← Math: from algebra to machine learning9 Oct 2026

Algebra · Part 3 of 10

Linear inequalities

Solve linear inequalities step by step, learn when the sign flips, and read answers as intervals and word problems.

Algebra16 min read

An equation tells you one exact value: "the ticket costs exactly $8." Real life is often looser: "I can spend at most $40," "I need at least 6 hours of sleep," "the elevator holds no more than 1000 kg." These are inequalities, and their answers are not single numbers but whole ranges of numbers.

Solving a linear inequality is almost identical to solving a linear equation: do the same operations to both sides to isolate the variable. There is exactly one new rule to learn, and it appears only when you multiply or divide by a negative number. This lesson shows when and why that rule kicks in, then practices reading and checking answers.

What an inequality solution looks like

The four inequality symbols are $<$ (less than), $>$ (greater than), $\leq$ (less than or equal to), and $\geq$ (greater than or equal to). A solution to an inequality is any value of the variable that makes the statement true, so solutions usually form an interval.

For example, $x < 3$ is true for $2.9$ , $2$ , $0$ , $- 100$ , and every other number below $3$ — but not for $3$ itself. We write the answer as $x < 3$ , or in interval notation $(- \infty, 3)$ , where a round parenthesis means the endpoint is excluded. For $x \leq 3$ the endpoint is included, written $(- \infty, 3]$ with a square bracket.

A good habit: pick one test value from your answer and substitute it into the original inequality. If $x = 0$ satisfies $x < 3$ (it does: $0 < 3$ ), your interval at least contains the right region.

The solution of x ≤ 3 is the whole ray up to 3, with a closed circle because 3 itself satisfies the inequality.

Example 1. Reading a solution as an interval Which values satisfy $x \leq 3$ ? Write the answer in interval notation and test $x = 3$ and $x = 4$ .

The endpoint $3$ is included because of the "or equal to" bar.

$x \leq 3 means (- \infty, 3]$

Test $x = 3$ : $3 \leq 3$ is true, so the endpoint belongs.

Test $x = 4$ : $4 \leq 3$ is false, so values above $3$ are excluded, as expected.

Solving inequalities like equations

Adding or subtracting the same number on both sides, and multiplying or dividing both sides by a positive number, never changes which values make the inequality true. So most of the work is exactly what you did with equations.

The one new rule: if you multiply or divide both sides by a negative number, the inequality sign flips ( $<$ becomes $>$ , $\leq$ becomes $\geq$ , and so on).

Why? Compare $2 < 5$ . Multiply both sides by $- 1$ : you get $- 2$ and $- 5$ , and $- 2 > - 5$ . Negatives reverse order, so the sign must flip to keep the statement true.

Strategy: move all variable terms to one side and constants to the other, combine like terms, and isolate $x$ . If a negative coefficient ends up in front of $x$ , divide by it and flip the sign. (Alternatively, you can avoid the flip entirely by moving $x$ -terms so $x$ ends up with a positive coefficient — both routes give the same answer.)

After dividing −2x > 10 by −2 and flipping the sign, the answer is an open circle at −5 with everything to the left shaded.

Example 2. A one-step inequality Solve $x + 4 \leq 9$ .

Subtract $4$ from both sides:

$x \leq 5$

Check with $x = 5$ : $5 + 4 = 9 \leq 9$ , true. And $x = 4$ gives $8 \leq 9$ , also true, so the interval $(- \infty, 5]$ behaves correctly.

Example 3. Dividing by a negative flips the sign Solve $- 2 x > 10$ .

Divide both sides by $- 2$ and flip the sign:

$x < - 5$

Check with $x = - 6$ : $- 2 (- 6) = 12 > 10$ , true.

Check the excluded endpoint $x = - 5$ : $- 2 (- 5) = 10$ , which is not greater than $10$ , so $- 5$ is correctly left out.

Example 4. Variables on both sides Solve $3 x - 7 \geq 5 x + 1$ .

Subtract $5 x$ from both sides:

$- 2 x - 7 \geq 1$

Add $7$ to both sides:

$- 2 x \geq 8$

Divide by $- 2$ and flip:

$x \leq - 4$

Check with $x = - 5$ : left side $3 (- 5) - 7 = - 22$ , right side $5 (- 5) + 1 = - 24$ , and $- 22 \geq - 24$ is true.

Check $x = - 3$ : left side $- 16$ , right side $- 14$ , and $- 16 \geq - 14$ is false — so values above $- 4$ are correctly excluded.

Inequalities with fractions and parentheses

Distribute first, clear fractions if you like, then collect terms. Clearing fractions by multiplying everything by a positive common denominator is safe — no flip needed since the multiplier is positive.

Example 5. Clearing a fraction Solve $\frac{x}{3} + 1 < \frac{x}{2}$ .

Multiply every term by $6$ (positive, so no flip):

$2 x + 6 < 3 x$

Subtract $2 x$ from both sides:

$6 < x$

That is the same as $x > 6$ .

Check with $x = 7$ : $\frac{7}{3} + 1 \approx 3.33$ and $\frac{7}{2} = 3.5$ , and $3.33 < 3.5$ is true.

Check $x = 6$ : $3 < 3$ is false, so $6$ is correctly excluded.

Compound inequalities

A compound inequality like $- 3 \leq 2 x + 1 < 7$ says two things at once: $- 3 \leq 2 x + 1$ and $2 x + 1 < 7$ . The trick is to operate on all three parts at once, keeping the chain valid.

If the two halves of a compound inequality contradict each other (for example it would require $x < 1$ and $x > 4$ at once), there is no solution.

The compound inequality −3 ≤ 2x + 1 < 7 reduces to a closed circle at −2 and an open circle at 3: included on the left, excluded on the right.

Example 6. Solving a chain Solve $- 3 \leq 2 x + 1 < 7$ .

Subtract $1$ from all three parts:

$- 4 \leq 2 x < 6$

Divide all three parts by $2$ :

$- 2 \leq x < 3$

In interval notation: $[- 2, 3)$ .

Check $x = 0$ : $2 (0) + 1 = 1$ , and $- 3 \leq 1 < 7$ is true.

Check $x = 3$ : $7 < 7$ is false, so the right endpoint is correctly excluded.

Inequalities in word problems

Translate the words into a symbol: "at most" means $\leq$ , "at least" means $\geq$ , "more than" means $>$ , "less than" means $<$ . Then solve as usual and interpret the answer in context — often the answer must be a whole number, so you round the right way.

Example 7. A budget limit A taxi charges $ $3$ to start plus $ $2$ per kilometer. You have $ $25$ . At most how many whole kilometers can you ride?

Let $k$ be the number of kilometers. The cost must satisfy:

$3 + 2 k \leq 25$

Subtract $3$ :

$2 k \leq 22$

Divide by $2$ :

$k \leq 11$

The largest whole number of kilometers is $11$ .

Check: $3 + 2 (11) = 25 \leq 25$ , exactly on budget; $12$ km would cost $ $27$ , too much.

Common mistakes

Forgetting to flip the inequality sign when dividing by a negative: $- 2 x > 10$ gives $x < - 5$ , not $x > - 5$ .
Flipping the sign when multiplying or dividing by a positive number — the flip happens only for negatives.
Treating $\leq$ like $<$ and excluding an endpoint that should be included: $x \leq 5$ includes $x = 5$ .
In a compound inequality, operating on only two of the three parts instead of applying each step to all three.
In word problems, rounding the wrong way: if $k \leq 11.5$ kilometers, the largest whole number is $11$ , not $12$ .

Practice

Solve the inequality $x + 7 > 12$ .
Solve the inequality $- 3 x \leq 15$ .
Solve the inequality $4 x - 3 < x + 9$ .
Solve the compound inequality $- 5 \leq 3 x + 4 \leq 10$ .
A gym charges a $ $20$ membership fee plus $ $4$ per visit. If you have $ $60$ to spend in total, what is the largest number of whole visits you can pay for?

Answers

$x > 5$ , since subtracting $7$ from both sides gives $x > 5$ .
$x \geq - 5$ , since dividing by $- 3$ flips the sign.
$x < 4$ , since $3 x < 12$ after collecting terms.
$- 3 \leq x \leq 2$ , after subtracting $4$ from all parts and dividing by $3$ .
$10$ visits, because $20 + 4 v \leq 60$ gives $v \leq 10$ .

Next, you will meet systems of linear equations, where two equations with two variables must be satisfied at the same time.