Equations and Inequalities

Equations and inequalities are the backbone of algebra. An equation pins down an exact value. An inequality describes a range. Both use the same core solving steps — with one critical rule that applies only to inequalities. This guide covers everything from linear equations and inequalities through compound inequalities and absolute value, with worked examples, practice problems, and a breakdown of the mistakes that cost students the most points.

The Core Rules

Equations give exact answers: Solve for the specific value (or values) that make the equation true. The answer is one or more numbers.

Inequalities give solution sets: Solve for the range of values that satisfy the inequality. Write the answer in inequality or interval notation.

The sign-flip rule: When you multiply or divide both sides of an inequality by a negative number, you must reverse the inequality symbol. Adding or subtracting never flips it.

Absolute value splits into cases: Isolate the absolute value first, then write two equations or a compound inequality depending on the symbol used.



1) Equations vs. Inequalities — Key Differences

An equation uses an equals sign (=) to state that two expressions have exactly the same value. Solving an equation means finding the specific value or values that make that statement true.

An inequality uses a comparison symbol to state that one expression is greater than, less than, or not equal to another. Solving an inequality produces a range of values — a solution set rather than one exact answer.

Reference card showing all five inequality symbols — less than, greater than, less than or equal to, greater than or equal to, and not equal to — with their names, plain-English meanings, and open or closed dot notation for number lines.
Strict inequalities use open dots on a number line because the boundary value is not part of the solution. Non-strict inequalities use closed dots because the boundary value is included.

Equation: exact answer

2x + 3 = 11 → x = 4. One specific value satisfies the equation.

Inequality: solution set

2x + 3 < 11 → x < 4. Every number less than 4 satisfies it.

Interval notation quick reference

x < 4 → (−∞, 4) parenthesis: 4 not included
x ≤ 4 → (−∞, 4] bracket: 4 is included
x > −1 → (−1, ∞) always parenthesis with ∞
x ≥ −1 → [−1, ∞)

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2) Solving Linear Equations — Quick Review

Solving a linear equation means isolating the variable using inverse operations. The same process applies directly to inequalities, so this review is worth doing before moving on. For a full treatment, see our solving linear equations guide.

The process step by step

Simplify both sides first — distribute, combine like terms, clear fractions if present. Then move all variable terms to one side and all constants to the other. Finally, divide (or multiply) to isolate the variable.

3x − 7 = 14
3x = 14 + 7 = 21
x = 21 ÷ 3 = 7
2(x + 4) − 3 = 11
2x + 8 − 3 = 11 ← distribute
2x + 5 = 11
2x = 6
x = 3

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3) Solving Linear Inequalities

Solving a linear inequality uses the exact same steps as solving an equation — with one critical exception that trips up students more than anything else in this topic.

The Sign-Flip Rule: When you multiply or divide both sides by a negative number, you must reverse the inequality symbol.

Adding or subtracting — even negative numbers — never flips the sign. Only multiplication and division by a negative trigger the flip.

Why does the sign flip?

Multiplying by −1 reverses the order of all numbers on the number line. Since 3 > 2, multiplying both by −1 gives −3 < −2. The relationship flips. The same thing happens to your inequality whenever you divide by a negative coefficient.

Example 1 — No sign flip needed

2x + 5 < 17
2x < 12
x < 6 ← divided by +2, no flip

Answer: x < 6, or (−∞, 6)

Example 2 — Sign flip required

−3x + 6 > 12
−3x > 6
x < −2 ← divided by −3, FLIP the sign

Answer: x < −2, or (−∞, −2)

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4) Compound Inequalities

A compound inequality joins two inequalities with the word and or the word or. The choice between those two words completely changes the solution set — getting them confused is one of the most common errors in this topic.

AND — Intersection

Both conditions must be true at once. The solution is the overlap — a single bounded region. Written as a three-part statement: −2 < x < 5.

OR — Union

At least one condition must be true. The solution is the combined set — two separate outer regions. Written as: x < −2 or x > 5.

Two number line diagrams. The top shows an AND compound inequality with the shaded region between two boundary values representing the intersection. The bottom shows an OR compound inequality with shaded regions extending outward from two boundary values representing the union.
AND shades the middle region between the two boundary values. OR shades the two outer regions beyond the boundaries.

Solving AND compound inequalities

Write the AND inequality as a single three-part statement and perform the same operation on all three parts simultaneously.

Solve: −1 ≤ 2x + 3 < 9

−4 ≤ 2x < 6 ← subtract 3 from all parts
−2 ≤ x < 3 ← divide all parts by 2

Answer: [−2, 3)

Solving OR compound inequalities

Solve each inequality separately, then join the two solution sets with “or.”

Solve: 3x − 1 < −4 or 2x + 5 > 11

Left: 3x < −3 → x < −1
Right: 2x > 6 → x > 3

Answer: (−∞, −1) ∪ (3, ∞)

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5) Absolute Value Equations

The absolute value of a number is its distance from zero. Distance is always non-negative, so |x| = 5 is satisfied by x = 5 and x = −5. That two-case structure is the foundation of every absolute value equation.

The absolute value rule for equations

If |A| = b and b > 0: write A = b and A = −b
If |A| = 0: one solution only, A = 0
If |A| = negative number: no solution

Always isolate the absolute value expression before splitting into two cases.

Example 1 — Standard case

|2x − 3| = 7

Case 1: 2x − 3 = 7 → x = 5
Case 2: 2x − 3 = −7 → x = −2

Answer: x = 5 or x = −2

Example 2 — Isolate first

3|x + 1| − 6 = 9
3|x + 1| = 15 ← add 6
|x + 1| = 5 ← divide by 3

Case 1: x + 1 = 5 → x = 4
Case 2: x + 1 = −5 → x = −6

Answer: x = 4 or x = −6

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6) Absolute Value Inequalities

Absolute value inequalities combine the distance-from-zero concept with compound inequalities. The direction of the symbol determines which compound form you write — this is one of the most testable patterns in algebra.

“Less than” → AND

|A| < b becomes −b < A < b

Lock it in the middle. One bounded region.

“Greater than” → OR

|A| > b becomes A < −b or A > b

Go to the outside. Two separate regions.

Decision flowchart for absolute value problems. Isolate the absolute value. Check if the right side is negative — if yes, no solution. Then branch by symbol: equals splits into two equations, less than becomes an AND compound inequality, greater than becomes an OR compound inequality.
Always isolate the absolute value before branching. Checking for a negative right-hand side prevents one of the most common errors on exams.

Example 1 — Less than

|x − 4| ≤ 3
−3 ≤ x − 4 ≤ 3 ← AND form
1 ≤ x ≤ 7 ← add 4 to all parts

Answer: [1, 7]

Example 2 — Greater than

|2x + 1| > 5
2x + 1 < −5 or 2x + 1 > 5 ← OR form
x < −3 or x > 2

Answer: (−∞, −3) ∪ (2, ∞)

Special cases to know:

|A| < negative number → No solution. Distance cannot be less than a negative.

|A| > negative number → All real numbers. Distance is always ≥ 0, so always greater than any negative.

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7) Graphing Inequalities on a Number Line

Every inequality solution can be graphed on a number line. Two conventions must be applied correctly every time: the type of dot at each endpoint, and which direction to shade.

Closed Dot • — used with ≤ or ≥

The endpoint is included in the solution set. Matches bracket notation [ ] in interval form.

Open Dot ° — used with < or >

The endpoint is not included in the solution set. Matches parenthesis notation ( ) in interval form.

x < 3 → open dot at 3, shade LEFT
x ≥ −1 → closed dot at −1, shade RIGHT
−2 < x ≤ 5 → open dot at −2, closed dot at 5, shade BETWEEN
x < −1 or x > 4 → open dots at −1 and 4, shade both OUTER regions

When in doubt, test a point. Pick any number from the region you shaded and plug it into the original inequality. If it makes the inequality true, you shaded the right direction.

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8) Common Mistakes

These are the errors that show up most consistently on algebra exams and homework platforms. Most come down to one skipped step or one forgotten rule.

Forgetting to flip the inequality sign

The most common error in this entire topic. After dividing or multiplying by a negative, circle the symbol and confirm it reversed. It is easy to do the arithmetic correctly and still write the wrong answer because you forgot the flip.

Confusing AND vs. OR in compound inequalities

AND narrows the solution to an intersection; OR widens it to a union. Pay close attention to the connecting word. A three-part statement like −2 < x < 5 is always AND.

Not isolating the absolute value before splitting

In 2|x − 1| + 4 = 10, you must get |x − 1| = 3 before writing two cases. Splitting before isolating produces wrong answers every time.

Wrong dot type on number line graphs

Strict inequalities (< and >) get open dots. Non-strict (≤ and ≥) get closed dots. Mixing these up costs easy points on exams and is one of the first things instructors check.

Assuming absolute value always gives two solutions

|A| = 0 has exactly one solution. |A| = negative number has no solution at all. Always check the right-hand side before writing any cases.

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9) Practice Problems with Solutions

Work through each problem before opening the solution. They cover every major concept from this guide.

Problem 1 — Linear Inequality

Solve: 5 − 4x ≥ 13. Write in interval notation.
Show Solution
5 − 4x ≥ 13
−4x ≥ 8
x ≤ −2 ← divided by −4, sign flips

Answer: (−∞, −2]

Problem 2 — Compound Inequality (AND)

Solve: −3 < 2x − 1 ≤ 7. Write in interval notation.
Show Solution
−3 < 2x − 1 ≤ 7
−2 < 2x ≤ 8 ← add 1 to all parts
−1 < x ≤ 4 ← divide all parts by 2

Answer: (−1, 4]

Problem 3 — Compound Inequality (OR)

Solve: x + 4 < 1 or 3x − 2 > 10
Show Solution
Left: x + 4 < 1 → x < −3
Right: 3x − 2 > 10 → x > 4

Answer: (−∞, −3) ∪ (4, ∞)

Problem 4 — Absolute Value Equation

Solve: |3x + 6| = 15
Show Solution
Case 1: 3x + 6 = 15 → x = 3
Case 2: 3x + 6 = −15 → x = −7

Answer: x = 3 or x = −7

Problem 5 — Absolute Value Inequality (Less Than)

Solve: |5x − 2| < 8. Write in interval notation.
Show Solution
Less than → AND form:
−8 < 5x − 2 < 8
−6 < 5x < 10
−6/5 < x < 2

Answer: (−1.2, 2)

Problem 6 — Absolute Value Inequality (Greater Than)

Solve: |4 − x| ≥ 6. Write in interval notation.
Show Solution
Greater than or equal → OR form:
4 − x ≤ −6 or 4 − x ≥ 6
x ≥ 10 or x ≤ −2 ← divide by −1, signs flip

Answer: (−∞, −2] ∪ [10, ∞)

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10) How FMMC Can Help

Equations and inequalities appear in every algebra module and on nearly every proctored exam. If you are behind on assignments or need help across a full algebra course, FMMC’s experts handle every module with an A/B guarantee. This topic connects directly to our solving linear equations guide and our factoring guide.

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FAQ: Equations and Inequalities

When do you flip the inequality sign?

You flip the inequality sign whenever you multiply or divide both sides by a negative number. Adding or subtracting — even negative values — never flips the sign. For example, dividing both sides of x > 4 by −1 gives x < −4. The relationship reverses because multiplying by a negative number reverses the order of all real numbers.

What is the difference between interval notation and inequality notation?

They describe the same solution set in different formats. Inequality notation uses symbols: x < 4 or −1 ≤ x < 7. Interval notation writes the same answers as (−∞, 4) or [−1, 7). Parentheses exclude the endpoint; brackets include it. Infinity always gets a parenthesis.

Can an absolute value equation have no solution?

Yes. If the absolute value is set equal to a negative number — for example |x + 3| = −5 — there is no solution. Absolute value represents distance from zero and is always non-negative. Always check the right-hand side before writing two cases. If it is negative, write “no solution” and stop.

How do I know if a compound inequality is AND or OR?

Look at the connecting word. If the problem says “and” or is written as a three-part expression like −3 < x < 7, it is an AND inequality — the solution is the intersection where both conditions hold at once. If it says “or,” the solution is the union of values satisfying at least one condition.

Why does |x| < a give an AND inequality but |x| > a gives an OR?

|x| < a means x is within a units of zero, which confines x to a bounded region: −a < x < a. Both conditions hold at once — that is AND. |x| > a means x is more than a units from zero, which describes two outer regions: x < −a or x > a. Either region works — that is OR. The same logic applies when the absolute value contains an expression rather than just x.

Can FMMC help with equations and inequalities homework?

Yes. FMMC handles algebra homework across MyMathLab, ALEKS, WebAssign, and Hawkes Learning. See our algebra homework help page or get a free quote.

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