Linear Inequalities in One Variable

When the answer is “anything up to…”

Back to the phone plan one last time: $$20$ a month plus $$3$ per gigabyte, and this month your budget is $$35$. How much data can you afford? Not “exactly how much” — up to how much. The question itself isn’t an equation; it’s a constraint:

$20 + 3g \le 35$

An inequality replaces $=$ with $\lt$, $\gt$, $\le$ (“at most”), or $\ge$ (“at least”), and its answer isn’t a single number but a whole range. Real life runs on these: speed limits, minimum heights, passing grades, budgets. Solve this one with exactly the A1 balance moves — subtract $20$ from both sides ($3g \le 15$), divide by $3$ ($g \le 5$) — and the answer is every usage up to five gigabytes. On a number line, that’s not a dot; it’s a shaded ray.

Almost everything transfers from A1 unchanged. One rule is new, and it’s the whole ballgame.

The flip rule — and why you have no choice

Multiply or divide both sides by a negative number and you must flip the inequality sign. Here’s the picture that makes it obvious rather than arbitrary: multiplying by $-1$ reflects the whole number line through zero. Reflections reverse left and right — $2$ sits left of $4$, but their mirror images land the other way around: $-2$ sits right of $-4$. So any true "$\lt$" between two numbers must become "$\gt$" between their negatives. The numbers have no choice, so neither do you.

Why is this the single most-forgotten rule in algebra? Because equations trained you that negatives are harmless — in A1 you divided by $-2$ all day and equality never cared, since a mirror image of “equal” is still “equal.” Order is the thing mirrors break, and inequalities are made of order. When in doubt, run the two-second check: $-2 \lt 4$ is true; divide both by $-2$ and keeping the sign claims $1 \lt -2$ — false; flip it, $1 \gt -2$ ✓.

The rule also gets over-applied, for the same fuzzy reason (“negative… flip something?”). Adding or subtracting a negative never flips — sliding the whole line left or right keeps everyone’s order — and dividing by a positive never flips. Only a negative ×/÷ holds up the mirror.

The solver opens on $-2x + 1 \ge 9$. Predict where the flip will happen before you read the steps ($-1$ first — no flip; then $\div(-2)$ — flip). Then use the tester: drop in $-4$, then $-5$, then $0$, and watch which ones land inside the shaded range.

Graphing: open ○ vs closed ●

To draw $x \le 5$: mark the boundary, shade the true side, and let the circle at the boundary say whether $5$ itself belongs. An “or equal to” sign ($\le$, $\ge$) includes the boundary — closed ●. A strict sign ($\lt$, $\gt$) excludes it — open ○. Strict ranges have a strange, useful property: $x \lt 6$ contains $5.9$, $5.99$, $5.999$… but no largest solution, because the boundary itself is the one point missing.

Set the boundary to $6$ with $\lt$, then switch to $\le$ and watch the endpoint fill in — one pixel of ink, one number of difference.

Compound inequalities — between two bounds

Some constraints are two-sided: a package ships only if its weight $w$ satisfies $1 \lt w \le 20$ kilograms. A chain like $-5 \le 2t + 1 \lt 11$ is just two inequalities sharing the middle expression — "$-5 \le 2t+1$" and "$2t + 1 \lt 11$" — so any balance move must hit all three parts to keep both statements true at once. Subtract $1$ from all three: $-6 \le 2t \lt 10$. Divide all three by $2$: $-3 \le t \lt 5$ — a segment, closed at one end and open at the other. (And if you ever divide a chain by a negative, both signs flip and the chain reverses direction.) Try these in the Compound tab — its example chips include a $\div(-2)$ chain.

When x vanishes

Just like equations, the $x$-terms can cancel entirely. Read what remains: a statement that’s always true ($5 \gt 1$) means all real numbers satisfy it; an impossible one ($5 \lt 1$) means no solution. The logic is A1’s, wearing an inequality sign.

The one thing to remember

An inequality is solved like an equation, but its answer is a range, and ranges care about order — so the one new law is the mirror: multiplying or dividing both sides by a negative reflects the number line and flips the sign. Boundary circles say whether the edge itself counts, and a compound chain is two constraints that every move must respect at once.

What changes from equations

An inequality uses $\lt$ (less than), $\gt$ (greater than), $\le$ (at most), or $\ge$ (at least) instead of $=$. The answer isn’t one number — it’s a whole range of numbers.

You solve it almost exactly like an equation: distribute, clear fractions, collect like terms, and undo operations to isolate $x$. One rule is new, and it’s the whole ballgame.

Worked example — basic

$3x - 7 \lt 11$

Worked example — the flip

$-2x + 1 \ge 9$

A great safety check: pick a number inside your range and test it in the original. $x = -5$ is $\le -4$, and $-2(-5) + 1 = 11 \ge 9$ ✓.

Graphing on a number line

Mark the boundary, then shade the direction that’s true.

With $x$ on the left, greater ($\gt$ $\ge$) shades right, less ($\lt$ $\le$) shades left.

Compound inequalities

Two conditions at once.

• “And” (between): written as one chain, e.g. $-1 \le 2x + 3 \lt 9$. Do every operation to all three parts. Subtract $3$ from all: $-4 \le 2x \lt 6$. Divide all by $2$: $-2 \le x \lt 3$. The solution is the segment between $-2$ and $3$.
• “Or”: e.g. $x \lt -1$ or $x \gt 4$ — two separate rays going opposite directions.

Two special answers

Sometimes the $x$-terms cancel out completely. Look at what’s left: