Solving One-Variable Linear Inequalities

Hello again. In the previous lesson, you rearranged formulas by applying balanced operations until a chosen variable stood alone. Linear inequalities use the same core technique, but the result is usually a range of possible values rather than one exact value.

This lesson develops the method for solving one-variable linear inequalities, interpreting the solution, and handling the one essential new rule: when you multiply or divide by a negative number, you must reverse the inequality sign. This matters whenever a data rule sets a threshold, such as keeping a cost below a budget or a model score above a required level.

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An inequality describes a set of valid values

An equation such as

$$x=3$$

has one solution: exactly $3$.

An inequality such as

$$x<3$$

has infinitely many solutions: every number smaller than $3$, including $2$, $0$, $-10$, and $2.999$. But $3$ itself does not work, because the inequality is strict.

The four comparison signs are:

Sign	Meaning	Does the boundary value count?
$<$	less than	No
$>$	greater than	No
$\le$	less than or equal to	Yes
$\ge$	greater than or equal to	Yes

For example:

$$x\le 3$$

includes $3$, as well as every smaller number.

On a number line, a strict inequality uses an open circle at the boundary, while an inclusive inequality uses a filled circle. The solution is shaded to the left for “less than” and to the right for “greater than.”

Khan Academy’s short video gives a useful visual interpretation of an inequality as a region of the number line, then introduces the sign-reversal rule.

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Solve them like equations—until a negative operation appears

A one-variable linear inequality has a variable only to the first power. Typical examples include:

$$4x+1<13$$ $$-3x+8\le 20$$ $$7-2x>x+1$$

The objective is the same as for a linear equation: isolate the variable. You may safely:

• add the same quantity to both sides;
• subtract the same quantity from both sides;
• multiply or divide both sides by a positive number;
• simplify expressions on either side.

Those operations preserve the direction of the comparison.

Consider:

$$4x+1<13$$

Subtract $1$ from both sides:

$$4x<12$$

Then divide both sides by $4$, which is positive:

$$x<3$$

The sign remains $<$ because dividing by a positive value preserves order.

This is the same balancing logic used for equations. The difference is that instead of finding one value, you identify all values that make the comparison true.

A compact workflow is:

1. Simplify each side if needed.
2. Gather variable terms on one side and constants on the other.
3. Isolate the variable.
4. Reverse the inequality sign only if the final multiplication or division is by a negative quantity.
5. Check one value that should work and one value that should not.

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Why a negative reverses the sign

The crucial rule is:

When multiplying or dividing both sides of an inequality by a negative number, reverse the inequality sign.

For instance, $3<7$ is true. Multiply both values by $-1$:

$$-3>-7$$

The original smaller number, $3$, becomes the larger number after negation, $-3$. Multiplying by a negative reflects numbers across zero, reversing their order.

Now solve:

$$-3x+8\le 20$$

First subtract $8$ from both sides:

$$-3x\le 12$$

Next divide both sides by $-3$. Because $-3$ is negative, reverse $\le$ to $\ge$:

$$x\ge -4$$

That is the solution.

A quick check confirms it:

• Try $x=0$, which satisfies $x\ge -4$:

$$-3(0)+8\le 20$$ $$8\le 20$$

This is true.

• Try $x=-5$, which is outside the solution set:

$$-3(-5)+8\le 20$$ $$23\le 20$$

This is false.

The sign reversal happens when you multiply or divide by a negative. It does not happen merely because a negative number appears in the expression.

For example, from

$$-2x-3<5$$

add $3$ to both sides:

$$-2x<8$$

The sign has not changed: you added a positive $3$, rather than multiplying or dividing by a negative. Only in the next step does it reverse:

$$x>-4$$

because dividing by $-2$ is required.

OpenStax provides a concise statement of the comparison rules and then applies them to practical constraints.

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A multi-step inequality with variables on both sides

The same approach works when $x$ appears on both sides. Consider:

$$7-2x>x+1$$

First collect the variable terms on one side. Subtract $x$ from both sides:

$$7-3x>1$$

Then remove the constant $7$ from the left by subtracting $7$:

$$-3x>-6$$

Finally, divide by $-3$. This reverses the sign:

$$x<2$$

A check makes the result more trustworthy. Test $x=0$, which lies below $2$:

$$7-2(0)>0+1$$ $$7>1$$

True.

Test $x=3$, which lies outside the solution:

$$7-2(3)>3+1$$ $$1>4$$

False.

Checking values is particularly useful while learning because it can catch the most common error: forgetting to reverse the sign after division by a negative coefficient.

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Interpreting threshold rules in data work

Inequalities express constraints. A constraint says which inputs, outputs, or parameter values are acceptable.

Suppose a simple estimate of a storage requirement, in gigabytes, is

$$s=0.8n+2$$

where $n$ is the number of data units being stored. If the available storage is at most $10$ GB, the constraint is

$$0.8n+2\le 10$$

Solve it:

$$0.8n\le 8$$ $$n\le 10$$

So the model says that $10$ data units is the largest allowable value. Since $n$ counts units, values such as $10.5$ may not make sense in context; a real application may require a whole-number decision.

In programming, the inequality itself can be evaluated directly. But algebra tells you the entire range of valid inputs before you test individual cases.

def storage_is_within_limit(n):
    storage_gb = 0.8 * n + 2
    return storage_gb <= 10

for n in [8, 10, 11]:
    print(n, storage_is_within_limit(n))

The output should show that $8$ and $10$ satisfy the limit, while $11$ does not. This matches the solved condition $n\le 10$.

Later in machine learning, inequalities will appear in filtering data, defining decision thresholds, and specifying optimization constraints. For now, the important skill is recognizing that a solved inequality gives a range, not a single prediction or parameter value.

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Common pitfalls

Forgetting the sign reversal

From

$$-5x\ge 15$$

the correct result is

$$x\le -3$$

because division by $-5$ reverses $\ge$ to $\le$.

Reversing the sign after addition or subtraction

From

$$x-4>-2$$

add $4$ to both sides:

$$x>2$$

The sign does not reverse. Addition and subtraction, even subtraction of a negative number, do not reverse the comparison.

Treating a strict boundary as included

If

$$x<3$$

then $x=3$ is not a solution. By contrast, $x=3$ does satisfy

$$x\le 3$$

Dividing by a variable of unknown sign

At this stage, avoid dividing an inequality by an unknown variable such as $b$. In

$$bx<3b$$

the correct direction depends on whether $b$ is positive or negative, and division is impossible if $b=0$. The inequalities in this lesson use numerical coefficients, so you can always determine whether the sign must reverse.

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Key takeaways

A linear inequality is solved with the same balanced-operation method used for linear equations, but its answer is a set of possible values.

• Add or subtract the same quantity on both sides without changing the sign.
• Multiply or divide by a positive number without changing the sign.
• Multiply or divide by a negative number and reverse the sign.
• Use $<$ and $>$ for strict bounds; use $\le$ and $\ge$ when the boundary is allowed.
• Check a value inside the solution set and one outside it when you want to verify your work.

Next, the course shifts briefly from comparisons to compact notation for repeated addition: finite sums written with sigma notation.