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Compound Inequality Calculator

Solve AND/OR compound inequalities and three-part inequalities instantly - get step-by-step solutions with intersection and union explained, dual number line graphs, and interval notation. Free, no sign-up required.

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Why this page is different

Separates AND, OR, and three-part logic instead of forcing one generic workflow.
Shows both branches and the final result on a dedicated dual number line.
Pairs interval notation with verification and set-logic visualization so the answer stays teachable.

Compound Inequality Solver

Built for AND, OR, and three-part inequalities with step-by-step logic, dual number lines, Venn-style set intuition, interval notation, and direct value verification.

Structured input

Use the mode-specific fields below if you prefer guided entry.

AND

Structured preview

x > 1 AND x < 5

Live parse preview
x > 1 and x < 5

Supported Input Styles

  • x > 1 and x < 5 for an AND intersection.
  • x < -2 or x > 4 for an OR union.
  • x > 1 && x < 5 and x < -2 || x > 4 are normalized automatically.
  • -3 <= 2x + 1 < 7 is read as a three-part inequality.
  • x > 5 and x < 2 shows the no-solution special case.

Math Keyboard

Tap AND/OR logic, symbols, and numbers for fast compound-inequality input.

Result

Both inequalities must be true, so the final answer is the intersection of the two solution sets.

Typeand
Condition1 < x < 5
Interval(1, 5)
AND COMPOUND INEQUALITY
Both conditions must be true. Find the intersection on the final number line.
x>1  AND  x<5x > 1 \; \text{AND} \; x < 5

Original

x>1  AND  x<5x > 1 \; \text{AND} \; x < 5

Equivalent Form

x>1  AND  x<5x > 1 \; \text{AND} \; x < 5

Condition

1<x<51 < x < 5

Interval

(1,5)(1, 5)
1

Step 1

Identify the Type

This is an AND compound inequality, so both conditions must be true. That means the final answer is the intersection of both solution sets.

Before

x>1  AND  x<5x > 1 \; \text{AND} \; x < 5

After

x>1  AND  x<5x > 1 \; \text{AND} \; x < 5
2

Step 2

Solve inequality 1: Move all terms to one side

Rewrite the inequality so the right-hand side is zero.

Before

x>1  AND  x<5x > 1 \; \text{AND} \; x < 5

After

x1>0x - 1 > 0
3

Step 3

Solve inequality 1: Isolate x

Add or subtract the constant term so that the x-term stands alone.

Before

x1>0x - 1 > 0

After

1x>11x > 1
4

Step 4

Solve inequality 1: Divide by the coefficient of x

Dividing both sides by a positive number keeps the inequality direction the same.

Before

1x>11x > 1

After

x>1x > 1
5

Step 5

Solve inequality 2: Move all terms to one side

Rewrite the inequality so the right-hand side is zero.

Before

x>1x > 1

After

x5<0x - 5 < 0
6

Step 6

Solve inequality 2: Isolate x

Add or subtract the constant term so that the x-term stands alone.

Before

x5<0x - 5 < 0

After

1x<51x < 5
7

Step 7

Solve inequality 2: Divide by the coefficient of x

Dividing both sides by a positive number keeps the inequality direction the same.

Before

1x<51x < 5

After

x<5x < 5
8

Step 8

Find the Intersection (AND)

Keep only the values that survive both branches.

Before

x<5x < 5

After

(1,)(,5)=(1,5)(1, \infty) \cap (-\infty, 5) = (1, 5)

Calculator Types

Switch to another inequality tool in one tap

Zone 4

How to Use the Compound Inequality Calculator

This page is designed for compound-inequality intent rather than generic inequality traffic. That distinction matters because students rarely struggle with the symbol itself. They struggle with the logic word in the middle. The difference between AND and OR changes the entire shape of the answer, the number-line shading, the interval notation, and even the language you use to explain why the answer works.

You can work in two ways. If you already know the full problem, paste it directly as x > 1 and x < 5, x < -2 or x > 4, or -3 <= 2x + 1 < 7. If you want more structure, switch modes and enter each branch separately. AND mode uses two linked inequalities that must both be true. OR mode uses two linked inequalities where either branch can be true. Three-Part mode keeps the shared middle expression and lets you operate on all three parts at once.

Once the input is recognized, the experience follows a clean teaching order. First identify the type. Next solve each branch or the three-part chain. Then read the final overlap or union on the dual number line. After that, switch to interval notation and verification so the answer is no longer abstract. The goal is not only to produce a correct interval. The goal is to make the logic visible enough that the next compound inequality feels familiar instead of arbitrary.

01

Choose AND mode, OR mode, or Three-Part mode, or paste a full expression such as x > 1 and x < 5.

02

Read the branch steps to see each inequality solved separately and how the final intersection or union is formed.

03

Use the dual number line, Venn diagram, interval notation, and verify tab to confirm what the answer means.

What Is a Compound Inequality?

A compound inequality combines two comparisons into one statement. The connection can be logical, as in x > 1 and x < 5 or x < -2 or x > 4, or it can be written as a chained statement such as -3 <= 2x + 1 < 7. In every case, the final solution is a set of real numbers, not a single isolated answer. That set may be a bounded interval, a union of two intervals, all real numbers, no solution, or a single point.

There are two main families. AND compound inequalities require both conditions to hold at the same time. That makes the final answer the intersection of the two solution sets. OR compound inequalities require at least one condition to hold. That makes the final answer the union of the two solution sets. A three-part inequality is best understood as an AND statement with a shared middle expression, which is why it often produces a bounded interval after the variable is isolated.

Compound inequalities matter because real problems often create more than one restriction at once. A score may need to stay above a minimum and below a maximum. A measurement may need to be outside a forbidden region. A formula may create a left boundary and a right boundary in the same sentence. If a student does not recognize whether the answer should narrow inward or open outward, even correct algebra can end in the wrong graph. That is why this page treats the logic word as a first-class concept instead of a side note.

AND example

x>1  AND  x<5x > 1 \; \text{AND} \; x < 5

OR example

x<2  OR  x>4x < -2 \; \text{OR} \; x > 4

Three-part example

32x+1<7-3 \le 2x + 1 < 7

AND Compound Inequalities: Intersection Explained

AND means both inequalities must be true for the same x-value. In set language, that is an intersection. If one branch says x > 1 and the other says x < 5, the final answer is the overlap between those two sets. Values such as 2, 3, and 4 survive because they satisfy both conditions. Values such as 0 fail the first condition. Values such as 6 fail the second. The overlap is the only part that matters.

On a number line, AND usually narrows the solution. That is why teachers often describe it as the middle-interval pattern. When the branches point toward each other, the answer is usually a bounded interval such as (1, 5) or [-2, 3). When both branches point right, the stricter lower bound wins. When both branches point left, the stricter upper bound wins. When the two branches never meet, the answer is no solution because there is no shared region at all.

A useful mental model is constraint stacking. Each new AND condition removes values that fail the new rule. The solution set can only stay the same size or get smaller. It cannot grow. That is why AND problems often produce the tighter bound, the smaller interval, or the empty set. If you remember that AND narrows the surviving set, the final graph becomes easier to predict before you even finish the algebra.

x > a AND x > b

x > max(a, b)

Both conditions point right, so the tighter lower bound survives.

x < a AND x < b

x < min(a, b)

Both conditions point left, so the tighter upper bound survives.

x > a AND x < b

a < x < b when a < b

The final answer is the middle overlap between the two branches.

OR Compound Inequalities: Union Explained

OR means at least one inequality must be true. In set language, that is a union. If one branch says x < -2 and the other says x > 4, the final answer includes both surviving regions because a value is allowed to satisfy either branch. A number such as -5 works because it satisfies the left inequality. A number such as 7 works because it satisfies the right inequality. The fact that neither number satisfies both conditions does not matter. OR never requires both to be true at the same time.

On a number line, OR usually opens the solution outward. That is why many OR answers look like two rays extending toward negative infinity and positive infinity. If the two branches overlap or even touch, the union can merge into one larger interval. If the branches cover the full number line together, the answer becomes all real numbers. In other words, OR is not always two separated pieces, but it almost always keeps more values than either branch alone.

A useful memory line is that OR widens instead of narrows. Each new OR condition gives x another way to survive. The solution set can only stay the same size or get larger. It cannot shrink. That is why OR problems often produce the looser bound, the larger interval, or the entire real line. If you think of OR as opening doors rather than closing them, the final interval notation becomes much more intuitive.

x > a OR x > b

x > min(a, b)

Either branch can work, so the looser lower bound survives.

x < a OR x < b

x < max(a, b)

Either branch can work, so the looser upper bound survives.

x > a OR x < b

All reals if b >= a, otherwise two outer rays

If the rays overlap enough to cover the line, the union becomes all real numbers.

How to Solve Compound Inequalities Step by Step

The first solving decision is structural, not algebraic. Before moving any terms, identify whether the problem is AND, OR, or three-part. That choice tells you whether the final operation will be an intersection, a union, or a simultaneous chain simplification. Skipping this step is one of the most common reasons students turn a correct branch solution into the wrong final interval.

For AND and OR problems, the safest workflow is to solve each inequality separately first. Rewrite each branch until x is isolated, convert each branch into interval notation, and then combine the intervals with the correct set operation. For AND, keep only the overlap. For OR, keep every region that survives at least one branch. If the intervals are adjacent or overlapping in an OR problem, combine them into a single interval before writing the final answer.

For three-part inequalities, the best workflow is different. Because the middle expression is shared, you can often add, subtract, multiply, or divide all three parts at the same time. This is faster and cleaner than splitting the chain immediately. The one rule that still deserves extra attention is the sign-flip rule: if you multiply or divide all three parts by a negative number, both inequality signs reverse direction. After that, write the final condition in standard order and convert it into interval notation.

Input compound inequality
Identify the structure: AND, OR, or three-part
Solve each branch or operate on all three parts
Combine with intersection or union
Check special cases
Write interval notation

Three-Part Compound Inequalities

A three-part compound inequality has the form a [sign] expression [sign] b. A common example is -3 <= 2x + 1 < 7. This format is compact, but it is not mysterious. It is equivalent to two inequalities at once: -3 <= 2x + 1 and 2x + 1 < 7. Because the middle expression is shared, the final answer comes from the overlap of both comparisons, which is why three-part problems belong to the AND family.

The main advantage of the three-part format is efficiency. If you subtract 1 from the middle expression, you also subtract 1 from both outer parts. If you divide by 2, you divide all three parts by 2. That lets you isolate x without splitting the chain immediately. Students who understand this move gain speed and reduce transcription errors because they keep the structure intact while solving.

The main danger is forgetting that a negative multiplier changes both inequality signs. Consider -4 < -2x + 6 <= 10. After subtracting 6, you get -10 < -2x <= 4. Dividing by -2 gives 5 > x >= -2, and both signs flip. If you miss even one flip, the final interval is wrong. That is why this page treats three-part solving as a visible sequence of operations rather than a one-line answer.

Equivalent AND form

32x+1  AND  2x+1<7-3 \le 2x + 1 \; \text{AND} \; 2x + 1 < 7
Divide or multiply all three parts together. If the factor is negative, both inequality signs flip.

Compound Inequality Examples with Solutions

The eight examples below cover bounded intersections, outer unions, three-part chains, coefficient-based branches, and the most important special cases. Each one is analyzed with the same logic used by the interactive calculator above.

Example 1

Basic AND

Bounded middle interval created by intersection.

Problem

x>1  AND  x<5x > 1 \; \text{AND} \; x < 5

Final interval

(1,5)(1, 5)

Both inequalities must be true, so the final answer is the intersection of the two solution sets.

Example 2

Basic OR

Two outer intervals created by union.

Problem

x<2  OR  x>4x < -2 \; \text{OR} \; x > 4

Final interval

(,2)(4,)(-\infty, -2) \cup (4, \infty)

At least one inequality must be true, so the final answer is the union of the two solution sets.

Example 3

Three-Part

Shared middle expression with mixed endpoints.

Problem

32x+1<7-3 \le 2x+1 < 7

Final interval

[2,3)[-2, 3)

A three-part inequality behaves like an AND statement, so the final answer is the overlap that survives both comparisons.

Example 4

AND with coefficients

Solve both branches first, then intersect.

Problem

2x1>3  AND  x+4102x-1 > 3 \; \text{AND} \; x+4 \le 10

Final interval

(2,6](2, 6]

Both inequalities must be true, so the final answer is the intersection of the two solution sets.

Example 5

OR with coefficients

Two solved branches that remain separated after union.

Problem

3x+2<4  OR  2x5>13x+2 < -4 \; \text{OR} \; 2x-5 > 1

Final interval

(,2)(3,)(-\infty, -2) \cup (3, \infty)

At least one inequality must be true, so the final answer is the union of the two solution sets.

Example 6

AND contradiction

No-solution case caused by empty intersection.

Problem

x>5  AND  x<2x > 5 \; \text{AND} \; x < 2

Final interval

\varnothing

There is no solution because the two conditions never overlap.

Example 7

OR full coverage

All-real-numbers case caused by complete union coverage.

Problem

x<8  OR  x>1x < 8 \; \text{OR} \; x > -1

Final interval

(,)(-\infty, \infty)

Every real number satisfies at least one branch, so the union covers the entire number line.

Example 8

Three-Part sign flip

Dividing all three parts by a negative flips both signs.

Problem

4<2x+610-4 < -2x+6 \le 10

Final interval

[2,5)[-2, 5)

A three-part inequality behaves like an AND statement, so the final answer is the overlap that survives both comparisons.

Graphing Compound Inequalities on a Number Line

Graphing is where compound-inequality logic becomes obvious. In an AND problem, draw the first branch, draw the second branch, and then keep only the region where the shadings overlap. In an OR problem, draw both branches and keep every highlighted region. The dual number line on this page makes that comparison explicit by showing branch one, branch two, and the final result on separate but synchronized tracks.

This matters because many wrong answers look plausible in algebra form but fall apart on a graph. A student may correctly solve x > 1 and x < 5, then accidentally shade two rays instead of the middle region. Another student may solve x < 8 or x > -1 and fail to notice that those two rays cover the entire number line. The graph acts as a second proof. If the picture does not match the logic word, the final answer deserves another look.

When graphing, pay attention to endpoint style as well as shading direction. Strict inequalities such as < and > use open circles because the endpoint is excluded. Inclusive inequalities such as <= and >= use closed circles because the endpoint is part of the solution. In a compound problem, the final endpoint style depends on the surviving interval after the intersection or union is formed, not only on one branch in isolation.

AND
--o====o--
OR
====o--o====
Three-part
--*====o--

Interval Notation for Compound Inequalities

Interval notation compresses the final answer into a precise form that is easy to scan. A bounded AND answer such as 1 < x < 5 becomes (1, 5). A mixed-endpoint three-part answer such as -2 <= x < 3 becomes [-2, 3). An OR answer such as x < -3 or x > 8 becomes (-infinity, -3) union (8, infinity). The notation is compact, but every bracket and parenthesis still carries meaning.

The fastest way to avoid mistakes is to connect interval notation to the graph. Open circle means parenthesis. Closed circle means bracket. Infinity always uses parentheses because infinity is never an included endpoint. If the solution contains two separated pieces, the notation must use a union symbol. If the graph is empty, the notation is the empty set. If the graph is a single point, set notation such as {3} can be clearer than interval notation.

On this page, the condition form, interval notation, and set notation are displayed together because each representation reinforces the others. The condition form is easiest for beginners to read. Interval notation is shortest for homework and tests. Set notation is useful when you want to state the rule verbally. Switching among the three is not busywork. It is a way to verify that the same mathematical idea survives every notation system.

Condition formInterval notationGraph cueType
a < x < b(a, b)--o====o--AND
a <= x <= b[a, b]--*====*--AND or three-part
a <= x < b[a, b)--*====o--Three-part
x < a or x > b(-infinity, a) union (b, infinity)====o--o====OR
All real numbers(-infinity, infinity)<========>OR special case
No solutionempty set(empty)AND special case

Special Cases: No Solution & All Real Numbers

Some compound inequalities are valuable precisely because they expose edge cases. In an AND problem such as x > 5 and x < 2, the two branches never overlap, so there is no solution. The empty set is not a failure of algebra. It is the correct description of a contradictory requirement. Likewise, x >= 3 and x <= 3 does not produce a wider interval. It collapses to the single point x = 3 because only one value satisfies both bounds at once.

OR problems have their own signature special case: all real numbers. If one branch says x < 8 and the other says x > -1, every real number satisfies at least one branch because the two surviving regions overlap enough to cover the full number line. This is one of the clearest moments where a graph prevents mistakes. The union does not merely create two rays. It erases the gap completely.

Three-part inequalities can also become contradictory. A statement such as 5 < x < 2 asks x to be greater than 5 and less than 2 at the same time. There is no real number that can do that, so the answer is no solution. These edge cases matter because they teach students not to assume every compound inequality must produce a standard interval. A strong solver checks the final set before stopping.

No Solution

x > 5 and x < 2

The two branches point toward a gap with no overlap, so the intersection is empty. This is the signature AND contradiction.

All Real Numbers

x < 8 or x > -1

The two OR branches cover the entire number line together, so every real x satisfies at least one branch.

Single Point

x >= 3 and x <= 3

The overlap collapses to exactly one included value, so the answer is the single point x = 3.

Frequently Asked Questions

What is a compound inequality?

A compound inequality combines two comparisons into one statement. The comparisons may be linked by AND, linked by OR, or written as a three-part chain such as -3 <= 2x + 1 < 7.

What is the difference between AND and OR compound inequalities?

AND compound inequalities require both conditions to be true, so the final answer is the intersection of both solution sets. OR compound inequalities require at least one condition to be true, so the final answer is the union of both solution sets.

What is a three-part compound inequality?

A three-part inequality has the form a [sign] expression [sign] b. It is equivalent to two linked inequalities with the same middle expression, so it behaves like an AND statement.

How is a compound inequality different from a simple inequality?

A simple inequality uses one comparison. A compound inequality uses two comparisons that must be combined with intersection logic, union logic, or simultaneous chain operations.

What does AND mean in a compound inequality?

AND means both inequalities must be true for the same x-value. The final answer keeps only the overlap shared by both branches.

What does OR mean in a compound inequality?

OR means at least one inequality must be true. The final answer keeps every value that survives either branch.

How do you know if a compound inequality is AND or OR?

Read the logic word in the statement. If the problem says AND, use intersection. If it says OR, use union. A chained three-part statement is usually an AND form with a shared middle expression.

What is the intersection of two inequalities?

The intersection is the set of x-values that satisfy both inequalities at the same time. On a number line, it is the shared overlap.

What is the union of two inequalities?

The union is the set of x-values that satisfy at least one inequality. On a number line, it includes every highlighted region from both branches.

Why does AND give a middle interval but OR gives outer intervals?

AND usually narrows the surviving values to the overlap between two branches, which often sits in the middle. OR usually widens the surviving values by keeping either outer branch.

How do you solve a compound inequality step by step?

First identify whether the problem is AND, OR, or three-part. Then solve each branch separately or operate on all three parts together, combine the results with intersection or union, and write the final answer in interval notation.

How do you solve an AND compound inequality?

Solve each inequality separately, graph both solution sets, and keep only the overlap. That overlap is the final answer.

How do you solve an OR compound inequality?

Solve each inequality separately, graph both solution sets, and keep every region that survives either branch. Merge overlapping regions when possible.

How do you solve a three-part compound inequality?

Apply the same algebraic operation to all three parts of the chain until x is isolated. If you multiply or divide by a negative number, flip both inequality signs.

What happens when you multiply or divide by a negative in a three-part inequality?

Both inequality signs reverse direction because the order of the numbers changes when you multiply or divide by a negative value.

When does a compound inequality have no solution?

A compound inequality has no solution when the surviving branches never overlap in an AND problem, or when neither branch leaves any valid values in an OR problem.

When does a compound inequality have all real numbers as solution?

This happens most often in OR problems when the union of both branches covers the entire number line.

Can a compound inequality have a single point as solution?

Yes. For example, x >= 3 and x <= 3 leaves only x = 3, so the solution is a single point.

How do you graph a compound inequality on a number line?

Graph each branch first, then take the overlap for AND or the full combined shading for OR. Use open circles for strict endpoints and closed circles for inclusive endpoints.

How do you show AND and OR on a number line?

AND is shown by the shared overlap between the branch graphs. OR is shown by keeping every highlighted branch region.

How do I use this compound inequality calculator?

Paste a full expression or choose a mode, enter the inequalities, then review the steps, dual number line, interval notation, and verification tabs.

Is this compound inequality calculator free?

Yes. The calculator is free to use, includes steps and visual explanations, and does not require sign-up.

Related Calculators

If you want to move from compound logic into a specific family of inequalities, these related tools cover the most common next steps.