Algebraic solutions of linear inequalities

Three rules, one extra care point. You may add or subtract the same quantity from both sides of an inequality freely, exactly as with an equation. You may multiply or divide both sides by the same positive number without touching the inequality sign. You may multiply or divide both sides by the same negative number too — but only if you also reverse the direction of the inequality. These three rules, applied step by step, are the whole method.

Isolating the variable. For a general single inequality of the form $ax+b<c$ with $a\neq0$, subtract $b$ and then divide by $a$:

where: $a$ is the coefficient of $x$, $b$ and $c$ are constants, and the direction of the final inequality depends only on the sign of $a$ — the divisor.

Stating the answer three ways. A solved inequality such as $x<5$ can equivalently be written in set-builder form, $\{x \in \mathbb{R} : x<5\}$, or in interval notation, $(-\infty,5)$. Strict inequalities ($<,>$) give round brackets (the endpoint is excluded); $\le$ or $\ge$ give square brackets (the endpoint is included). Fluency in translating between all three forms matters because a question may demand any one of them, and marks are lost on presentation alone if the wrong form is given.

Systems of inequalities. When two or more linear inequalities in the same variable must hold simultaneously, solve each one separately first, then take the intersection of the individual solution sets — the values that satisfy every condition at once, not just one of them. If the individual ranges do not overlap anywhere, the system has no solution.

The picture below shows why intersection, not union, is the right operation: only the values lying on both number lines survive.

Solve the system of inequalities $2x+5<9$ and $3x-4\ge-1$, and state the combined solution set. This goal sentence spells out exactly what "solving a system" demands: two separate linear inequalities in the same variable $x$ must be reduced to their individual solution sets by the standard add-subtract-multiply-divide steps, with the sign flipped only when dividing by a negative number, and then the two resulting ranges must be combined by intersection into one final interval that is checked by substituting a test value back into both original statements.

Step 1. Solve the first inequality. Subtract $5$ from both sides: $2x<4$. Divide by $2$, a positive number, so the direction is unchanged: $x<2$.

Step 2. Solve the second inequality. Add $4$ to both sides: $3x\ge3$. Divide by $3$, again positive: $x\ge1$.

Step 3. Combine the two solution sets by intersection — the system demands both $x<2$ and $x\ge1$ hold at once, so overlay the two ranges and keep only where they coincide.

Step 4. The overlap runs from $1$ (included, since $\ge$) up to but not including $2$ (excluded, since $<$):

Check with $x=1.5$: $2(1.5)+5=8<9$ ✓ and $3(1.5)-4=0.5\ge-1$ ✓, confirming the interval. The solution set is $[1,2)$.

Competitive stretch

Clearing a fractional coefficient is a very common first move before the same three rules apply. For $\dfrac{5-2x}{3}<4$, multiply both sides by the positive denominator $3$: $5-2x<12$. Subtract $5$: $-2x<7$. Now divide by $-2$, a negative number, so the sign reverses: $x>-\dfrac{7}{2}$. Nothing about the method changed — only the bookkeeping of the fraction did.

• Forgetting to reverse the sign when dividing by a negative coefficient — e.g. solving $-3x<12$ as $x<-4$ instead of the correct $x>-4$.
• Taking the union instead of the intersection in a system problem — the answer is where all the inequalities hold together, not where any single one holds.
• Mixing up open and closed brackets — strict $<$ or $>$ gives a round bracket (endpoint excluded); $\le$ or $\ge$ gives a square bracket (endpoint included).
• Stopping after solving each inequality separately without combining them — a system question wants one final overlapping interval, not two disconnected answers.

Ask the tutor: Want to try a system of three inequalities together, or have the AI tutor check whether your interval brackets are right on a tricky fractional-coefficient problem?