Inequalities in One Variable

Write down the true statement $4 < 10$. Now multiply both sides by $-1$ and write what you get. Is the new statement true? If not, what would you have to change to make it true, and what does that tell you about multiplying an inequality by a negative?

Solving an inequality uses the same steps as solving an equation, with one exception: multiplying or dividing both sides by a negative reverses the inequality sign. The answer is a region on the number line, drawn with an open or closed circle at the boundary.

Almost every step used on an equation works unchanged on an inequality. You may add the same thing to both sides, subtract the same thing, and multiply or divide both sides by a positive number.

$$3x - 5 \leq 7 \ \Rightarrow \ 3x \leq 12 \ \Rightarrow \ x \leq 4$$

The exception is multiplying or dividing by a negative, which reverses the direction of the sign.

$$-2x > 6 \ \Rightarrow \ x < -3$$

Why? Because a negative multiplier flips the number line over. Start from a true statement, $4 < 10$. Multiplying both sides by $-1$ gives $-4$ and $-10$, and $-4$ is the larger of those, since it sits further right. So the correct statement is $-4 > -10$: the sign has turned round.

Test your answer whenever you are unsure. For $-2x > 6$, try $x = -5$: $-2(-5) = 10 > 6$, true, and $-5 < -3$, consistent. Try $x = 0$: $0 > 6$ is false, and $0$ is correctly excluded. Two substitutions settle the direction with certainty.

Consider $\dfrac{6}{x} > 2$. It is tempting to multiply both sides by $x$, but that step is not available.

The reason is that you do not know the sign of $x$. If $x$ is positive, the sign stays; if negative, it reverses. Since both are possible, the step has two contradictory outcomes and cannot be taken.

The same applies to a bracket. From $(x-3)y < (x-3)z$ you may not cancel $(x-3)$, because it may be positive, negative or zero.

Two safe alternatives:

Split into cases. Handle $x > 0$ and $x < 0$ separately, since within each case the sign is known.

Rearrange to compare with zero. Move everything to one side and reason about the sign of the resulting expression.

At Year 10 the inequalities you meet will be linear, so this rarely arises. It is worth stating because the habit of "multiply both sides to clear the fraction", learned on equations, transfers silently and wrongly.

The solution of an inequality in one variable is a region of the number line, and the drawing has two parts to get right.

The circle at the boundary. A strict sign, $<$ or $>$, means the boundary is not a solution, so the circle is open. An inclusive sign, $\leq$ or $\geq$, means it is, so the circle is filled.

The direction of the thick line. Shade towards the values that satisfy the inequality. For $x > 2$ that is to the right; for $x \leq -1$ it is to the left.

A quick check on direction: pick any number in the shaded part and substitute it. For $x > 2$ shaded to the right, $x = 5$ gives $5 > 2$, true.

The same solution can be written three ways, exactly as domains were in Lesson 4:

$$x > 2 \qquad \{x : x > 2\} \qquad (2, \infty)$$

The open circle and the round bracket are the same information, drawn and written. A filled circle corresponds to a square bracket.

Two conditions can be combined, and the word joining them decides the picture.

Both at once ("and"). Written as a single chain:

$$-1 \leq x < 3$$

which means $x \geq -1$ and $x < 3$ simultaneously. The region is the overlap: a single segment from $-1$ to $3$, with a filled circle at $-1$ and an open one at $3$. In interval notation, $[-1, 3)$.

To solve a chain, do the same thing to all three parts:

$$-4 < 2x - 6 \leq 8 \ \Rightarrow \ 2 < 2x \leq 14 \ \Rightarrow \ 1 < x \leq 7$$

Either one ("or"). Written as two separate statements:

$$x < -2 \quad \text{or} \quad x \geq 5$$

The region is two pieces with a gap between them, and in interval notation it needs a union: $(-\infty,-2) \cup [5,\infty)$.

The distinction matters because "and" regions can be empty while "or" regions never are. There is no number satisfying both $x > 5$ and $x < 2$, so that solution set is empty; but $x > 5$ or $x < 2$ describes most of the number line.

Five items on one-variable inequalities. Work each one, then reveal the answer.

1. 1 Solve $2x + 5 > 13$.

   Show answer
   Subtract $5$, divide by the positive $2$.$x > 4$

2. 2 Solve $-x \geq 3$.

   Show answer
   Multiplying by $-1$ reverses the sign.$x \leq -3$

3. 3 Solve $\dfrac{x}{-4} < 2$.

   Show answer
   Multiplying by $-4$ reverses the sign.$x > -8$

4. 4 Which circle does $x \geq 6$ take at the boundary?

   Show answer
   The sign is inclusive, so the boundary is a solution.Filled

5. 5 Write $-2 < x \leq 7$ in interval notation.

   Show answer
   Round for strict, square for inclusive.$(-2, 7]$