Inequalities in one variable

An inequality compares two expressions using $<$ (less than), $>$ (greater than), $\le$ (less than or equal to), or $\ge$ (greater than or equal to). Because so many values can satisfy it, the answer is described as a solution set, not a single value.

Three properties govern every step of solving one. Adding or subtracting the same quantity from both sides never changes the direction:

where: $a$ and $b$ are the two sides being compared, and $c$ is any real number added or subtracted equally on both sides. Multiplying or dividing both sides by the same positive number also preserves the direction:

But multiplying or dividing both sides by the same negative number reverses it:

where: $c$ is the negative multiplier or divisor — the sign flip is the one place this chapter differs from equation-solving.

Solving method: treat the inequation exactly like an equation — expand any brackets, then transpose terms so the variable collects on one side and constants on the other. The only extra rule is to flip the inequality sign the moment you multiply or divide both sides by a negative quantity.

Representing the solution on a number line: a filled (solid) dot marks a value that is included — used for $\le$ or $\ge$ — while an open circle marks a value that is excluded — used for $<$ or $>$. An arrow extends from that point in the direction of every included value. If the solution must come from $N$, $W$, or $Z$, only the discrete points belonging to that set are marked, not a continuous shaded line.

Solve $2x-5\le 9$ for $x\in Z$ (integers), and represent the solution on a number line.

Step 1. Transpose $-5$ to the right-hand side: $2x\le 9+5$, so $2x\le 14$.

Step 2. Divide both sides by $2$. Since $2$ is positive, the inequality sign is unchanged: $x\le 7$.

Step 3. Since $x\in Z$, the solution set is $\{\ldots,-2,-1,0,1,2,3,4,5,6,7\}$ — every integer less than or equal to $7$, extending infinitely to the left.

Step 4. On the number line, mark a solid dot at $7$ (included, because of $\le$) and draw the arrow extending left through every smaller integer, exactly as pictured above.

Answer: the solution set is $\{x\in Z : x\le 7\}$.

Second case: where the reversal rule bites

Solve $\dfrac{3-2x}{4}>1$ for $x\in N$ (natural numbers).

Step 1. Multiply both sides by $4$, a positive number, so the sign is unchanged: $3-2x>4$.

Step 2. Transpose $3$: $-2x>4-3$, so $-2x>1$.

Step 3. Divide both sides by $-2$. Because $-2$ is negative, the sign flips: $x<-\tfrac12$.

Step 4. No natural number ($1,2,3,\ldots$) is less than $-\tfrac12$, so the solution set is empty — there is no natural-number solution. (Had the domain been $Z$ or $R$ instead, the answer would be every integer or real number strictly less than $-\tfrac12$, shown with an open circle there and an arrow extending left.)

Final answer: no solution exists in $N$.

• Forgetting to flip the inequality sign when multiplying or dividing both sides by a negative number — the single most common error in this chapter.
• Using a solid (filled) dot for a strict inequality ($<$ or $>$) on the number line, when an open circle is required because that endpoint is excluded.
• Ignoring the stated domain (such as $N$ or $W$) and giving a continuous range of real numbers when only whole- or natural-number solutions were asked for.
• Assuming a solution set always has a start or end point — for domains like $Z$ or $R$ the set often extends infinitely in one direction, so it must be described with an inequality symbol rather than listed in full.

Ask the tutor: Want to try flipping the sign correctly across a few more inequations, or practise plotting solution sets on a number line? Ask the AI tutor to work through extra examples with you.