Topic · Algebra

Inequalities

An inequality is an equation's looser cousin: instead of saying two things are equal, it says one is bigger, smaller, or at most as big. The solution to an inequality is usually not a single number but a whole region of the number line — and almost everything you know about solving equations carries over, with one famous exception.

What you'll leave with

  • The four inequality symbols and what each really means.
  • Solving linear inequalities the same way you solve equations — with one small but loud catch.
  • The flip rule: multiplying or dividing by a negative reverses the direction of the inequality.
  • Compound inequalities (AND, OR) and how to graph the resulting regions on a number line.

1. What an inequality says

An equation ($=$) says two expressions describe the same number. An inequality says they don't, and how they relate.

SymbolRead asSolutions look like
$x < a$"strictly less than"Everything to the left of $a$ on the number line, not including $a$.
$x > a$"strictly greater than"Everything to the right of $a$, not including $a$.
$x \leq a$"less than or equal to"Everything to the left, including $a$.
$x \geq a$"greater than or equal to"Everything to the right, including $a$.

The solution set of an inequality is the set of all values of the variable that make it true. Where the solution set of $x + 3 = 10$ is a single number ($x = 7$), the solution set of $x + 3 < 10$ is a whole region of the number line ($x < 7$ — every number less than $7$, with $7$ itself excluded).

Strict vs. non-strict

The symbols $<$ and $>$ are strict — they exclude equality. $\leq$ and $\geq$ include it. On a number-line graph this difference shows up as an open versus filled circle at the endpoint, which we'll see in §5.

2. Solving linear inequalities

Solving a linear inequality is almost identical to solving a linear equation. You isolate the variable using the same toolkit:

  • Add or subtract the same number on both sides.
  • Multiply or divide both sides by the same positive number.
  • Simplify each side.

Solve $3x + 5 < 17$:

$$ 3x + 5 < 17 \quad\Longrightarrow\quad 3x < 12 \quad\Longrightarrow\quad x < 4. $$

The solution set is "all $x$ less than $4$" — infinitely many numbers, with $4$ itself excluded.

Check by trying a value inside the set: at $x = 0$, $3(0) + 5 = 5$, and $5 < 17$ ✓. Try a value outside: at $x = 5$, $3(5) + 5 = 20$, and $20 < 17$ is false ✓.

3. The flip rule for negatives

Here is the one place inequalities differ from equations:

If you multiply or divide both sides of an inequality by a negative number, you must flip the direction of the inequality.

Why? Multiplying by a negative reflects across zero, and reflection swaps "left" and "right." If $3 < 5$, then negating both gives $-3$ and $-5$ — and now $-3 > -5$, because $-3$ is to the right of $-5$ on the number line. The relationship has flipped.

Concrete example: solve $-2x \geq 8$.

Divide both sides by $-2$ — and flip $\geq$ to $\leq$:

$$ -2x \geq 8 \quad\Longrightarrow\quad x \leq -4. $$

Check: $x = -5$ should be in the solution set. $-2 \cdot (-5) = 10$, and $10 \geq 8$ ✓.

The single most common inequality error

Forgetting to flip when multiplying or dividing by a negative. Whenever the coefficient of $x$ comes out negative, take a moment to write the flip explicitly. If you stayed inside the positives all the way through, no flip is needed.

Adding or subtracting a negative does not require flipping — the rule is strictly about multiplication and division. $x - 3 < 5$ becomes $x < 8$ without any flipping, even though "$-3$" is involved.

4. Compound inequalities

Two inequalities joined into one statement. There are two flavors, "AND" and "OR," and they describe different regions.

AND (intersection)

$2 < x < 7$ means $x > 2$ and $x < 7$ — both must be true. The solution set is the strip between $2$ and $7$, exclusive of the endpoints.

To solve compound AND inequalities written as a "sandwich," do the same operation to all three pieces. $-1 < 2x - 3 \leq 5$:

Add $3$ throughout: $2 < 2x \leq 8$.

Divide by $2$ throughout: $1 < x \leq 4$.

OR (union)

$x < -1$ or $x \geq 4$ is satisfied by either condition — values to the left of $-1$, or values at or to the right of $4$. The solution set is two disjoint pieces of the number line.

OR inequalities are written as two separate inequalities joined by the word "or." There's no compact sandwich notation; each piece is solved independently.

AND can be empty

An AND like $x > 5$ AND $x < 2$ has no solutions — there's no number bigger than $5$ and smaller than $2$. The solution set is the empty set, $\emptyset$. OR compounds, by contrast, almost always cover something.

5. Graphing solutions on the number line

The standard convention:

  • Filled (closed) circle at the endpoint: the endpoint is included ($\leq$, $\geq$).
  • Open circle at the endpoint: the endpoint is not included ($<$, $>$).
  • Shaded line: the set of solutions extending from the endpoint(s).
  • Arrow at the end of the shaded region: the solution extends to infinity in that direction.
x < 4 -4 -2 0 2 4 6 x ≥ 4 -4 -2 0 2 4 6

For a compound inequality like $1 < x \leq 4$, you'd draw an open circle at $1$, a closed circle at $4$, and shade the segment between.

Solution sets can also be written in interval notation, which is a compact textual form of the same idea: parentheses for excluded endpoints, brackets for included ones. So $x < 4$ is $(-\infty, 4)$; $x \geq 4$ is $[4, \infty)$; $1 < x \leq 4$ is $(1, 4]$. Infinity always gets a parenthesis — it's not a number to include.

6. Playground: graph an inequality

Build $ax + b \;\square\; c$ by dragging the sliders and picking an operator. The solution is computed live — and when $a$ is negative, you'll see the flip rule fire automatically.

2x + 1 > 5
Solution: x > 3
2
-1
5
Operator
Try it

Slide $a$ down through zero into the negatives. The solution direction flips, and the note appears. Now try $a = 0$ — the inequality has no variable left to solve for, and the playground reports whether the statement is universally true or universally false.

7. Common pitfalls

Forgetting to flip

Dividing both sides of $-3x < 12$ by $-3$ gives $x > -4$, not $x < -4$. The flip is the most-missed step in inequality work. Quick sanity check: pick a number you think is in your solution set and plug it back into the original; if it doesn't satisfy the inequality, you forgot to flip.

Flipping when subtracting

$x - 3 < 5$ becomes $x < 8$ without any flipping. You only flip when multiplying or dividing by a negative — not when adding or subtracting one. Subtracting $3$ moves both sides by the same amount; their relative order doesn't change.

Endpoint inclusion

"$x \leq 5$" includes $5$; "$x < 5$" does not. On a graph, this is the closed-vs-open-circle distinction. In interval notation, it's bracket vs parenthesis. The single character makes a real mathematical difference, especially when interpreting word problems ("at most $5$" → $\leq$; "less than $5$" → $<$).

Confusing AND with OR

"AND" requires both conditions; the solution set is the intersection. "OR" is satisfied by either; the solution set is the union. They give different pictures: AND is one strip, OR is two pieces. Make sure you know which one a word problem is asking before you graph.

8. Worked examples

Example 1 · Solve $4x - 7 \leq 17$

Add $7$ to both sides: $4x \leq 24$. Divide by $4$ (positive — no flip): $x \leq 6$.

Solution: $\boxed{x \leq 6}$, or in interval notation $(-\infty, 6]$.

Example 2 · Solve $-3x + 1 > 10$

Subtract $1$: $-3x > 9$. Divide by $-3$ and flip the inequality: $x < -3$.

Solution: $\boxed{x < -3}$. Sanity check: at $x = -4$, $-3(-4) + 1 = 13 > 10$ ✓.

Example 3 · Solve the compound $-1 < 2x - 3 \leq 5$

Add $3$ throughout: $2 < 2x \leq 8$. Divide by $2$ throughout: $1 < x \leq 4$.

Solution: $\boxed{1 < x \leq 4}$, or $(1, 4]$. Open circle at $1$, closed at $4$, shaded between.

Example 4 · Solve $5 - 2x \geq 11$

Subtract $5$: $-2x \geq 6$. Divide by $-2$, flipping: $x \leq -3$.

Solution: $\boxed{x \leq -3}$, or $(-\infty, -3]$.

Example 5 · Solve the OR compound $2x + 1 < -3$ or $x - 4 \geq 1$

Solve each piece separately:

  • $2x + 1 < -3 \Rightarrow 2x < -4 \Rightarrow x < -2$.
  • $x - 4 \geq 1 \Rightarrow x \geq 5$.

Solution: $\boxed{x < -2 \text{ or } x \geq 5}$. In interval notation: $(-\infty, -2) \cup [5, \infty)$. The set is two disjoint pieces.

Sources & further reading

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