces back — the sign does NOT change at this zero. Example: (x − 3)² is always ≥ 0; the sign is + on both sides.
Example: x(x − 2)² > 0. Sign does not change at x = 2 (even multiplicity), but does change at x = 0 (odd). Sign chart: negative on (−∞, 0), positive on (0, 2) and (2, ∞) — answer: (0, 2) ∪ (2, ∞), simplified as (0, ∞) excluding x = 2 if the inequality is strict.
Including Undefined Points in Sign Analysis
A rational inequality involves a fraction with a variable in the denominator. The key complication: values that make the denominator zero are critical points even though they are not solutions — they create vertical asymptotes and cannot be included in the answer.
Move everything to one side
Get zero on the right side. Do NOT cross-multiply — you do not know the sign of the denominator, so you cannot safely move it.
Combine into a single fraction
Find a common denominator if needed and write the left side as one fraction. Factor both numerator and denominator.
Find zeros of the numerator
Set the numerator equal to zero. These x-values make the fraction equal to zero and may or may not be included (depending on the inequality symbol).
Find zeros of the denominator
Set the denominator equal to zero. These x-values make the fraction undefined. NEVER include them in the solution, even with ≤ or ≥.
Build a sign chart with all critical points
Plot all zeros of numerator and denominator on a number line. These divide the line into intervals.
Test each interval
Evaluate the sign of the fraction in each interval. You can track numerator and denominator signs separately and then divide.
Write the solution
Select intervals where the fraction has the required sign. Use open endpoints at denominator zeros always; open or closed at numerator zeros depending on the inequality.
⚠ Never Cross-Multiply a Rational Inequality
Cross-multiplying requires knowing the sign of both sides. In a rational inequality, the denominator changes sign depending on x. If you multiply by a quantity that is sometimes negative, you would need to flip the sign — but you cannot do this once for the entire solution. Instead, always bring everything to one side and use the sign chart on the combined fraction.
Example 1 — (x + 2)/(x − 3) > 0
Numerator zero: x + 2 = 0 → x = −2
Denominator zero: x − 3 = 0 → x = 3 (undefined, excluded)
Critical points: −2 and 3
| Interval | Test x | (x+2) | (x−3) | Fraction |
|---|---|---|---|---|
| (−∞, −2) | −3 | − | − | + ✓ |
| (−2, 3) | 0 | + | − | − |
| (3, ∞) | 4 | + | + | + ✓ |
Strict inequality (>), so x = −2 is excluded. x = 3 is always excluded (denominator zero).
Solution: (−∞, −2) ∪ (3, ∞)
Example 2 — (2x − 1)/(x + 3) ≤ 2
(2x − 1)/(x + 3) − 2 ≤ 0 (move 2 to left side)
(2x − 1 − 2(x + 3))/(x + 3) ≤ 0
(2x − 1 − 2x − 6)/(x + 3) ≤ 0
−7/(x + 3) ≤ 0
−7 is a negative constant. The fraction −7/(x+3) is ≤ 0 when (x+3) > 0, i.e., when x > −3. Note x = −3 is excluded (undefined), and the numerator is never zero so no numerator zeros.
Solution: (−3, ∞)
x = −3 excluded because the original expression is undefined there.
Example 3 — x/(x − 1) ≥ x/(x + 2)
x/(x−1) − x/(x+2) ≥ 0
x(x+2)/[(x−1)(x+2)] − x(x−1)/[(x−1)(x+2)] ≥ 0
[x(x+2) − x(x−1)] / [(x−1)(x+2)] ≥ 0
[x²+2x − x²+x] / [(x−1)(x+2)] ≥ 0
3x / [(x−1)(x+2)] ≥ 0
Zeros of numerator: x = 0. Zeros of denominator: x = 1, x = −2. Critical points: −2, 0, 1. Sign chart gives positive on (−2, 0) and (1, ∞). Include x = 0 (numerator zero, ≥); exclude x = −2 and x = 1 (denominator zeros).
Solution: (−2, 0] ∪ (1, ∞)
Intersections and Regions Above / Below the x-axis
The algebraic sign chart method and the graphical method are two sides of the same coin. Understanding both deepens intuition and provides a visual check on algebraic work.
Move everything to one side: f(x) > 0. Graph y = f(x). The solution to f(x) > 0 is the set of x-values where the graph lies above the x-axis. The solution to f(x) < 0 is where the graph lies below. Zeros of f(x) are where the graph crosses the x-axis.
Graph both sides as separate functions: y = f(x) and y = g(x). The solution to f(x) > g(x) is where the graph of f lies above the graph of g. Find intersection points first — those are the critical numbers.
f(x) > 0: The x-values where the curve is above the x-axis (y > 0). Look for the portions of the graph with positive y-coordinates.
f(x) < 0: The x-values where the curve is below the x-axis (y < 0). Look for the portions of the graph with negative y-coordinates.
f(x) > g(x): The x-values where the graph of f is higher than the graph of g. The intersection points are the boundaries of the solution intervals.
At x-intercepts: The function equals zero. Include these x-values if the inequality symbol is ≤ or ≥; exclude them if it is < or >. For rational functions, vertical asymptotes (where the function is undefined) always split intervals but are never included.
Brief Overview
A system of inequalities consists of two or more inequalities that must be satisfied simultaneously. In one variable, the solution is found by intersecting the individual solution sets. In two variables (Algebra 2 / Precalculus), the solution is a region in the coordinate plane.
Solve each inequality and intersect the solution sets.
System:
2x − 3 > 1
x < 7
Solutions:
x > 2 → (2, ∞)
x < 7 → (−∞, 7)
Intersection: (2, 7)
Graph each inequality on the coordinate plane. The solution region is the intersection of all shaded half-planes.
System:
y ≥ 2x − 1
y < −x + 4
Graph both lines. Shade above the first line (≥) and below the second (<). The overlapping region is the solution. Used in linear programming.
Budget Constraints, Distance Problems, and More
Inequalities model real-world situations where you need a range of acceptable values rather than a single answer. The word “at least,” “at most,” “between,” “no more than,” and “no fewer than” are signals to write an inequality.
Problem: You have a budget of at most $120 for supplies. You must spend $35 on required materials. Each additional item costs $8.50. How many additional items can you buy?
35 + 8.50n ≤ 120
8.50n ≤ 85
n ≤ 10
You can buy at most 10 additional items.
Solution in interval notation for n: [0, 10] (assuming n must be a non-negative whole number in context).
Problem: A car travels at 60 mph. How long must it travel to cover between 90 and 150 miles?
90 ≤ 60t ≤ 150
1.5 ≤ t ≤ 2.5
The car must travel between 1.5 and 2.5 hours.
Solution: t ∈ [1.5, 2.5]
Problem: A part must be manufactured to within 0.005 cm of the target length of 12.000 cm. Express the acceptable length range using absolute value, then solve.
|L − 12| ≤ 0.005
−0.005 ≤ L − 12 ≤ 0.005
11.995 ≤ L ≤ 12.005
Acceptable length: [11.995, 12.005] cm
This is the standard absolute value tolerance model used in engineering and manufacturing contexts.
Problem: Revenue is R(x) = 50x and cost is C(x) = 20x + 1200, where x is units sold. For what values of x is the company profitable (revenue exceeds cost)?
R(x) > C(x)
50x > 20x + 1200
30x > 1200
x > 40
The company is profitable when more than 40 units are sold.
Break-even point: x = 40. Profitable region: (40, ∞).
Problem: A student has scores of 72, 85, 68, and 90 on four tests. What score does the student need on the fifth test to have an average of at least 80?
(72 + 85 + 68 + 90 + x)/5 ≥ 80
(315 + x)/5 ≥ 80
315 + x ≥ 400
x ≥ 85
The student needs at least 85 on the fifth test.
Since test scores are bounded above by 100, the full solution in context is 85 ≤ x ≤ 100, or [85, 100].
The Most Tested Traps
Mistake 1: Forgetting to flip the inequality sign when dividing by a negative
WRONG
−3x > 12 → x > −4
CORRECT
−3x > 12 → x < −4 (divide by −3, FLIP sign)
Tip: Every time you divide or multiply by a negative, ask yourself: did I flip? Make it a habit to circle the negative coefficient before dividing.
Mistake 2: Including denominator zeros in rational inequality solutions
WRONG
(x−2)/(x+1) ≥ 0 → Solution includes x = −1
CORRECT
x = −1 makes the denominator zero → always excluded with a parenthesis, even for ≥
Tip: When building the solution of a rational inequality, write a parenthesis (never bracket) at every denominator zero before you even start the sign chart.
Mistake 3: Cross-multiplying in a rational inequality
WRONG
(x+3)/(x−2) > 1 → x+3 > x−2 → 3 > −2 (seems like all reals, wrong!)
CORRECT
Move everything to one side: (x+3)/(x−2) − 1 > 0 → 5/(x−2) > 0 → x > 2
Tip: Rational inequality? Never cross-multiply. Always subtract one side and combine into a single fraction first.
Mistake 4: Using the wrong bracket type for absolute value inequalities
WRONG
|x − 2| < 5 → [−3, 7] (brackets used for strict inequality)
CORRECT
|x − 2| < 5 → (−3, 7) (strict < uses open endpoints)
Tip: Check the original inequality symbol: strict (< or >) uses parentheses; non-strict (≤ or ≥) uses brackets at the numerical boundary.
Mistake 5: Forgetting to check if the expression inside an absolute value could equal zero
WRONG
Treating |E| > 0 as always solved by x > 0 or x < 0
CORRECT
|E| > 0 means E ≠ 0 — exclude only the value(s) where E = 0
Tip: |E| > 0 is satisfied for all x except where E = 0. The solution is all reals minus the zeros of E.
Mistake 6: Writing compound AND solutions as union instead of intersection
WRONG
x > 2 AND x < 5 → (−∞, 2) ∪ (5, ∞)
CORRECT
x > 2 AND x < 5 → (2, 5) (intersection, the overlap)
Tip: AND = intersection = overlap. OR = union = everything in either set. Draw both solution sets on a number line and shade the overlap (AND) or everything (OR).
Mistake 7: Distributing a negative incorrectly when moving terms
WRONG
5 − (x + 3) ≤ 0 → 5 − x + 3 ≤ 0 → 8 − x ≤ 0
CORRECT
5 − (x + 3) ≤ 0 → 5 − x − 3 ≤ 0 → 2 − x ≤ 0 → x ≥ 2
Tip: When distributing a minus sign, it negates every term inside the parentheses. A common source of sign errors — write out the distribution step explicitly.
Mistake 8: Stopping after finding the zeros but not testing signs
WRONG
x² − 4 > 0 → zeros are ±2 → solution is (−2, 2)
CORRECT
Sign chart shows x² − 4 > 0 when x < −2 or x > 2 → (−∞, −2) ∪ (2, ∞)
Tip: Always test a value in each interval. The zeros divide the number line, but the direction of the inequality (which side is positive) still requires verification.
| Type | Key Rule | Solution Form |
|---|---|---|
| Linear | Flip sign when ÷ or × by negative | Ray or bounded interval |
| Compound AND | Intersection of both solution sets | Single interval or ∅ |
| Compound OR | Union of both solution sets | Union of intervals or (−∞, ∞) |
| Abs Val < a | AND: −a < expr < a | (−a, a) — bounded |
| Abs Val > a | OR: expr < −a or expr > a | (−∞,−a) ∪ (a,∞) — unbounded |
| Quadratic | Factor → zeros → sign chart (3 intervals) | Interval(s) or ∅ or ℝ |
| Polynomial | Factor → all real zeros → sign chart | Union of intervals |
| Rational | One side = 0; never cross-multiply; exclude denom zeros | Union of intervals (open at denom zeros) |
What is the most important rule when solving inequalities?
The most critical rule is: whenever you multiply or divide both sides of an inequality by a negative number, you must reverse the inequality symbol. For example, −2x > 6 becomes x < −3 after dividing both sides by −2. Forgetting to flip the sign when dividing by a negative is the single most common error in inequality problems.
How do you convert between inequality notation and interval notation?
Use parentheses for strict inequalities (< or >) where the endpoint is excluded, and square brackets for non-strict inequalities (≤ or ≥) where the endpoint is included. Infinity always requires a parenthesis. For example: x > 5 becomes (5, ∞); x ≤ 3 becomes (−∞, 3]; −2 ≤ x < 7 becomes [−2, 7). On a number line, an open circle corresponds to a parenthesis and a filled circle corresponds to a bracket.
What is the difference between an absolute value less-than and greater-than inequality?
For |x| < a: the solution is −a < x < a — a bounded interval (AND compound inequality). For |x| > a: the solution is x < −a OR x > a — an unbounded union. Memory trick: Less than = AND = between. Greater than = OR = outside.
How does the sign chart method work for polynomial inequalities?
A polynomial can only change sign at its real zeros. Factor the polynomial, find all real zeros, mark them on a number line (dividing it into intervals), then test one value in each interval. The sign is constant throughout each interval. Select the intervals where the sign matches the inequality.
Why can you never cross-multiply in a rational inequality?
Cross-multiplying works only when both denominators are known to be positive. In a rational inequality, the denominator may be positive for some x and negative for others. Multiplying by an unknown sign might require flipping the inequality — which you cannot do globally. The safe method: bring everything to one side as a single fraction, then apply the sign chart.
How do you solve a quadratic inequality like x squared minus 5x plus 6 is less than 0?
Factor: (x − 2)(x − 3) < 0. Find zeros: x = 2 and x = 3. Test intervals: in (2, 3), the product is negative (check with x = 2.5). In (−∞, 2) and (3, ∞) the product is positive. Solution: (2, 3). Strict inequality means open endpoints.
What is set-builder notation and how does it relate to interval notation?
Set-builder notation describes a set by the property its members satisfy: the format is {x | condition⌉. For example, {x | x > 3⌉ is written (3, ∞) in interval notation. Interval notation is preferred for connected intervals due to its brevity. Set-builder notation is more flexible for complex conditions like x ≠ 0.
How do you solve a compound AND inequality like negative 3 is less than 2x plus 1 which is less than or equal to 7?
Apply every operation to all three parts simultaneously. Subtract 1: −4 < 2x ≤ 6. Divide by 2: −2 < x ≤ 3. Solution: (−2, 3]. Open circle at −2 (strict <), closed circle at 3 (non-strict ≤).
Master linear, compound, absolute value, polynomial, and rational inequalities with adaptive practice problems. NailTheTest gives you instant feedback and step-by-step solutions matched to your level.
Linear Equations
Solving for x — the foundation before inequalities.
Absolute Value
Distance interpretation and equation solving.
Quadratic Equations
Factoring, completing the square, quadratic formula.
Polynomial Division
Long division and synthetic division for factoring.
Rational Functions
Asymptotes, holes, and graphing behavior.
Completing the Square
Vertex form and solving unfactorable quadratics.