Inequalities

Tap a term to take it off both sides and watch the number line. The answer is a region rather than a point, and one of the moves turns the relation around.

Worked example — Inequalities

Boundary	4 (excluded)

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Lesson

The theory — Inequalities259 words

Solving an inequality means naming the set of values that satisfy it. That set is decided by two separate things: where the boundary is, and whether the boundary point is a member.

What each symbol means

<: a strict relation. The boundary is excluded, and the solution set is open at that end.
≤: a non-strict relation. The boundary is included, and the set is closed at that end.
sup: the supremum — the least value that nothing in the set exceeds. It always exists for a set bounded above, whether or not the set contains it.

Assumes: That the boundary is where the two sides are exactly equal, which is why solving an inequality uses the arithmetic of solving an equation. The relation is not doing arithmetic; it is choosing a side and deciding membership.
Breaks when: The strict-or-not choice decides something larger than membership: whether a largest value exists at all. On x ≤ 5 the largest solution is 5, and you can hand it to somebody. On x < 5 there is no largest solution — name any candidate below 5 and the midpoint between it and 5 is a larger one still in the set, for ever. The set still has a least upper bound of 5; it simply does not contain it. This is why an optimisation over a strict constraint can fail to have an answer while the same problem with ≤ has one, and why the distinction that looks like punctuation in a school exercise is load-bearing the moment anything is being maximised.

One move reverses it, and you already know why 🖖

Every move that is legal on an equation is legal here. Add to both sides, subtract, multiply, divide — all fine, all reversible, none of them touching which numbers satisfy the statement. Except that multiplying by a negative reverses the relation, and that comes straight off the number line: −5 is less than −3, but 5 is greater than 3. Negating reflects the line about zero, and reflecting swaps left and right. Press The relation flips and skip that reversal in your head: you get x < −4 instead of x > −4, which is the other half of the line.

The answer is a region, and the endpoint is a decision 🖖

An equation has a solution. An inequality has a set of them, so solving it means finding two things: a boundary and a side. The boundary is where the two expressions are exactly equal, which is why the same arithmetic works — and then the relation decides whether that single point belongs. Press The endpoint counts: −4x ≥ 12 gives x ≤ −3, and at x = −3 both sides come to 12. Since 12 ≥ 12 is true, −3 is in. Change the ≥ to a > and the dot goes hollow while nothing else moves. One point, and it is the only thing that distinguishes the two symbols.

The move that is not available 🖖

Press Multiply by x. The tool will not do it. Multiplying by a negative reverses the relation, multiplying by a positive does not, and x is the thing you have not found yet. There is no single answer to give. This is why x/(x−2) > 1 cannot be cleared by multiplying through, and why anyone who does it gets an answer that is right on one side of 2 and wrong on the other. Equations forgive this — an equation multiplied by an unknown gains solutions you can spot and discard by checking. An inequality does not offer that check in the same form, because the answer is a region rather than a short list of candidates you can test one at a time.

Problems solved in full

A delivery costing 5 plus 2 per kilometre with a budget under 20 7 steps
A delivery costs 5 plus 2 per kilometre. Your budget is under 20. How far can you go — and what happens to the algebra if the fare were quoted as a discount instead?
1. Write it as an inequality rather than an equation, because the question is not what distance costs exactly 20.
  
  $5 + 2 x < 20$
2. Subtract 5 from both sides. Adding and subtracting never touch the relation, whatever the sign of what you add.
  
  $2 x < 15$
3. Divide by 2. It is positive, so the relation stays as it is, and the answer is a region open at the top end.
  
  $x < 7.5$
4. Under 7.5 kilometres. The endpoint is excluded because the budget was strictly under 20, and at exactly 7.5 the fare is exactly 20 — which is over budget by the wording, if not by much.
  
  $5 + 2 (7.5) = 20 \neq < 20$
5. Now the discount version. Suppose instead each kilometre took 2 off a 20 credit, and you need to keep the balance above 5.
  
  $20 - 2 x > 5$
6. Dividing by −2 reverses the relation, and the answer is the same region as before. It has to be: it is the same journey described from the other end. Skipping the reversal would have given x > 7.5, which is the unaffordable side of the boundary instead of the affordable one.
  
  $- 2 x > - 15 \Rightarrow x < 7.5$
7. And here the arithmetic stops helping. Both forms give x < 7.5, and neither of them settles whether 7.5 itself is affordable. Whether "under 20" means 19.99 or means 20 is a question about the contract, not about the inequality, and the strict symbol is doing work the algebra cannot check.
  
  $x < 7.5 vs x \leq 7.5$
Answer

Under 7.5 kilometres, with 7.5 itself excluded. Written as a discount instead, the same journey needs a division by −2 and therefore a reversal, and forgetting it hands you the unaffordable side instead.
The inequality with no x left in it 6 steps
This preset is 2x + 5 ≥ 2x + 5. Solve it, then decide whether an inequality that loses its x has failed to produce an answer or has produced a very large one.
1. Subtract 2x from both sides. Adding the same quantity to both sides preserves the relation whatever that quantity is, so the move is safe even though it involves x — which is exactly what multiplying by x is not.
  
  $2 x + 5 \geq 2 x + 5 - 2 x$
2. What is left is 5 ≥ 5. Both sides now carry a coefficient of zero on x, so there is nothing to isolate and nothing to divide by.
  
  $0 \cdot x + 5 \geq 5$
3. The boundary formula shows the same obstruction arithmetically. Solving ax + b ≥ cx + d gives x = (d − b)/(a − c), which here is 0 ÷ 0. The panel prints a dash in the boundary row because there is no boundary to print.
  
  $x = \frac{d - b}{a - c} = \frac{5 - 5}{2 - 2} = \frac{0}{0}$
4. 5 ≥ 5 is true, and it is true without mentioning x. So every real number satisfies the original inequality, and the answer row says so.
  
  $5 \geq 5 is true and holds no x ⟹ every x is a solution$
5. Make the relation strict and 5 > 5 is false: nothing works. Nudge the right-hand constant to 5.001 instead and the statement becomes 0 ≥ 0.001, false again. Almost the same four numbers, and the solution set goes from the whole line to nothing at all.
  
  $2 x + 5 > 2 x + 5 \Rightarrow 5 > 5 false \Rightarrow \emptyset; 2 x + 5 \geq 2 x + 5.001 \Rightarrow 0 \geq 0.001 false \Rightarrow \emptyset$
6. That jump looks like a discontinuity and is not one. Hold the right-hand side at 5.001 and let the two x coefficients differ by ε instead of matching. The boundary sits at 0.001/ε: at ε = 0.1 the answer is x ≥ 0.01, at ε = 0.001 it is x ≥ 1, at ε = 0.0001 it is x ≥ 10. The half-line retreats towards infinity as ε closes, which is the empty set arriving in slow motion.
  
  $a - c = ε, d - b = 0.001 : x \geq \frac{0.001}{ε} : ε = 0.1 \Rightarrow x \geq 0.01, 0.001 \Rightarrow x \geq 1, 0.0001 \Rightarrow x \geq 10$
Answer

Every x. The empty answer is its twin rather than its opposite: when the two x coefficients match, the x term cancels and the two constants decide alone, so the statement is either true and every number solves it or false and none does. Neither case has a boundary, which is why the boundary row blanks in both and the answer row is the only one worth reading. The strict-or-not choice that moves a single point on every other preset moves the entire real line on this one.

$a \neq = c : x ≷ \frac{d - b}{a - c}; a = c : the constants alone decide, and the answer is R or \emptyset$

Learning path

Solving for x, from the minus sign upward

Leads to Rearranging formulas the exception, and with it a test.

References (1)

Where the guarantee comes from that a least upper bound always exists, even when no largest member does: R. Dedekind, Stetigkeit und irrationale Zahlen. Vieweg, Braunschweig, 1872.

By the 42Math team · Updated 2026-08-26

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

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Example problems

Nothing flips - 3x − 5 > 7 becomes x > 4 with no surprises: add 5, divide by 3, and 3 is positive so nothing reverses. The endpoint 4 is hollow, because x = 4 gives exactly 7 and 7 is not greater than 7.
The relation flips - −2x + 1 < 9 becomes x > −4. Dividing by −2 turned the < into a >, and if you skip that step you get x < −4, which is exactly the wrong half of the line.
The endpoint counts - −4x ≥ 12 gives x ≤ −3, and here the endpoint is filled in. At x = −3 both sides come to 12, and 12 ≥ 12 is true, so −3 belongs to the answer. That filled dot is the whole difference between ≥ and >.
Always true - 2x + 5 ≥ 2x + 5 is the same expression on both sides. The x terms cancel and 5 ≥ 5 remains, which is true whatever x is. The line is shaded end to end.