Linear inequalities

Swap the " $=$ " in a linear equation for $<,\ \le,\ >,$ or $\ge$ and the whole character of the problem changes: instead of one exact answer, you get an entire range of values that work. A shopkeeper deciding how many items keep her profit above a target, or an engineer keeping a stress below a safety limit, is really solving a linear inequality. The algebra looks almost identical to solving equations — add, subtract, multiply, divide — with one sharp exception: multiplying or dividing by a negative number reverses the direction of the inequality. Miss that single rule and every other step can be flawless while the final answer is still wrong. This section builds the solving rules from first principles, shows how a solution set is written as an interval and pictured on a number line, and extends the method to compound inequalities where two conditions must hold at once.

The core insight

A linear inequality is solved like an equation except that multiplying or dividing both sides by a negative number flips the inequality sign, and the resulting solution set is a range of values written in interval notation and pictured on a number line.

The core idea

A linear inequality in one variable has the form $ax+b<c$ (or with $\le,>,\ge$ ), where $a\neq0$ . It behaves like a linear equation with a range of solutions instead of a single one, and three rules govern how it can be manipulated:

Adding or subtracting the same quantity from both sides never changes the direction of the inequality.
Multiplying or dividing both sides by a positive number never changes the direction.
Multiplying or dividing both sides by a negative number reverses the direction — $<$ becomes $>$ , $\le$ becomes $\ge$ , and vice versa.

For instance, solving $-2x<6$ by dividing both sides by $-2$ gives $x>-3$ , with the sign flipping from $<$ to $>$ exactly because the divisor is negative.

Writing the solution as an interval. Once a variable is isolated, the solution set is written using interval notation and drawn on a number line:

x>-3 \iff (-3,\infty), \qquad x\ge2 \iff [2,\infty), \qquad -1<x\le4 \iff (-1,4].

where: a round bracket $($ or $)$ (matched by an open circle on the line) marks a boundary that is excluded, and a square bracket $[$ or $]$ (matched by a filled circle) marks a boundary that is included. The smaller number always sits on the left of the interval.

Figure

Number line showing the solution x less than or equal to negative five as a solid ray shaded to the left of a filled circle at negative five

Compound inequalities. An expression like $2<3x-1\le11$ is two inequalities joined together, and the same operation must be applied to all three parts at once to isolate $x$ in the middle.

Systems of inequalities. When two or more linear inequalities in the same variable must hold simultaneously, solve each one separately, then take the intersection of the resulting intervals as the combined solution.

Derivation

Derivation AWhy multiplying or dividing an inequality by a negative number reverses its direction

The goal is to show, using nothing beyond the definition of "greater than" as a statement about position on the number line, that multiplying both sides of a true inequality by a negative number forces the inequality sign to flip if the new statement is to remain true, and that this single rule — not an arbitrary convention — is what makes every flip-the-sign step in solving inequalities logically necessary rather than optional.

Step 1. Start from a true, simple inequality: $2<5$ . On the number line, $2$ sits to the left of $5$ , which is exactly what " $2<5$ " means.

Step 2. Multiply both sides by $-1$ . The numbers $2$ and $5$ become $-2$ and $-5$ . Multiplying by $-1$ reflects every point on the number line through zero, so a point that was to the left of another point is now to the right of it: $-2$ lies to the right of $-5$ on the line.

Step 3. Read off the correct relation. Since $-2$ is to the right of $-5$ , the true statement is $-2>-5$ , not $-2<-5$ . The multiplication by a negative number reversed left–right order, so the inequality symbol had to reverse too, from $<$ to $>$ , for the statement to stay true.

Step 4. Generalise. For any real numbers $p<q$ and any negative number $k$ , multiplying both sides by $k$ reflects both points through zero and swaps their order, so $kp>kq$ . The same reasoning with division follows because dividing by $k$ is multiplying by $1/k$ , which has the same sign as $k$ . This is why the flip rule is not memorised as an isolated fact — re-deriving the small example $2<5\Rightarrow-2>-5$ is the fastest way to recover it whenever it is forgotten. Verified: substituting $p=2,\ q=5,\ k=-1$ gives $kp=-2$ , $kq=-5$ , and $-2>-5$ holds, confirming the flip.

Worked example

Solve the compound inequality $-3\le\dfrac{2x-1}{4}<2$ and write the solution in interval notation, then check it on the number line.

Step 1. Clear the fraction by multiplying all three parts by $4$ , a positive number, so no sign flips: $-12\le2x-1<8$ .

Step 2. Add $1$ to all three parts to isolate the $x$ -term: $-11\le2x<9$ .

Step 3. Divide all three parts by $2$ , again positive, so no flip: $-\dfrac{11}{2}\le x<\dfrac92$ .

Step 4. Convert to interval notation and check with an interior value. The solution set is $\left[-\dfrac{11}{2},\ \dfrac92\right)$ — a closed (included) boundary at $-\dfrac{11}{2}$ and an open (excluded) boundary at $\dfrac92$ . Testing $x=0$ , which lies inside the interval: $\dfrac{2(0)-1}{4}=-\dfrac14$ , and indeed $-3\le-\dfrac14<2$ holds.

The solution is $x\in\left[-\dfrac{11}{2},\ \dfrac92\right)$ .

Competitive stretch

If the same problem asked for integer solutions only, list every integer in $\left[-\dfrac{11}{2},\ \dfrac92\right)=[-5.5,\ 4.5)$ : that is $-5,-4,-3,-2,-1,0,1,2,3,4$ — ten integers, with $-5.5$ and $4.5$ themselves excluded since neither is an integer anyway.

Common traps

Forgetting to flip the inequality sign when multiplying or dividing by a negative number — the single most common error in this topic.
Treating $<$ and $\le$ as interchangeable when writing the final interval — mixing an open and closed bracket changes whether the boundary point is actually a valid solution.
In a compound inequality, applying an operation to only one part instead of all three parts at once, which breaks the equivalence of the whole chain.
Writing the interval backwards, such as $(5,-\infty)$ instead of $(-\infty,5)$ — the smaller number always comes first.

Ask the tutor: If the flip rule keeps tripping you up, ask the AI tutor for a few practice inequalities with negative coefficients and have it check whether your final interval brackets are open or closed.

Why does multiplying by a negative number flip the inequality but adding or subtracting never does?

Adding or subtracting the same number shifts every point on the number line by the same amount in the same direction, so the left-to-right order of any two numbers is preserved. Multiplying by a negative number reflects every point through zero, which reverses left-to-right order — a point that was smaller becomes larger after reflection. Since the inequality symbol is really just a statement about order on the number line, reversing the order forces the symbol to reverse too. Try it with $2<5$ : multiply by $-1$ and you get $-2>-5$ , not $-2<-5$ .

How is solving an inequality different from solving the corresponding equation?

The algebraic steps — isolate the variable by adding, subtracting, multiplying, dividing on both sides — are identical to solving an equation, with one extra rule: flip the inequality sign whenever you multiply or divide by a negative number. The other difference is the answer itself. An equation like $3x-7=5x+3$ has one solution, a single number. The inequality $3x-7\ge5x+3$ has infinitely many solutions, an entire interval, because any value of $x$ that keeps the left side at least as large as the right side qualifies, not just one.

Do I always need a number line, or is interval notation enough?

For most written answers, interval notation such as $(-\infty,-5]$ is sufficient and is the standard way to state a final solution. The number-line sketch is mainly a checking tool and is explicitly asked for in some questions ("represent on a number line"), where open circles mark excluded boundaries and filled circles mark included ones. Drawing it even when not required is a good habit, because a quick glance often reveals an error such as an accidentally flipped or unflipped sign.