Advanced level

Linear inequalities

A linear inequality has not one solution but a whole set of them. Learn the operations that keep the inequality sign as it is, memorise the single one that flips it, and learn to mark the solution set on a number line.

Before you start

This topic builds on earlier ideas. Before you start, it's worth working through the lessons below — they'll make everything click:

All formulas

  • A linear inequality

    ax+b>c(a0)ax + b > c \quad (a \neq 0)

    one of <, >, ≤, ≥ in place of the equals sign

  • Adding a number to both sides

    a>b    a+c>b+ca > b \iff a + c > b + c

    the inequality sign stays as it is

  • Multiplying by a positive number

    a>b    ac>bc(c>0)a > b \iff ac > bc \quad (c > 0)

    the inequality sign stays as it is

  • Multiplying by a negative number

    a>b    ac<bc(c<0)a > b \iff ac < bc \quad (c < 0)

    the inequality sign FLIPS

  • Interval notation

    x>3    x(3,)x > 3 \iff x \in (3, \infty)

    a round bracket — the endpoint is not included

  • A non-strict inequality

    x3    x[3,)x \ge 3 \iff x \in [3, \infty)

    a square bracket — the endpoint is included

A linear inequality looks like a linear equation, except that instead of an equals sign it carries one of <<, >>, \le, \ge:

ax+b>c(a0)ax + b > c \quad (a \neq 0)

The difference in the answer is fundamental. A linear equation usually has one solution; an inequality has a whole set of them. The answer is not a number but a condition such as x>3x > 3.

What you may do to an inequality

Inequalities are rearranged just like equations, with one exception covered below. The following keep the inequality sign unchanged:

a>b    a+c>b+ca > b \iff a + c > b + c a>b    ac>bc(c>0)a > b \iff a \cdot c > b \cdot c \quad (c > 0)

So terms may be moved to the other side (changing sign, as in equations), and both sides may be divided by a positive number.

The one exception: multiplying and dividing by a negative number

When both sides are multiplied or divided by a negative number, the inequality sign must be flipped:

a>b    ac<bc(c<0)a > b \iff a \cdot c < b \cdot c \quad (c < 0)

Why? Because multiplying by a negative number reflects the number line about zero and reverses the order of numbers. 2<52 < 5 is true, but after multiplying both sides by 1-1 we have 2-2 and 5-5, and 2>5-2 > -5. If the sign stayed as it was, a true statement would turn false.

Solve: −3x + 1 < 10.

Note that the flip is caused by multiplying or dividing by a negative number, not by a minus appearing somewhere in the problem. Moving 3-3 to the other side changes nothing about the sign.

The solution set on a number line

The solution of an inequality is easiest to see on a number line: mark the boundary number and take the whole ray on the side where the solutions lie.

−2−1012345678910x > 33
The solutions of x > 3 are all the numbers to the right of 3; 3 itself is not one of them.

Whether the boundary belongs to the set is decided by the kind of inequality:

  • strict (<<, >>) — the boundary is not included; on a hand drawing it gets an open circle,
  • non-strict (\le, \ge) — the boundary is included; a filled circle.

The same set is written more briefly as an interval:

x>3    x(3,)x3    x[3,)x > 3 \iff x \in (3, \infty) \qquad x \ge 3 \iff x \in [3, \infty)

A round bracket marks an end not in the set, a square bracket one that is. Infinity always takes a round bracket, because infinity is not a number.

Solve and write as an interval: 4x − 5 ≥ 2x + 7.

Inequalities in practice

Inequalities answer questions of the "at most" and "at least" kind. A delivery costs 40 plus 6 per parcel, and the budget is 100. How many parcels fit the budget?

40+6x1004040 + 6x \le 100 \quad \big| -40 6x60:66x \le 60 \quad \big| : 6 x10x \le 10

The solution set is every x10x \le 10, but a number of parcels is a non-negative whole number — so the answer is: at most ten. That is typical of word problems: mathematics gives the solution set, and the meaning of the problem narrows it down.

In the exercises below, enter the whole solution set, for example x > 3 or x ≥ 3 (x >= 3 works too). The boundary number alone is not enough — the direction counts as well.

Practice

Work through a set of exercises — they get harder as you go. At the end you'll see your score and the mistakes worth reviewing.

Exercise 1 of 8Score: 0
4x + 1 > 21

Common mistakes

  • Not flipping the sign after dividing by a negative number — the single most common mistake here.
  • Flipping the sign when moving a term across — moving is addition, so the sign stays.
  • Mixing up strict and non-strictx>3x > 3 excludes 3, x3x \ge 3 includes it.
  • Answering with the boundary alone33 is not the answer; x>3x > 3 is.
  • Ignoring what the problem is about — a number of parcels or people is never a fraction or negative.

Formula card

Topic: Linear inequalities

  • A linear inequality

    ax+b>c(a0)ax + b > c \quad (a \neq 0)

    one of <, >, ≤, ≥ in place of the equals sign

  • Adding a number to both sides

    a>b    a+c>b+ca > b \iff a + c > b + c

    the inequality sign stays as it is

  • Multiplying by a positive number

    a>b    ac>bc(c>0)a > b \iff ac > bc \quad (c > 0)

    the inequality sign stays as it is

  • Multiplying by a negative number

    a>b    ac<bc(c<0)a > b \iff ac < bc \quad (c < 0)

    the inequality sign FLIPS

  • Interval notation

    x>3    x(3,)x > 3 \iff x \in (3, \infty)

    a round bracket — the endpoint is not included

  • A non-strict inequality

    x3    x[3,)x \ge 3 \iff x \in [3, \infty)

    a square bracket — the endpoint is included

−2−1012345678910x > 33
The solution set of x > 3: every number to the right of 3, but not 3 itself.

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