Enter two linear conditions joined by AND or OR, or write a chained bound. The calculator solves each part before showing where the answer sets overlap or combine.
Result
Values greater than -1 and at most 5 satisfy both bounds
−1<x≤5⟺(−1,5]
Conditions
Both comparisons are interpreted together, not as separate answer choices.
The left endpoint is open because the first comparison is strict.
The right endpoint is closed because the second comparison includes equality.
Steps
Keep both bounds attached The chain requires the middle expression to be above 1 and at most 7.1<x+2≤7
Subtract from every part Subtract 2 from the left bound, the middle, and the right bound.−1<x≤5
Write the intersection Keep the overlap of x > -1 and x ≤ 5, including only the right endpoint.(−1,5]
✓
Independent check
At x = -1 the left comparison becomes 1 < 1 and fails; at x = 5 both comparisons hold. x = 0 passes both, while x = 6 fails the upper bound.
Scope
What this compound inequality covers
AND keeps only values that satisfy both conditions, so take an intersection. OR keeps values that satisfy at least one, so take a union. A chain such as 1 < x + 2 ≤ 7 means both comparisons must hold at once.
Intersect conditions joined by AND
Solve each linear comparison and keep only values satisfying both; detect when the overlap is empty.
Examples: x > 2 and x ≤ 6, x < -1 and x > 4
Unite conditions joined by OR
Keep values satisfying either comparison; combine touching or overlapping intervals when they cover one continuous region.
Examples: x < -3 or x ≥ 1, x ≤ 0 or x ≥ 0
Solve a chained bound
Treat a three-part chain as two simultaneous comparisons, applying any balancing operation to all three parts.
Examples: 1 < x + 2 ≤ 7, -4 ≤ 2x + 2 < 8
Show exact set notation and endpoints
Use an interval for an intersection and a union for separated OR branches; mark strict endpoints open and inclusive endpoints closed.
Examples: (-1, 5], (-∞, -3) ∪ [1, ∞)
How to use it
Enter enough information for one clear task
1
Choose the intended connector
Type AND when both conditions must hold or OR when either condition is enough. A chained comparison is an AND statement.
2
Solve each side before combining
Isolate x in each ordinary linear inequality. If you divide by a negative number, reverse that comparison before combining the answer sets.
3
Inspect the final overlap or union
Check the boundary points and a sample from each visible region against both original conditions using the selected connector.
Worked inputs
Examples to try
Use these examples to recognize the method, compare equivalent forms, and check your own work.
Chain with one open and one closed end
Subtract 2 from all three parts and preserve each comparison symbol.
1<x+2≤7
Expected result
(−1,5]
Two conditions joined by AND
Take the intersection: values must pass both conditions.
x>2andx≤6
Expected result
(2,6]
Two separated OR branches
Take the union: either separate ray is accepted.
x<−3orx≥1
Expected result
(−∞,−3)∪[1,∞)
Chain after scaling
Subtract 2 throughout, then divide throughout by positive 2.
−4≤2x+2<8
Expected result
[−3,3)
An empty intersection
No real value can be below -1 and above 4 simultaneously.
x<−1andx>4
Expected result
∅
An OR union covering every value
Negative, positive, and zero values all satisfy at least one branch.
x≤0orx≥0
Expected result
(−∞,∞)
Two solved branches with different coefficients
Solve x < 2 and x ≥ -2, then keep the overlap.
3x−1<5andx+2≥0
Expected result
[−2,2)
Complete example
Combine two solved branches with OR
The two comparisons must be solved independently before OR can join their answer sets.
2x−1<−5or3x+4≥10
1
Solve the left branch
Add 1, then divide by positive 2.
2x<−4⟹x<−2
2
Solve the right branch
Subtract 4, then divide by positive 3.
3x≥6⟹x≥2
3
Keep either passing region
OR joins the left ray and the right ray; the left boundary is open and the right is closed.
(−∞,−2)∪[2,∞)
x<−2orx≥2
Verification: x = -3 passes the left branch; x = 2 passes the right branch exactly; x = 0 passes neither. At x = -2 the left branch is equality and the right branch is false, so -2 is excluded.
Avoidable errors
Common mistakes and how to fix them
Reading AND as OR
Problem: For x > 2 and x ≤ 6, retain either ray.
Why it matters: A value on only one ray does not satisfy both original conditions.
Better approach: Draw or compare both sets, then keep only their overlap (2, 6].
Reading OR as AND
Problem: For x < -3 or x ≥ 1, report no solution because the rays do not overlap.
Why it matters: OR asks whether at least one branch passes, so disjoint regions are allowed.
Better approach: Keep both valid regions and write their union.
Changing only the middle of a chain
Problem: Subtract 2 from x + 2 but leave the outer bounds unchanged.
Why it matters: Each bound compares with the same middle expression; an operation must preserve both comparisons.
Better approach: Apply the same addition or subtraction to all three parts of a chain.
Treating a touching OR union as disconnected
Problem: For x ≤ 0 or x ≥ 0, leave a gap at zero.
Why it matters: Zero is included by both comparisons, and every other real number lies on one of the two sides.
Better approach: Check open and closed endpoints before deciding whether the union has a gap.
Trust the result for the right reasons
Checks, assumptions, and limits
How results are checked
Both constituent comparisons are solved before applying the connector.
AND is set intersection; OR is set union; a chained comparison is treated as AND.
Each endpoint is tested against the original joined statement for inclusion.
A sample from the overlap, from each outer branch, and from rejected regions confirms the displayed set.
When to stop and revise the input
Supported input is exactly two linear inequalities in one real variable joined by AND or OR, or one three-part linear chain.
Nested Boolean groups, three or more separate branches, quadratic branches, and multiple variables are outside this tool's scope.
Use the ordinary inequality calculator for one comparison and the absolute-value calculator for one distance condition.
Do not infer a connector from punctuation: enter AND, OR, or a complete chain explicitly.
Common questions
Solve a compound inequality FAQ
What is the difference between AND and OR?
AND requires both comparisons to be true for the same x, so take their intersection. OR requires at least one true comparison, so take their union.
Is a chained inequality an AND statement?
Yes. For example, 1 < x + 2 ≤ 7 means 1 < x + 2 and x + 2 ≤ 7 simultaneously. Solving all three parts together is a compact way to find that intersection.
Can a compound inequality have no real solution?
Yes. An AND statement can demand incompatible regions, such as x < -1 and x > 4. Its intersection is empty.
Can an OR statement equal all real numbers?
Yes. The union x ≤ 0 or x ≥ 0 covers every real number, including zero. Endpoint inclusion decides whether a touching union leaves a gap.
Why do some answers use a union symbol?
A union joins two acceptable sets. It is especially useful when an OR statement produces separated intervals, such as one left ray and one right ray.
Sources and curriculum alignment
This page follows standard introductory mathematics notation and learning sequences. Use these references to continue with a complete course treatment.
