Linear Inequalities

A linear inequality compares two expressions using $<$ , $>$ , $\le$ or $\ge$ . Its solution is usually a set of values rather than a single number.

Most equation-solving operations still apply, but one rule is different: multiplying or dividing both sides by a negative number reverses the inequality sign because it reverses order on the number line.

On this page (8 sections)

Solving a basic inequality

A one-variable linear inequality compares expressions whose variable terms are linear. Unlike an equation with one solution, it usually describes an interval of acceptable values.

Solve $3x-5<10$

$3x<15$
Add 5 to both sides.
$x<5$
Divide by positive 3; the direction stays the same.

All values to the left of 5, with 5 excluded

The open circle excludes 5; the coloured arrow continues towards smaller values.

Diagram explained

The solution is the open ray x < 5. There is no finite left endpoint.

The solution is $(-\infty,5)$ . An open endpoint excludes $5$ ; the arrow represents all smaller numbers, not just the values visible in the drawing.

Why a negative multiplier reverses the sign

The comparison $2<5$ becomes $-2>-5$ after multiplication by $-1$ . Reflection across zero reverses order on the number line. This is why a negative multiplier or divisor reverses the inequality.

Solve $-2x+3\ge11$

$-2x\ge8$
Subtract 3 from both sides.
$x\le-4$
Divide by negative 2 and reverse the sign.

The boundary belongs to the solution: $-2(-4)+3=11$ . The larger input $x=0$ fails, which is consistent with the solution extending to the left.

Compound inequalities

A compound inequality can place an input between two boundaries. Apply an operation to all three parts, not only the middle.

1<2x+3\le9

-2<2x\le6

-1<x\le3

The solution is $(-1,3]$ : greater than $-1$ and at most $3$ . The word “and” means both conditions must hold.

A union is different from a bounded interval

The condition $x<-1$ or $x>3$ describes two separate rays, written $(-\infty,-1)\cup(3,\infty)$ . It does not describe the numbers between the endpoints.

Inequalities in context

A hypothetical purchase has a fixed charge of $12$ dollars plus $8$ dollars per item. With a budget of at most $80$ dollars, let $u$ count the items.

12+8u\le80\quad\Longleftrightarrow\quad8u\le68\quad\Longleftrightarrow\quad u\le8.5

The formula alone allows negative and fractional values, but this context requires $u$ to be a non-negative integer. Therefore $u\in\{0,1,2,3,4,5,6,7,8\}$ . At most 8 items can be purchased.

Eight items cost $76$ dollars; nine cost $84$ dollars. Rounding $8.5$ up would violate the budget, so the context determines the rounding direction.

Checking a solution set

Test one value inside the proposed solution and one value outside it. This is not a general proof, but it is an efficient error check, especially after dividing by a negative number.

Practice questions

Solve $3x – 5 < 10$ .
Solve $-2x + 3 \ge 11$ .
Write $-1 < x \le 3$ in interval notation.

Check your answers

$3x<15$ , so $x<5$ . The interval is $(-\infty,5)$ .
$-2x\ge8$ , so division by $-2$ gives $x\le-4$ .
$(-1,3]$ . The left endpoint is excluded and the right endpoint is included.

Sources and further reading

OpenStax: Linear inequalities. Definitions, conventions and related examples.
OpenStax Elementary Algebra 2e. Language of algebra, equations, inequalities and factoring.
OpenStax Intermediate Algebra 2e. Quadratic equations, factoring and algebraic techniques.
Wolfram Language: Algebraic Transformations. Symbolic simplification, expansion and factoring reference.