6.04a - Linear inequalities

What an inequality describes

How can one statement describe every number of tickets you can afford? Suppose tickets cost £5 each, there is one £2 booking fee and you can spend at most £22. If you buy $n$ tickets, the condition is $5n+2\le22$. The symbol matters: spending exactly £22 is allowed.

Start with a simpler condition, $x<3$. It accepts $2$, $0$, $-8$ and $2.99$, among infinitely many others, but rejects $3$. Changing it to $x\le3$ includes the boundary value as well.

The solutions are all values that make the statement true. A strict inequality, using $<$ or $>$, excludes equality. A non-strict inequality, using $\le$ or $\ge$, includes it. You can reverse the order of the written values if you also turn the symbol: $3>x$ says the same thing as $x<3$.

[DIAGRAM: asset_name: Strict and inclusive boundaries; asset_slug: inequality-endpoints; caption: An open circle excludes three; a filled circle includes it.; description: Two aligned number lines show x less than three and x less than or equal to three. Both solution arrows extend left from three. The first has an open circle at three and the second a filled circle.]

Unless a question limits $x$ to integers, decimals and fractions between the marked numbers are included too. A ticket count is different: it must be a whole number, and buying tickets requires a positive count.

Check a value by reading the whole condition, including equality.

Solving while preserving order

Use the balancing moves you know from equations, but keep an inequality symbol throughout the solution.

Adding or subtracting the same number on both sides preserves order: if one runner is ahead by 4 metres, moving both runners 2 metres forwards leaves the same gap. Multiplying or dividing both sides by the same positive number also preserves order. These reversible moves keep exactly the original solutions.

Change only the final comparison to include equality. The calculation stays the same; the boundary decision changes.

Brackets and an unknown on both sides work in the same way. For example, expanding $2(x+1)\le x+5$ gives $2x+2\le x+5$. Subtract $x$, then subtract 2, to get $x\le3$. A fraction with a fixed positive denominator can be cleared by positive multiplication: $\frac{x+1}{2}>4$ becomes $x+1>8$, hence $x>7$.

Now keep track of an inclusive boundary through two moves.

Multiplying or dividing by a negative

Negative multiplication changes the order, so it needs a different rule.

You know $2<5$. Multiplying both numbers by $-1$ gives $-2$ and $-5$, and $-2>-5$. Reflection in zero swaps left and right on a number line. Multiplying by any negative number combines this reflection with a positive scaling, so it reverses order. Negative division does the same.

Read the coefficient's sign before dividing.