Solving and Graphing Linear Inequalities

Hello. In the previous lesson, you solved linear equations by keeping both sides balanced, simplifying carefully, and using inverse operations. Linear inequalities use nearly the same algebra, but their answers are usually sets of numbers rather than one number.

In this lesson, you will solve one-variable linear inequalities, graph their solution sets on a number line, and write those sets in interval notation. The one new algebra rule is crucial: when you multiply or divide by a negative number, you must reverse the inequality symbol.

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Inequalities describe a range of possible values

An equation such as

$$x+4=9$$

has one solution:

$$x=5$$

An inequality such as

$$x+4>9$$

means that $x+4$ is greater than $9$. Its solution is:

$$x>5$$

This includes $6$, $10$, $5.2$, and infinitely many other numbers. It does not include $5$, because $5+4=9$, not a number greater than $9$.

Here are the four inequality symbols:

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

A useful way to read the symbols is to notice that the wider opening faces the larger number. For example, $8>3$ because $8$ is larger than $3$.

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Graphing the solution set

Once the variable is isolated, the number line shows every value that works.

There are two decisions:

1. Is the endpoint included?

   • Use an open circle for $<$ or $>$.
   • Use a closed circle for $\le$ or $\ge$.

2. Which direction contains the solutions?

   • “Less than” means shade left.
   • “Greater than” means shade right.

The same ideas appear in interval notation:

Inequality	Number-line graph	Interval notation
$x>a$	Open endpoint at $a$, shade right	$(a,\infty )$
$x\ge a$	Closed endpoint at $a$, shade right	$[a,\infty )$
$x<a$	Open endpoint at $a$, shade left	$(-\infty ,a)$
$x\le a$	Closed endpoint at $a$, shade left	$(-\infty ,a]$

Parentheses mean “not included,” while brackets mean “included.” Always use parentheses with $\infty$ and $-\infty$, because infinity is not an actual endpoint number that can be included.

Example: solve, graph, and write an interval

Solve:

$$4x-7\le 13$$

Add $7$ to both sides:

$$4x-7+7\le 13+7$$ $$4x\le 20$$

Divide both sides by positive $4$:

$$x\le 5$$

Because we divided by a positive number, the inequality symbol stays the same.

The graph has a closed circle at $5$, because $5$ is included, with shading to the left.

The interval notation is:

$$(-\infty ,5]$$

A quick check confirms the endpoint: substitute $x=5$ into the original inequality.

$$4(5)-7\le 13$$ $$13\le 13$$

This is true, so including $5$ makes sense.

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Solving works like equations — until a negative multiplier appears

The basic strategy is the same one you used for multistep equations:

1. Simplify each side: distribute and combine like terms if necessary.
2. Get variable terms on one side.
3. Get constants on the other side.
4. Divide by the coefficient of the variable.
5. Graph and state the solution set.

You can add or subtract the same number on both sides without changing the inequality direction. You can also multiply or divide by a positive number without changing it.

The exception is multiplying or dividing by a negative number.

Why does the symbol reverse?

Start with a true statement:

$$2<5$$

Multiply both sides by $-1$:

$$-2>-5$$

The values switch positions on the number line when their signs change. Since $-2$ is greater than $-5$, the original less-than symbol must reverse to remain true.

Rule: Multiply or divide an inequality by a negative number, and reverse the inequality symbol.

Do not reverse a symbol merely because you see a negative somewhere in the problem. The reversal happens only when the operation performed is multiplication or division by a negative number.

Example: a negative coefficient

Solve:

$$-3x+8>-7$$

Subtract $8$ from both sides:

$$-3x+8-8>-7-8$$ $$-3x>-15$$

Now divide by $-3$. Because this is division by a negative number, reverse the symbol:

$$x<5$$

The graph has an open circle at $5$ and shading to the left. In interval notation:

$$(-\infty ,5)$$

Check one number that should work, such as $x=0$:

$$-3(0)+8>-7$$ $$8>-7$$

True. Now check a value outside the set, such as $x=6$:

$$-3(6)+8>-7$$ $$-10>-7$$

False. These checks support the solution $x<5$.

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Multistep inequalities and variables on both sides

An inequality can have parentheses and variables on both sides, just as equations can. Keep the same orderly approach: simplify first, then isolate the variable.

Solve:

$$5(x-2)+3x\ge 2x+18$$

First distribute:

$$5x-10+3x\ge 2x+18$$

Combine like terms on the left:

$$8x-10\ge 2x+18$$

Subtract $2x$ from both sides:

$$6x-10\ge 18$$

Add $10$ to both sides:

$$6x\ge 28$$

Divide by positive $6$:

$$x\ge \frac{14}{3}$$

The solution includes $\frac{14}{3}$, so graph a closed circle there and shade right. In interval notation:

$$\left[\frac{14}{3},\infty \right)$$

Notice that subtracting $2x$ did not reverse the sign. Although the expression contains a variable, this step was subtraction, not multiplication or division by a negative number.

Frequent errors to avoid

• Forgetting to distribute:

  $$5(x-2)\ne 5x-2$$

  Instead,

  $$5(x-2)=5x-10$$

• Flipping after adding or subtracting a negative: Adding $-4$ is still addition, so no flip is needed.

• Forgetting to flip after dividing by a negative:

  $$-2x\ge 10$$

  becomes

  $$x\le -5$$

  not $x\ge -5$.

• Using the wrong endpoint type: Strict symbols, $<$ and $>$, use open circles. Inclusive symbols, $\le$ and $\ge$, use closed circles.

• Shading based on the original inequality rather than the final one: Always graph the inequality after the variable has been isolated.

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When the variable disappears

Sometimes simplifying causes all variable terms to cancel. At that point, decide whether the remaining statement is always true or always false.

All real numbers

Consider:

$$3(x+2)-x<2x+9$$

Distribute and combine:

$$3x+6-x<2x+9$$ $$2x+6<2x+9$$

Subtract $2x$ from both sides:

$$6<9$$

This is always true. Every real number satisfies the original inequality.

The solution can be written as:

$$(-\infty ,\infty )$$

On a number line, the entire line is shaded.

No solution

Now consider:

$$2(x-1)+x\ge 3x+5$$

Distribute and combine:

$$2x-2+x\ge 3x+5$$ $$3x-2\ge 3x+5$$

Subtract $3x$ from both sides:

$$-2\ge 5$$

This is false. No number can make the original inequality true.

The solution set is:

$$\varnothing$$

The graph is an empty number line: no point is shaded.

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Key takeaways

A one-variable linear inequality has a set of solutions, usually represented with an inequality statement, a graph, or interval notation.

• Solve inequalities using the same inverse-operation reasoning used for equations.
• Add or subtract on both sides without changing the symbol.
• Multiply or divide by a positive number without changing the symbol.
• Multiply or divide by a negative number and reverse the inequality symbol.
• Use an open endpoint for $<$ or $>$, and a closed endpoint for $\le$ or $\ge$.
• If the variable cancels, a true statement means all real numbers work; a false statement means no solutions.

Next, you will use the same algebraic discipline to rearrange literal equations—formulas with several variables—so that a chosen variable is isolated.