How this instrument works
A linear inequality like ax + b < c is solved with exactly the moves used on the equation ax + b = c: subtract b from both sides, then divide by a. What gets isolated at the end is not a single answer but a boundary — every number on one side of that point satisfies the inequality, and every number on the other side does not. This calculator carries out that algebra and returns the boundary point, (c − b) ⁄ a, ready to be dropped into interval notation.
Once that boundary is known, writing the final interval takes one more decision: which direction the inequality opens. A 'less than' relation opens toward negative infinity, giving (−∞, boundary); a 'greater than' relation opens the other way, giving (boundary, ∞). For a strict inequality, one using < or > rather than ≤ or ≥, the boundary point itself is always excluded, so that end of the interval gets a parenthesis rather than a bracket, no matter which direction it runs.
It helps to keep the algebra and the notation as two separate jobs. This page handles the first one: isolating x the same way you would for the equation ax + b = c. A sibling calculator on this site, which tests a candidate number against the original inequality, answers a different question — whether one particular value sits inside the solution set — rather than finding the boundary that defines that set in the first place.
- Enter the coefficient on x into the a field, matching the ax + b < c form of your inequality.
- Enter the constant being added into b, and the right-hand value into c.
- Read Boundary point (x <): the value x is being compared against, once isolated.
- Write the solution as (−∞, boundary) if the original inequality used <, or (boundary, ∞) if it used >.
- Keep the boundary itself out of the interval with a parenthesis, since a strict inequality never includes its own boundary value.
Worked example — solving 2x + 3 < 11
Start with 2x + 3 < 11. Subtract 3 from both sides, the same first move used on the equation 2x + 3 = 11, leaving 2x < 8. Divide both sides by 2 and the boundary appears on its own: x < 4. That matches the formula directly, since boundary = (c − b) ⁄ a = (11 − 3) ⁄ 2 = 8 ⁄ 2 = 4. Because the original relation used a strict less-than and opens toward smaller numbers, the solution set in interval notation is (−∞, 4), with a parenthesis at 4 rather than a bracket.
The same two moves handle any values of a, b, and c. For x − 5 < 0, treat it as a = 1, b = −5, c = 0: boundary = (0 − (−5)) ⁄ 1 = 5, so the interval is (−∞, 5). For 3x < 9, a = 3, b = 0, c = 9: boundary = (9 − 0) ⁄ 3 = 3, giving (−∞, 3). Subtract, then divide, and the boundary falls out the same way every time.
Questions
How is a linear inequality like ax + b < c solved?
The same way as the matching equation: subtract b from both sides to get ax < c − b, then divide by a to isolate x, giving x < (c − b) ⁄ a. For 2x + 3 < 11 that works out to x < (11 − 3) ⁄ 2 = 4. Dividing by a negative a would flip the direction of the inequality, though this calculator assumes a positive coefficient.
How do I turn a solved inequality into interval notation?
Once the boundary is known and you know which way the inequality points, the interval writes itself. A 'less than' result opens toward negative infinity: (−∞, boundary). A 'greater than' result opens the other way: (boundary, ∞). For x < 4 that gives (−∞, 4).
Why is the boundary point excluded from the interval?
Because the inequality is strict — it uses < rather than ≤ — so the boundary itself never actually satisfies it. Plugging x = 4 into 2x + 3 < 11 gives 2(4) + 3 = 11, which is not less than 11, only equal to it. That is why the interval (−∞, 4) uses a parenthesis at 4 instead of a bracket.
What if the inequality uses ≤ instead of <?
The algebra to find the boundary stays identical, subtract, then divide, but the boundary point itself would now belong to the solution set, since a ≤ relation is satisfied exactly at that point. The interval would then use a bracket at that end, such as (−∞, 4], instead of a parenthesis.
How does this differ from testing one specific candidate value?
This page solves for the boundary itself, the single point where the matching equation would balance, useful for writing the full solution set. A sibling calculator instead plugs one number into the original inequality and reports whether that number satisfies it, a narrower, one-point check rather than a full boundary search.
Does this work if a, b, or c are decimals rather than whole numbers?
Yes, the same subtract-then-divide steps apply regardless of whether a, b, and c are whole numbers or decimals. Solving 0.5x + 1.5 < 4, for instance, follows the identical path: subtract 1.5, then divide by 0.5, arriving at the boundary the same way as with tidier figures.