Inequalities

Intuition

An equation like $x + 3 = 7$ asks a sharp question: which single number makes the two sides equal? The answer is just $x = 4$ . But a lot of real questions are not that exact. A ride says you must be at least $48$ inches tall. A budget says you can spend no more than $20$ dollars. A thermometer warns of temperatures below freezing. None of these names one number; each names a whole stretch of numbers that count as allowed. To write claims like ” $x$ is at least $48$ ” we need symbols for “less than” and “greater than.” We must also picture the answer not as a dot but as an entire region of the number line. That is an inequality, and the good news is that you solve one with the very same balance moves you already use on equations. There is a single new rule to watch for as well.

Learning goals

Read the four symbols, and say which include the boundary
Describe a solution as a range rather than one value
Graph with an open or closed circle, shading the right way
Solve with the same balance moves used for equations
Flip the symbol when multiplying or dividing by a negative
Test one value inside the range and one outside

The four inequality symbols

An inequality compares two expressions that need not be equal. In place of the equals sign it uses one of four symbols:

Symbol	Read as	Meaning
$<$	”less than”	strictly smaller, not equal
$>$	”greater than”	strictly larger, not equal
$\le$	”less than or equal to”, “at most”	smaller, or exactly equal
$\ge$	”greater than or equal to”, “at least”	larger, or exactly equal

So $5 < 8$ says five is less than eight, and $8 > 5$ says eight is greater than five: the same fact read from each end. The symbol always opens toward the larger amount and narrows to a point at the smaller one, like a mouth that wants the bigger meal. A quick check is that the small pointed end aims at the small number: in $5 < 8$ the point is on the $5$ .

The two symbols with a line underneath, $\le$ and $\ge$ , allow equality as well. The statement $x \le 6$ is true when $x$ is $6$ and also true when $x$ is anything below $6$ . In words, $\le$ means “at most” (no bigger than) and $\ge$ means “at least” (no smaller than). The bar is the difference between letting the boundary number in and shutting it out. We call $<$ and $>$ strict inequalities, because they exclude equality, and $\le$ and $\ge$ non-strict, because they include it.

A solution is a range, not a point

For an equation, a solution is usually one number. For an inequality, a solution is every number that makes the statement true, and that is almost always a whole range of them. Take

x > 3.

Is $4$ a solution? Yes, since $4 > 3$ . Is $10$ ? Yes. Is $3.1$ , or $3.5$ , or a thousand? Every one of them is greater than $3$ , so every one is a solution. What about $3$ itself? No: $3 > 3$ is false, because $3$ is not strictly greater than $3$ . And $2$ fails too, since $2$ is not greater than $3$ . So the solutions are exactly the numbers to the right of $3$ , with $3$ left out. There are infinitely many of them, which is why we describe the answer as a range rather than listing it.

Compare that with $x \ge 3$ . Now $3$ itself does count, because the bar under the symbol allows equality, so the solution is $3$ together with everything above it. The only difference between $x > 3$ and $x \ge 3$ is that one boundary number, $3$ . Keeping straight whether the boundary belongs to the solution is the whole art of reading an inequality, and the number line makes it visible.

Graphing the solution on a number line

A picture of all the solutions at once is a graph on the number line. You mark the boundary number, decide whether it belongs, and shade the direction that holds the rest of the solutions. Two rules cover every case:

The circle at the boundary shows whether the boundary itself is a solution. Use an open (hollow) circle for a strict inequality, $<$ or $>$ , because the boundary is excluded. Use a closed (filled) circle for a non-strict inequality, $\le$ or $\ge$ , because the boundary is included.
The shaded arrow points in the direction of all the solutions. For “greater” relations ( $>$ or $\ge$ ) the variable is larger than the boundary, so shade to the right. For “less” relations ( $<$ or $\le$ ) shade to the left. The arrowhead signals that the solutions continue forever that way.

The variable will not always sit on the left. If you see $3 < x$ , first read what it says about $x$ : $3 < x$ means the same thing as $x > 3$ , the boundary is just written first. So it still gets an open circle on $3$ , shaded right.

Here is $x > 3$ . The circle on $3$ is open, because $3$ is not a solution, and the shading runs right toward the larger numbers.

The graph of x > 3. The open circle on 3 marks it as excluded, and the shaded arrow runs right to cover every number greater than 3.

Now compare $x \ge 3$ . Everything is the same except the circle on $3$ is filled, because this time $3$ is part of the solution.

The graph of x is greater than or equal to 3. The closed circle on 3 marks it as included, and the shading again runs right.

A “less” relation shades the other way. Here is $x \le 1$ : the circle on $1$ is filled because the bar includes it, and the arrow runs left toward the smaller numbers.

The graph of x is less than or equal to 1. The closed circle on 1 includes it, and the shading runs left to cover every number below 1.

Check your understanding

Which number line shows the solution of $x < 2$ ?

A an open circle on

2

, shaded left B a closed circle on

2

, shaded left C an open circle on

2

, shaded right D a closed circle on

2

, shaded right

Solving inequalities with the balance moves

Here is the payoff for everything you learned about equations. An inequality is solved by isolating the variable, using the same inverse operations on both sides. Adding or subtracting the same amount from both sides keeps an inequality true. So does multiplying or dividing both sides by the same positive number. The reasoning is the balance principle again, only now the two sides need not be equal. Whatever comparison held before the move still holds after it: if one side was strictly larger, it stays larger; if the two sides were tied at an included $\le$ or $\ge$ boundary, they stay tied.

Why does adding the same amount to both sides keep the inequality? Picture the number line. Saying $a < b$ means $a$ sits to the left of $b$ . Suppose you slide both points the same distance in the same direction, by adding the same number to each. Their order does not change: $a$ is still on the left. So $a < b$ leads to $a + c < b + c$ for any number $c$ , and subtracting works the same way, since subtracting is just adding a negative. Multiplying or dividing both sides by a positive number stretches or shrinks every distance from zero by the same factor. A positive multiplier or divisor never flips anything across zero, so again the left-right order is preserved.

That gives the one-step inequalities at once. They look exactly like one-step equations, only with an inequality symbol carried straight down each line.

Worked example 1 Solve $x + 4 < 9$ and graph it

The variable has $4$ added to it, so subtract $4$ from both sides, carrying the $<$ symbol straight down:

x + 4 - 4 < 9 - 4 \;\Rightarrow\; x < 5.

The solution is every number less than $5$ . To check the direction, test a number from inside the claimed range and one outside it. Try $x = 0$ : is $0 + 4 < 9$ ? That is $4 < 9$ , true, and $0$ is indeed less than $5$ . Try $x = 6$ , which is outside: $6 + 4 < 9$ is $10 < 9$ , false, and $6$ is not less than $5$ . The two tests agree with $x < 5$ .

Graph it with an open circle on $5$ , since $<$ excludes the boundary, and shade left:

The graph of x < 5: an open circle on 5, shaded left.

So the solution is $x < 5$ , every number to the left of $5$ , with $5$ itself left out.

Worked example 2 Solve $5x \ge 20$

The variable is multiplied by the positive coefficient $5$ , so divide both sides by $5$ . Dividing by a positive number does not disturb the symbol, so the $\ge$ comes straight down:

\frac{5x}{5} \ge \frac{20}{5} \;\Rightarrow\; x \ge 4.

The solution is $4$ together with every number above it. Check with a value inside the range and one outside. Test $x = 4$ , the boundary: $5(4) = 20$ , and $20 \ge 20$ is true, so the boundary belongs, which matches the closed circle. Test $x = 0$ , outside: $5(0) = 0$ , and $0 \ge 20$ is false, as expected.

So $x \ge 4$ . On a number line this is a closed circle on $4$ with the shading running right.

Check your understanding

Solve $x - 6 \le 1$ for $x$ .

x \le 7

x \ge 7

x \le -5

x \le 5

The one new rule: a negative flips the symbol

Everything above carried the symbol down unchanged. There is exactly one situation where that fails, and it is the heart of this lesson. When you multiply or divide both sides of an inequality by a negative number, you must reverse the inequality symbol. A $<$ becomes a $>$ , a $\le$ becomes a $\ge$ , and so on. This is the step almost everyone forgets, so it is worth seeing exactly why it is forced.

Start with a true statement that needs no variables, say

2 < 3.

Two really is less than three. Now multiply both sides by $-1$ . The left side becomes $-2$ and the right side becomes $-3$ . If the symbol stayed the same we would be claiming $-2 < -3$ , but that is false. On the number line $-2$ sits to the right of $-3$ , so $-2$ is the larger number. The true statement is

-2 > -3.

Multiplying by $-1$ took each number to its opposite, which reflects it across zero, and a reflection across zero swaps left and right. The number that was smaller is now the larger one. So the order has to reverse, and the only way to keep the statement true is to flip the symbol from $<$ to $>$ .

Multiplying by a negative reflects each number across zero. 2 is left of 3, but their opposites -2 and -3 land in swapped order: -2 is now to the right of -3, so the symbol reverses.

Why multiplying by a negative reverses the inequality#

Suppose $a < b$ , so $a$ is the smaller of the two numbers, and let $k$ be any positive number. We will show that multiplying both sides by $-k$ forces the symbol to flip, giving $-ka > -kb$ .

Begin with what $a < b$ tells us: the difference $b - a$ is a positive number, because $b$ is larger than $a$ . Multiply that positive difference by the positive number $k$ . A positive times a positive is positive, so $k(b - a)$ is still positive. Distributing the $k$ , this says $kb - ka$ is positive, which is the same as saying $kb - ka > 0$ .

Read $kb - ka > 0$ directly: a positive value of $kb - ka$ means $kb$ is larger than $ka$ , that is $ka < kb$ . (This is just the earlier fact that multiplying both sides of $a < b$ by the positive number $k$ keeps the order.) Finally, rewrite that same positive difference using opposites: $kb - ka$ is exactly $(-ka) - (-kb)$ . So $(-ka) - (-kb) > 0$ , and by the same reading as before, a positive difference means the left side is the larger one:

-ka > -kb.

That is exactly the claim: $a < b$ multiplied through by the negative number $-k$ yields $-ka > -kb$ , with the symbol reversed. The same argument runs with $\le$ in place of $<$ , giving $\ge$ , since the boundary case $a = b$ becomes $-ka = -kb$ either way. Division by a negative is the same move, because dividing by $-k$ is multiplying by the positive reciprocal $\tfrac{1}{k}$ and then by $-1$ . Only that final $-1$ flips the symbol. So every multiply-or-divide by a negative reverses the inequality, once and exactly once.

The rule is precise, so do not over-apply it. You flip the symbol only when the number you multiply or divide both sides by is negative. Adding or subtracting a negative number does not flip anything, and a negative appearing somewhere else in the problem is irrelevant. What matters is the sign of the number you multiply or divide both sides by.

Worked example 3 Solve $-2x < 8$ and graph it

The variable is multiplied by the negative coefficient $-2$ , so divide both sides by $-2$ . Because $-2$ is negative, reverse the symbol from $<$ to $>$ as you divide:

\frac{-2x}{-2} > \frac{8}{-2} \;\Rightarrow\; x > -4.

The flip is the whole point: dividing by a negative reverses the relation. Check it against the original, where the inequality is still $-2x < 8$ . Test $x = 0$ , which should satisfy $x > -4$ : $-2(0) = 0$ , and $0 < 8$ is true. Test $x = -5$ , which should fail $x > -4$ : $-2(-5) = 10$ , and $10 < 8$ is false, exactly as predicted. Both tests support $x > -4$ , not $x < -4$ .

Graph it with an open circle on $-4$ (the symbol is strict) and shade right:

The graph of x > -4: an open circle on -4, shaded right. Dividing -2x < 8 by the negative -2 reversed the symbol.

So $x > -4$ . Forgetting to flip would have given the wrong half of the line.

Check your understanding

Solve $-x \ge 3$ for $x$ .

x \le -3

x \ge -3

x \le 3

x \ge 3

Two-step inequalities

When two operations sit on the variable, undo them in the reverse of the order of operations. That is exactly what you did for two-step equations: clear the added or subtracted constant first, then deal with the coefficient. The only extra vigilance is the flip rule, and it can come up at the multiply-or-divide step alone, never at the add-or-subtract step.

Worked example 4 Solve $3x + 5 \le 17$

The constant $5$ is added on, so subtract $5$ from both sides. Adding or subtracting never touches the symbol:

3x + 5 - 5 \le 17 - 5 \;\Rightarrow\; 3x \le 12.

Now the variable is multiplied by the positive coefficient $3$ , so divide both sides by $3$ . Since $3$ is positive, the symbol stays as $\le$ :

\frac{3x}{3} \le \frac{12}{3} \;\Rightarrow\; x \le 4.

Check a value inside the range against the original inequality. Test $x = 4$ , the boundary: $3(4) + 5 = 17$ , and $17 \le 17$ is true. Test $x = 5$ , outside: $3(5) + 5 = 20$ , and $20 \le 17$ is false. The solution is

x \le 4,

a closed circle on $4$ shaded left.

Worked example 5 Solve $-2x + 1 > 9$

Subtract the constant $1$ from both sides first. This is a subtraction, not a multiply or divide, so the symbol does not move:

-2x + 1 - 1 > 9 - 1 \;\Rightarrow\; -2x > 8.

Now divide both sides by the coefficient $-2$ . The divisor is negative, so reverse the symbol from $>$ to $<$ :

\frac{-2x}{-2} < \frac{8}{-2} \;\Rightarrow\; x < -4.

The flip happens at the division step and only there. Check against the original: $x = -5$ should work, and $-2(-5) + 1 = 10 + 1 = 11 > 9$ is true; $x = 0$ should fail, and $-2(0) + 1 = 1 > 9$ is false. So

x < -4,

an open circle on $-4$ shaded left.

Worked example 6 Solve $8 - 3x \ge 2$

Read the left side as $8 + (-3x)$ . Subtract the constant $8$ from both sides, leaving the variable term alone; subtraction does not affect the symbol:

8 - 3x - 8 \ge 2 - 8 \;\Rightarrow\; -3x \ge -6.

Divide both sides by the coefficient $-3$ . It is negative, so flip $\ge$ to $\le$ . A negative divided by a negative is positive, the rule from the integers chapter:

\frac{-3x}{-3} \le \frac{-6}{-3} \;\Rightarrow\; x \le 2.

Check against the original: $x = 2$ gives $8 - 3(2) = 8 - 6 = 2 \ge 2$ , true, so the boundary is included, matching the closed circle; $x = 3$ gives $8 - 9 = -1 \ge 2$ , false. The solution is

x \le 2.

Check your understanding

While solving $-4x - 1 < 11$ , you reach $-4x < 12$ . What is the correct next step?

A divide both sides by

-4

and reverse to

x > -3

B divide both sides by

-4

and keep

x < -3

C subtract

4

from both sides D divide both sides by

4

and keep

x < 3

Checking a solution by testing the range

The balance moves above already give the exact answer, as long as every step was valid. Testing values against the original inequality is a fast way to catch a mistake, especially a missed sign flip; it does not replace the algebra. Test a value from inside the claimed range and a value from outside it. The inside value should make the original true; the outside value should make it false. If either test disagrees, go back and check the sign at every multiply-or-divide step.

Worked example 7 Solve $\dfrac{x}{2} - 3 \ge -1$ and check it with two test values

Add the constant $3$ to both sides; the symbol is unaffected by addition:

\frac{x}{2} - 3 + 3 \ge -1 + 3 \;\Rightarrow\; \frac{x}{2} \ge 2.

The variable is divided by the positive number $2$ , so undo that by multiplying both sides by $2$ . The symbol is unchanged, because $2$ is positive:

2 \cdot \frac{x}{2} \ge 2 \cdot 2 \;\Rightarrow\; x \ge 4.

Now check it. Pick $x = 6$ from inside the range $x \ge 4$ and substitute into the original:

\frac{6}{2} - 3 = 3 - 3 = 0, \qquad 0 \ge -1 \;\checkmark

Then pick $x = 0$ from outside the range:

\frac{0}{2} - 3 = -3, \qquad -3 \ge -1 \;\text{false}

The inside value passes and the outside value fails, exactly as the algebra predicts, so the check supports $x \ge 4$ .

Check your understanding

A classmate says the solution to $x > 3$ is $x > 2$ . Testing $x = 4$ gives true, and testing $x = 0$ correctly gives false. Does that prove the claim $x > 2$ is right?

A No:

x = 2.5

satisfies

x > 2

but not the original

x > 3

B Yes, because both test values worked as expected C Yes, because one point was inside and one was outside D No, because a correct check always needs three test values

Reading an inequality before you move

Every inequality in this lesson comes down to a short routine. First, isolate the variable with the same inverse operations you use on equations, clearing the constant before the coefficient. Second, at a multiply-or-divide step, check the sign: if you multiply or divide both sides by a negative number, reverse the symbol; otherwise leave it. Third, read off the graph: an open circle for $<$ or $>$ , and a closed circle for $\le$ or $\ge$ . Shade right for a “greater” relation and left for a “less” one. Finally, test a value from inside the range and one from outside it, against the original inequality, to catch a mistake before you move on. The table collects the symbol facts at a glance.

Symbol	In words	Boundary circle	Shade
$x > a$	greater than $a$	open	right
$x \ge a$	at least $a$	closed	right
$x < a$	less than $a$	open	left
$x \le a$	at most $a$	closed	left

The single fact that sets inequalities apart from equations is the flip. Multiplying or dividing both sides by a negative reverses the symbol, because taking opposites swaps the order of the numbers on the line. Hold onto that, and an inequality is no harder than the equation it resembles.

Common mistakes

Forgetting to flip the symbol. Multiplying or dividing both sides by a negative number reverses the inequality. Solving $-2x < 8$ by dividing by $-2$ gives $x > -4$ , not $x < -4$ . This is the single most common error across this lesson.
Flipping when you should not. The symbol reverses only for multiplying or dividing by a negative. Subtracting a number, even a negative one, never flips it, and a negative sitting elsewhere in the problem is irrelevant. In $-2x + 1 > 9$ , subtracting $1$ keeps $>$ ; only the divide-by- $-2$ step flips it.
Open versus closed circles. A strict symbol ( $<$ or $>$ ) gets an open circle, because the boundary is excluded. A non-strict symbol ( $\le$ or $\ge$ ) gets a closed circle, because the boundary is included. Drawing the wrong one misstates whether the boundary number is a solution.
Shading the wrong way. “Greater” relations ( $>$ , $\ge$ ) shade toward the larger numbers, to the right; “less” relations ( $<$ , $\le$ ) shade left. Solving and then graphing the opposite half is easy to do on autopilot, so let the symbol decide the direction.
Treating the answer as one number. A solution to an inequality is a whole range. $x > 3$ is not “the answer is $4$ ”; it is every number greater than $3$ , which is why we draw a ray, not a dot.
Misreading “at most” and “at least.” “At most $20$ ” means $x \le 20$ (no more than $20$ ), and “at least $48$ ” means $x \ge 48$ (no less than $48$ ). Both include the boundary, so both use a closed circle.

Key takeaways

An inequality compares two expressions with $<$ (less than), $>$ (greater than), $\le$ (at most), or $\ge$ (at least). The bar under $\le$ and $\ge$ includes the boundary; $<$ and $>$ exclude it.
A solution is every value that makes the inequality true, almost always a whole range of numbers, not a single value.
Graph the solution on a number line, with an open circle for $<$ or $>$ (boundary excluded). Use a closed circle for $\le$ or $\ge$ (boundary included), and shade right for “greater” relations, left for “less” relations.
Solve by isolating the variable with the same balance moves as equations: add or subtract the constant, then multiply or divide by the coefficient.
The one new rule: multiplying or dividing both sides by a negative number reverses the symbol. The reversal happens because taking opposites swaps the order of the numbers on the line. Adding or subtracting never flips it.
Check by testing a value from inside the claimed range, and that value should make the original true. Then test a value from outside the range, and that one should make the original false.

Inequalities

Learning goals

The four inequality symbols

A solution is a range, not a point

Graphing the solution on a number line

Solving inequalities with the balance moves

The one new rule: a negative flips the symbol

Two-step inequalities

Checking a solution by testing the range

Reading an inequality before you move

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice