What is the difference between < and ≤?

The line underneath ≤ and ≥ means 'or equal to', so the boundary value is included. are strict: the boundary value is excluded. 'x is at least 7' includes 7, so it is x ≥ 7. 'x is more than 7' excludes 7, so it is x > 7. The two differ only at the boundary value 7, and that difference is the whole point.

How do I list the integers for −3 ≤ x < 2?

Read each end separately. The lower end ≤ includes −3, so −3 is on the list. The upper end < excludes 2, so 2 is left off. The integers are −3, −2, −1, 0, 1. Dropping −3 or adding 2 is the boundary-swap error AQA examiners report.

When do you flip the inequality sign?

Only when you multiply or divide both sides by a negative number. Solve 5y + 14 ≥ 11 by dividing by +5: the sign stays, giving y ≥ −0.6. But −2x −3. Dividing by a positive never flips the sign.

GCSE Maths Foundation · AQA · Algebra

Inequalities: strict vs inclusive, and why an inequality is not an equation

Q: What is the difference between < and ≤?

The line underneath ≤ and ≥ means 'or equal to', so the boundary value is included. are strict: the boundary value is excluded. 'x is at least 7' includes 7, so it is x ≥ 7. 'x is more than 7' excludes 7, so it is x > 7. The two differ only at the boundary value 7, and that difference is the whole point.

Q: How do I list the integers for −3 ≤ x < 2?

Read each end separately. The lower end ≤ includes −3, so −3 is on the list. The upper end < excludes 2, so 2 is left off. The integers are −3, −2, −1, 0, 1. Dropping −3 or adding 2 is the boundary-swap error AQA examiners report.

Q: When do you flip the inequality sign?

Only when you multiply or divide both sides by a negative number. Solve 5y + 14 ≥ 11 by dividing by +5: the sign stays, giving y ≥ −0.6. But −2x −3. Dividing by a positive never flips the sign.

On the Foundation paper, inequalities trip students who treat the symbol as decorative — as if $x \geqslant 7$ were just $x = 7$ wearing a different hat. Three errors follow: the boundary value is included or excluded by guesswork, $\geqslant$ collapses to $=$ at the end of a solve, and dividing by a negative leaves the symbol untouched. AQA's reports flag exactly these: listing the integers for $-3 \leqslant x < 2$ with the wrong endpoints, and solving $5y + 14 \geqslant 11$ as the equation $y = -0.6$ .

The thirty-second fix: an inequality is not an equation. The symbol carries the meaning, and it lives at the boundary value: $\leqslant$ and $\geqslant$ include it, $<$ and $>$ exclude it. When you solve, keep the symbol all the way down, and flip it only when you divide or multiply by a negative.

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How to spot it in your own work

You wrote $x > 7$ for "at least 7" (or $x \geqslant 7$ for "more than 7"), guessing the boundary.
Listing integers for $-3 \leqslant x < 2$ , you left out $-3$ or included $2$ .
Solving an inequality, your last line was $y = -0.6$ instead of $y \geqslant -0.6$ .
You divided both sides by a negative number and kept the symbol the same way round.

An exam question that triggers it

Here is a canonical AQA Foundation trigger, in the shape of JUN24 Paper 1 Q26(b):

Solve $5y + 14 \geqslant 11$ .

The misconception answer is $y = -0.6$ , found by solving it like an equation and dropping the inequality at the end. The working is right — $5y \geqslant -3$ , then divide by 5 — but the symbol is the point: nothing here divides by a negative, so it must stay.

The correct answer is $y \geqslant -0.6$ (equivalently $y \geqslant -\tfrac{3}{5}$ ). It names a whole range of values, not the single number $-0.6$ .

Why students fall for this

Students meet equations long before inequalities, and the solving ritual — isolate the letter, get a number — is deeply grooved. An inequality looks like an equation with one odd symbol, so the brain runs the equation procedure and treats $<$ , $>$ , $\leqslant$ , $\geqslant$ as interchangeable noise. The symbol's one job — deciding what happens at the boundary value — never registers, so the boundary becomes a coin-flip.

The flip-on-negative rule is invisible for the same reason: if the symbol is just decoration, why would dividing by $-2$ change it? But $-2x < 6$ and $x < -3$ disagree (try $x = -4$ : the first is false, the second is true), while $-2x < 6$ and $x > -3$ agree. Multiplying by a negative reverses order, so the symbol must reverse with it.

The fix: The symbol decides the boundary; only a negative flips it

An inequality is not an equation. The symbol decides whether the boundary value is in or out. The line under $\leqslant$ and $\geqslant$ means "or equal to", so the boundary is included; $<$ and $>$ are strict, so it is excluded. The words pick the family: "at least" / "up to" include, "more than" / "less than" exclude.

When you solve, work exactly as for an equation but keep the symbol on every line — the answer is a range, never a single $=$ value. The symbol flips only when you multiply or divide both sides by a negative number.

Worked example

Solve $-2x < 6$ (the flip case examiners watch most closely).

Divide both sides by the coefficient. The coefficient is $-2$ , a negative number, so this is the move that flips the symbol.
Flip the symbol as you divide. $<$ becomes $>$ :
$-2x < 6 \;\Rightarrow\; x > \dfrac{6}{-2} = -3$
Answer. $x > -3$ .
Verify with test values. Try $x = 0$ : $-2(0) = 0 < 6$ is true, and $0 > -3$ is true — they agree. Try $x = -4$ : $-2(-4) = 8 < 6$ is false, and $-4 > -3$ is false — they agree. The flipped form is right; keeping $x < -3$ would have got both checks backwards.

Compare $5y + 14 \geqslant 11$ : $5y \geqslant -3$ , then divide by $+5$ — positive, so the symbol is kept — $y \geqslant -0.6$ . No negative, no flip.

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Common questions

What does "at least" mean as an inequality?: "At least 7" means 7 is allowed, so the boundary is included: $x \geqslant 7$ . "More than 7" means strictly bigger, so 7 is excluded: $x > 7$ . The two phrases disagree only about the number 7 itself.
Why is $-3 \leqslant x < 2$ the integers $-3, -2, -1, 0, 1$ ?: Read each end on its own. The lower end $\leqslant$ includes $-3$ , so it is on the list. The upper end $<$ excludes $2$ , so it is left off. That gives $-3, -2, -1, 0, 1$ . On a number line, a filled circle marks $-3$ and an open circle marks $2$ .
Do I always flip the inequality sign when I solve?: No. You flip it only when you multiply or divide both sides by a negative number. Dividing by a positive (like the $+5$ in $5y \geqslant -3$ ) keeps the sign: $y \geqslant -0.6$ . Dividing by a negative (like the $-2$ in $-2x < 6$ ) flips it: $x > -3$ .
Is $y = -0.6$ a correct answer to $5y + 14 \geqslant 11$ ?: No — that solves it as an equation and throws away the inequality. The answer is a range: $y \geqslant -0.6$ (i.e. $y \geqslant -\tfrac{3}{5}$ ). Every value of $y$ from $-0.6$ upwards works, not just $-0.6$ .

Related misconceptions

Solving equations: use the inverse, not the same operationThe companion solving skill: undo each operation with its inverse — the backbone of rearranging an inequality too.
Function machines: read left-to-right, invert to reverseAnother solving foundation: run a machine forwards, then invert each step in reverse order.
Decimal place valueWhere an inequality answer is a decimal like −0.6, place-value reasoning keeps the bound right.

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