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≤

First-degree inequalities

Level 5 · Ages 15–16

Quick recall· Let's wake up what you already know before we start.

Refresher: solve the equation 2x + 1 = 9.

The whole lesson

Read it again any time, including while you work through the exercises.

An equation asks: "for which value are these two quantities equal?" An inequality asks a broader question: "for which values does this one exceed that one?" A budget not to overshoot, a minimum grade to reach, a speed limit: life is full of inequalities, and algebra knows how to solve them.

First big difference: the solution is no longer one number, it is a whole territory. The inequality x + 3 > 5 is true for x = 2.1, for x = 7, for x = 1000... in short, for ALL numbers strictly greater than 2. We write: x > 2. Picture the number line with a shaded zone stretching on forever.

Good news: the equation techniques almost all carry over. Adding or subtracting the same number on both sides? Allowed, the inequality keeps its direction. Multiplying or dividing by a POSITIVE number? Allowed too. So you solve 3x - 5 < 7 exactly like an equation: 3x < 12, then x < 4.

But there is ONE new rule, and it is the star of this chapter: when you multiply or divide both sides by a NEGATIVE number, the inequality sign FLIPS. Why? Look: 2 < 5 is true. Multiply by -1: we get -2 and -5, and indeed -2 > -5. Taking opposites reverses the order, like a mirror. So -2x ≤ 8 gives x ≥ -4: the ≤ becomes ≥.

One last champion's reflex: check with a test value. You found x < 4? Take x = 0, plug it into the original inequality, and see whether it holds. This little check catches almost every direction mistake.

The worked example, step by step

Solve the inequality 3x - 5 < 7.

  1. 3x < 7 + 5, soit 3x < 12

    I add 5 to both sides: the inequality keeps its direction.

  2. x < 12 ÷ 3, soit x < 4

    I divide by 3, a POSITIVE number: the direction still does not change.

  3. test x = 0 : 3 × 0 - 5 = -5 < 7 ✓

    I check with a value from the zone I found: 0 is indeed less than 4 and makes the inequality true.

Result: The solutions are all numbers x such that x < 4.

Solve the inequality -2x + 3 ≤ 11.

  1. -2x ≤ 11 - 3, soit -2x ≤ 8

    I subtract 3 from both sides: no danger, the direction is preserved.

  2. x ≥ 8 ÷ (-2), soit x ≥ -4

    I divide by -2, a NEGATIVE number: the inequality flips, ≤ becomes ≥.

  3. test x = 0 : -2 × 0 + 3 = 3 ≤ 11 ✓

    I test x = 0 (which is indeed ≥ -4) in the original inequality: 3 ≤ 11, true. Flipping was the right call.

Result: The solutions are all numbers x such that x ≥ -4.