How this instrument works
Graphing an inequality like ax + b < c on a number line means shading every point that makes the statement true and leaving every other point blank, with an open or closed circle marking where the shading stops. Deciding whether any single point belongs inside that shaded stretch does not require redrawing the whole picture; it only requires plugging that point's value into the expression and checking the sign of the result.
This calculator computes test = ax + b − c for whatever candidate point is picked. A negative test value means the point makes the left side smaller than c, so it satisfies the original inequality and sits inside the shaded region. A positive value means the opposite: the point sits outside the shading, on the side of the number line that fails the statement. A test value of exactly zero means the candidate landed precisely on the boundary itself, the open circle at the edge of the shading.
A sibling page on this site, working the algebra of the same ax + b < c form, solves for that boundary point directly by isolating x. This page starts from the opposite direction: rather than finding the boundary, it takes a point already in mind and checks which side of that boundary it falls on, the exact move made with a pencil and a number line when testing a value to decide which way the shading runs.
- Enter the inequality's coefficients into a, b, and c, matching the ax + b < c form.
- Enter the number you want to check into Candidate point (x).
- Read the Test value: negative means x satisfies the inequality and falls inside the shaded region.
- A positive test value means x sits outside the shading, on the side that fails the inequality.
- A test value of zero means x sits exactly on the boundary, the open circle at the edge, not included since the inequality is strict.
Worked example — testing points against 2x + 3 < 11
Take the inequality 2x + 3 < 11 and test the candidate point x = 3. The test value is ax + b − c = 2(3) + 3 − 11 = 6 + 3 − 11 = −2. That's negative, so x = 3 satisfies the inequality and belongs inside the shaded region on the number line, everywhere left of the boundary.
Now test x = 5 against the same inequality: 2(5) + 3 − 11 = 10 + 3 − 11 = 2, a positive result, meaning x = 5 sits outside the shading. Testing x = 4 lands exactly on the boundary itself: 2(4) + 3 − 11 = 8 + 3 − 11 = 0. That zero marks the open circle at the edge of the shaded region; the boundary point is never included, since the original inequality is strict.
Questions
How do you test whether a point satisfies an inequality?
Plug the point's value into the expression on the inequality's smaller side and subtract the larger side: for ax + b < c, compute ax + b − c. A negative result means the point satisfies the inequality; a positive result means it does not; zero means the point sits exactly on the boundary.
What does a negative test value mean on a number line?
It means the candidate point falls inside the shaded region, the part of the number line where the inequality holds true. For 2x + 3 < 11 tested at x = 3, the result is −2, confirming that x = 3 belongs to the solution set being graphed.
What does it mean when the test value comes out to zero?
The candidate point sits exactly on the boundary of the inequality, the same value solving ax + b = c directly would give. Because a strict inequality like < never includes its own boundary, that point earns an open circle on the number line rather than shading.
Why test a point instead of just solving the whole inequality?
Testing a point answers a narrower, faster question: does this one number work, yes or no. Solving the inequality fully finds the boundary and the entire direction of the shading at once, useful when the complete solution set is needed rather than a single spot check.
Can this method work for 'greater than' inequalities too?
Yes, the sign just flips in meaning. Rearranging a 'greater than' inequality into the same ax + b − c form, a positive result would indicate the point satisfies it, and a negative result would mean it does not, the mirror image of how the 'less than' case reads here.
How does this relate to the page that solves for the boundary itself?
That page isolates x algebraically to report the single boundary point where the shading begins or ends. This page keeps that boundary in the background and instead answers a point-by-point question, letting you check any candidate number against the inequality without redoing the algebra each time.