➕ Algebra 1 · Inequalities

Solving Inequalities

Solve one- and multi-step inequalities, graph the answers on a number line, and learn why multiplying or dividing by a negative reverses the sign.

Grades 9, 10⏱ 16 min✔ AI-checkedCCSS A-REI.B.3CCSS A-CED.A.1CCSS 7.EE.B.4bCCSS 6.EE.B.8

About standards and checking

Written by AI, then independently re-solved and checked by separate AI reviewers and automated answer checks. Written by Claude (Sonnet 5.5) from the concept template. Automated gate: schema, notation and every math answer verified by evaluated expressions. Three blind re-solves by separate AI solvers with no access to the answer key: every keyed answer matched. Two prose audits by a separate, stronger AI reviewer (Claude Opus 5.5): every finding fixed, no wrong answer keys found. Final small edits re-checked by the automated gate (no human teacher review yet). Last checked 2026-11-07.

A-REI.B.3: Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters.
A-CED.A.1: Create equations and inequalities in one variable and use them to solve problems.
7.EE.B.4b: Solve word problems leading to inequalities of the form px + q > r or px + q < r, where p, q, and r are specific rational numbers. Graph the solution set of the inequality and interpret it in the context of the problem.
6.EE.B.8: Write an inequality of the form x > c or x < c to represent a constraint or condition in a real-world or mathematical problem. Recognize that inequalities of the form x > c or x < c have infinitely many solutions; represent solutions of such inequalities on number line diagrams.

By the end you can…

Read and write inequalities with <, >, ≤ and ≥, including phrases like “at least” and “at most”.
Graph the solutions of an inequality on a number line, using an open or a closed circle correctly.
Solve one-step and multi-step inequalities, and reverse the sign when you multiply or divide both sides by a negative number.
Check an answer with test values, and spot inequalities with no solution or with every real number as a solution.
Write and solve an inequality for a word problem, and say what the answer means in the story.

Lesson (school style)

An equation says two things are equal. An inequality says one thing is bigger or smaller than another, like x > 3. Because of that, an inequality usually has a whole range of solutions instead of just one. You solve it with the same kinds of moves you use on equations, plus one new rule: when you multiply or divide both sides by a negative number, the inequality sign reverses.

What an inequality says

An inequality compares two expressions with one of four symbols:

< means “is less than”.
> means “is greater than”.
≤ means “is less than or equal to”.
≥ means “is greater than or equal to”.

A solution is a value of the variable that makes the inequality true. For x > 3, the number 4 is a solution because 4 > 3 is true. So are 3.1 and 100. The number 3 is not a solution, because 3 > 3 is false. For x ≥ 3, the number 3 is a solution, because 3 ≥ 3 is true.

Unless a problem says otherwise, x can be any real number. Then x > 3 has infinitely many solutions: you cannot list them all, and there is no smallest one. The set of all solutions is the solution set. We describe it with an inequality like x > 3 and draw it on a number line.

3 < x and x > 3 say the same thing. When you can, put x on the left so the inequality reads like a sentence: “x is greater than 3.”

Graphing on a number line

To show a solution set, mark the boundary (the number where the inequality switches between true and false) and shade the numbers that work.

Use an open circle for < and >. The boundary is not a solution.
Use a closed (filled) circle for ≤ and ≥. The boundary is a solution.
Shade toward the solutions and end with an arrow, which means the shading goes on forever. Bigger numbers are on the right, so numbers greater than the boundary are shaded to the right, and numbers less than the boundary to the left.

For example, x > 3 is an open circle at 3 with shading to the right. And x ≤ −1 is a closed circle at −1 with shading to the left.

To choose the direction, read the inequality as a sentence that starts with x. “x is less than 3” means shade the numbers less than 3. If x is on the right, as in 3 < x, turn the sentence around first: x > 3.

Some books draw a parenthesis ( instead of an open circle and a square bracket [ instead of a closed circle. They mean the same thing.

Solving: the moves that keep the sign

To solve an inequality, get the variable alone by undoing operations, just as you did for two-step and multi-step equations. These moves keep the sign exactly as it is:

Add the same number to both sides.
Subtract the same number from both sides.
Multiply or divide both sides by the same positive number.

Each move gives a new inequality with exactly the same solutions. For example, subtract 4 from both sides of x + 4 > 9 to get x > 5. Multiply both sides of x/2 ≤ 3 by 2 to get x ≤ 6.

As with equations, never multiply both sides by 0. In this lesson you multiply or divide only by plain numbers, never by an expression that contains the variable.

The one new rule: negative multipliers

If you multiply or divide both sides of an inequality by a negative number, reverse the inequality sign: < becomes >, > becomes <, ≤ becomes ≥, and ≥ becomes ≤. This rule belongs to inequalities. With an equation, nothing ever reverses.

Why does it work? Start with a true statement: 2 < 5. Multiply both sides by −2. You get −4 on the left and −10 on the right. On the number line, −4 is to the right of −10, so −4 > −10. The statement −4 < −10 would be false. Multiplying by a negative number mirrors the number line across 0, so the left-to-right order of any two numbers swaps, and the sign has to swap with it. (Multiplying by −2 also stretches distances by a factor of 2, but it is the mirror that swaps the order.) Dividing by −2 is the same as multiplying by −1/2, which is also negative, so it also mirrors the number line, and the sign reverses.

Compare a plain addition or a positive multiplier. From 2 < 5 you get 2 + 1 < 5 + 1, which is 3 < 6, and 2(4) < 5(4), which is 8 < 20. The order is kept, so the sign stays.

Example: −3x < 12. Divide both sides by −3 and reverse the sign: x > −4. Test it. x = 0 gives 0 < 12 (true), and 0 > −4 is true too. x = −10 gives 30 < 12 (false), and −10 > −4 is false too. If you forgot to reverse the sign and wrote x < −4, then x = −10 would look like a solution, but it is not.

A negative number in the problem is not a reason to reverse the sign by itself. In x − 3 > −8 you add 3 to both sides and get x > −5, with no reversal. Only multiplying or dividing both sides by a negative number reverses the sign.

You can also avoid dividing by a negative. From −3x < 12, add 3x to both sides: 0 < 12 + 3x. Subtract 12: −12 < 3x. Divide by 3: −4 < x. That says the same thing as x > −4.

Multi-step, variables on both sides, fractions

Use the same plan as for multi-step equations:

Simplify each side (distribute, combine like terms).
Gather the variable terms on one side and the plain numbers on the other by adding or subtracting on both sides. The sign stays.
Divide (or multiply) both sides to free the variable. Reverse the sign only if the number you use is negative.
Check, and graph.

If the variable is on both sides, you can subtract a variable term from both sides, and the sign stays. If you keep the variable on the side where its coefficient comes out positive, you never have to divide by a negative number. Either side works, and both routes give the same solution set.

To clear fractions, multiply every term on both sides by the same positive common denominator. The sign stays. If you multiply by a negative number instead, reverse the sign.

Checking with test values

An inequality has many solutions, so you check with a few test numbers. Always put them into the original inequality.

Pick a number from the shaded part. It should make the original inequality true.
Pick a number from the unshaded part. It should make it false.
When your answer is a ray like x > −4, try the boundary too. It should make both sides equal. For ≤ or ≥ the boundary counts as a solution. For < or > it does not.

For −3x < 12 with the answer x > −4: x = 0 works, x = −10 fails, and x = −4 gives 12 < 12, which is false. So −4 is not a solution, which matches the open circle.

If a test number from the shaded part fails, the most common cause is a sign that should have been reversed. A few tests can catch a mistake, but they cannot prove the whole solution set is right. The algebra steps are what guarantee that.

No solution, or every real number

Sometimes the variable terms cancel and only a statement about plain numbers is left. Read that statement, as you did for equations.

If it is true, every real number is a solution. Example: x + 1 > x becomes 1 > 0 after you subtract x from both sides. That is true, so the original inequality is true for every real number x.
If it is false, there is no solution. Example: x + 1 < x becomes 1 < 0. That is false, so no number works.

Watch the difference between > and ≥. From x + 1 ≥ x + 1 you get 1 ≥ 1, which is true, so every real number works. From x + 1 > x + 1 you get 1 > 1, which is false, so there is no solution.

On a number line, “every real number” is the whole line shaded, and “no solution” is nothing shaded.

For equations, you called this case “infinitely many solutions”, because a linear equation with infinitely many solutions is true for every number. For inequalities, that phrase is not specific enough: x > 3 also has infinitely many solutions, but not every real number is one of them. So for inequalities, say “every real number is a solution” when the leftover is true, and “no solution” when it is false.

From words to inequalities

First say what the variable means: “Let p be the number of people.” Then translate the key phrase:

“more than”, “greater than”, “exceeds”: >
“fewer than”, “less than”, “below”: <
“at least”, “no less than”, “a minimum of”: ≥
“at most”, “no more than”, “a maximum of”: ≤

“At least 18” includes 18 itself, so the circle is closed. “More than 18” does not include 18, so the circle is open.

Watch the word order. “x is fewer than 5” is the inequality x < 5, but “5 fewer than x” is the expression x − 5.

A story can limit the variable. If m counts months, only whole numbers that are 0 or more make sense. Solve the inequality first, then interpret the answer by looking at the solved inequality, not at the words in the story. If it says m ≤ 5.4, the greatest whole number that works is 5 (round down). If it says m ≥ 5.4, the least whole number that works is 6 (round up). With a strict sign, check the boundary: for whole numbers, m < 5 allows at most 4.

The story’s words can mislead. “You have 30 dollars, a pass costs 4 dollars, and you want at least 5 dollars left” gives, with p passes, 30 − 4p ≥ 5, which solves to p ≤ 25/4, or p ≤ 6.25. The story says “at least”, but the most passes is 6 (round down). Then test the whole number you chose in the original story: 6 passes leave 6 dollars, which is at least 5, and 7 passes leave only 2. Say the answer in words, with units.

Key words

Inequality: A statement that compares two expressions with <, >, ≤ or ≥, such as x > 3.
Solution of an inequality: A value of the variable that makes the inequality true. For x > 3, the number 4 is a solution and the number 3 is not.
Solution set: All the solutions together, such as every real number greater than 3. It is usually shown on a number line.
Boundary: The number where an inequality switches between true and false, like 3 in x > 3. It is where the circle goes on the number line.
Open circle and closed circle: An open circle marks a boundary that is not a solution (< or >). A closed (filled) circle marks a boundary that is a solution (≤ or ≥).
At least, at most: “At least” means ≥ (the number itself counts). “At most” means ≤ (the number itself counts).
Every real number: The solution when the variable cancels and the leftover statement is true, like 1 > 0. Every point on the number line is shaded.
No solution: The result when the variable cancels and the leftover statement is false, like 1 < 0. No number makes the inequality true.

Worked examples

Example 1. Write an inequality for each sentence, and say whether the boundary number is a solution. (a) A water slide requires riders to be at least 54 inches tall. (b) An elevator may carry at most 8 people. (c) You need more than 70 points to pass. (d) The class has fewer than 30 students.

(a) Let h be a rider’s height in inches. “At least 54” means 54 or more: h ≥ 54. A rider exactly 54 inches tall may ride, so the boundary is a solution (closed circle).
(b) Let p be the number of people in the elevator. “At most 8” means 8 or fewer: p ≤ 8. A full elevator is allowed, so the boundary is a solution (closed circle).
(c) Let s be a score in points. “More than 70” means above 70: s > 70. Exactly 70 points is not more than 70, so the boundary is not a solution (open circle).
(d) Let c be the number of students. “Fewer than 30” means below 30: c < 30. Exactly 30 is not fewer than 30, so the boundary is not a solution (open circle).
Check: in (a), 54 ≥ 54 is true. In (c), 70 > 70 is false.

Answer: Answer: (a) h ≥ 54. (b) p ≤ 8. (c) s > 70. (d) c < 30.

Example 2. Graph each inequality on a number line: (a) x < 4, (b) x ≥ −6, (c) 10 ≥ x.

(a) The sign is <, so draw an open circle at 4. Numbers less than 4 are to the left, so shade left and end with an arrow.
(b) The sign is ≥, so draw a closed circle at −6. Numbers greater than −6 are to the right, so shade right.
(c) Turn it around so x comes first: 10 ≥ x says the same thing as x ≤ 10. Draw a closed circle at 10 and shade left.
Check (c) with test numbers. x = 0 is in the shaded part, and 10 ≥ 0 is true. x = 12 is not shaded, and 10 ≥ 12 is false.

Answer: Answer: (a) Open circle at 4, shade left. (b) Closed circle at −6, shade right. (c) Closed circle at 10, shade left.

Example 3. Solve 2x + 11 ≤ 1 and graph the answer.

Subtract 11 from both sides. The sign stays: 2x ≤ −10.
Divide both sides by 2. The number 2 is positive, so the sign stays: x ≤ −5.
Graph: a closed circle at −5, with shading to the left.
Check. Boundary: 2(−5) + 11 = 1, and 1 ≤ 1 is true, so −5 is included. Shaded part, x = −8: 2(−8) + 11 = −5, and −5 ≤ 1 is true. Unshaded part, x = 0: 11 ≤ 1 is false.

Answer: Answer: x ≤ −5

Example 4. Solve 5 − 4x ≥ 17.

Subtract 5 from both sides: −4x ≥ 12.
Divide both sides by −4. The divisor is negative, so reverse the sign: x ≤ −3.
Check. Boundary: 5 − 4(−3) = 17, and 17 ≥ 17 is true. Shaded part, x = −5: 5 − 4(−5) = 25, and 25 ≥ 17 is true. Unshaded part, x = 0: 5 ≥ 17 is false.
If you had forgotten to reverse the sign and written x ≥ −3, the test number x = 0 would have looked like a solution, but it fails.
Another route avoids the negative divisor. Add 4x to both sides: 5 ≥ 4x + 17. Subtract 17: −12 ≥ 4x. Divide by 4: −3 ≥ x, which is the same as x ≤ −3.

Answer: Answer: x ≤ −3

Example 5. Solve −(x/3) + 2 < 6.

The term −(x/3) means (−1/3)x. Subtract 2 from both sides: −(x/3) < 4.
Multiply both sides by −3. The multiplier is negative, so reverse the sign: x > −12.
Check. Boundary: −(−12/3) + 2 = 4 + 2 = 6, and 6 < 6 is false, so −12 is not included (open circle). Shaded part, x = 0: 2 < 6 is true. Unshaded part, x = −15: −(−15/3) + 2 = 7, and 7 < 6 is false.
If the fraction had been positive, as in x/3 + 2 < 6, you would multiply by 3 and keep the sign.

Answer: Answer: x > −12

Example 6. Solve 7x + 2 > 11x − 30.

The variable is on both sides. Keep the x-terms on the right, where the coefficient will be positive. Subtract 7x from both sides: 2 > 4x − 30.
Add 30 to both sides: 32 > 4x.
Divide both sides by 4. The sign stays: 8 > x, which is the same as x < 8.
Other route: subtract 11x from both sides instead. That gives −4x + 2 > −30. Subtract 2: −4x > −32. Divide by −4 and reverse the sign: x < 8. Same answer.
Check. Boundary: at x = 8, 7(8) + 2 = 58 and 11(8) − 30 = 58. The sides are equal, so 8 is not included. At x = 7: 51 > 47 is true. At x = 9: 65 > 69 is false.

Answer: Answer: x < 8

Example 7. Solve −2(x − 3) + 5 < 3x − 24.

Distribute −2 with its sign: −2(x − 3) = −2x + 6. The left side is −2x + 6 + 5 = −2x + 11.
The inequality is −2x + 11 < 3x − 24. Add 2x to both sides to keep the variable on the right: 11 < 5x − 24.
Add 24 to both sides: 35 < 5x. Divide by 5: 7 < x, which is the same as x > 7.
Check. Boundary: at x = 7, the left side is −2(4) + 5 = −3 and the right side is 3(7) − 24 = −3. They are equal, so 7 is not included. At x = 8: −5 < 0 is true. At x = 0: 11 < −24 is false.

Answer: Answer: x > 7

Example 8. Solve (a) 4(x + 1) > 4x + 9 and (b) 3(x − 2) ≤ 3x + 1.

(a) Distribute: 4x + 4 > 4x + 9. Subtract 4x from both sides: 4 > 9.
The variable vanished and 4 > 9 is false, so no real number is a solution. Test: x = 0 gives 4 > 9, false, and x = 100 gives 404 > 409, false.
(b) Distribute: 3x − 6 ≤ 3x + 1. Subtract 3x from both sides: −6 ≤ 1.
The variable vanished and −6 ≤ 1 is true, so every real number is a solution. Test: x = 0 gives −6 ≤ 1, true, and x = −10 gives −36 ≤ −29, true.

Answer: Answer: (a) No solution. (b) Every real number is a solution.

Example 9. A video-editing app charges a one-time fee of 15 dollars plus 12 dollars per month. Marcus can spend at most 80 dollars in total. How many whole months can he subscribe?

Let m be the number of months. The total cost is 12m + 15 dollars. “At most 80” means the cost is ≤ 80, so 12m + 15 ≤ 80.
Subtract 15 from both sides: 12m ≤ 65. Divide both sides by 12 (positive, so the sign stays): m ≤ 65/12. That is a little more than 5 (about 5.4).
Months are whole numbers, and cannot be negative, so the possible values are 0, 1, 2, 3, 4 and 5. The solved inequality m ≤ 65/12 says m can be at most 65/12, so the greatest whole number that works is 5 (round down): the most is 5 months.
Check. 5 months cost 12(5) + 15 = 75 dollars, which is within 80. 6 months cost 12(6) + 15 = 87 dollars, which is more than 80.

Answer: Answer: 12m + 15 ≤ 80 gives m ≤ 65/12, so Marcus can subscribe for at most 5 whole months.

In plain English

An equation says two things are exactly the same. An inequality says one is bigger or smaller. So instead of one answer, you usually get a whole crowd of answers.

A sign at a ride says, “You must be at least 120 centimeters tall.” Lots of heights work: 120, 121, 135, even 150.5. You cannot list them all, so you write a rule: height ≥ 120. On a number line, you color in every height that works.

You solve inequalities with the same moves as equations. Do the same thing to both sides. There is one twist. If you multiply or divide both sides by a negative number, the sign turns around. Here is why. We know 2 < 5. Multiply both sides by −1 and you get −2 and −5. Now −2 is bigger than −5, so you write −2 > −5.

Last, the circles. An open circle says the boundary number does not count. A filled circle says it counts. Then color in the numbers that work.

See it

An open circle leaves the boundary out (< and >). A closed circle keeps it (≤ and ≥).

Multiplying both sides of 2 < 5 by −2 swaps the order of the numbers, so the sign swaps too: −4 > −10.

The same idea, the fun way

Pick the one that clicks. Each one explains the same concept from a different angle.

Relationships

A promise to a friend

Your friend says, “I’ll be there in no more than 15 minutes.” If your friend arrives in exactly 15 minutes, the promise is kept. Now suppose the promise had been “in fewer than 15 minutes.” Arriving in exactly 15 minutes would break it. Let t be the minutes until your friend arrives. The first promise is t ≤ 15 and the second is t < 15.

How it connects: The two promises differ only at the boundary: t ≤ 15 has a closed circle at 15, and t < 15 has an open circle. The limit: time cannot be negative, so only t ≥ 0 makes sense, and a friend who arrives at 15 minutes and 1 second has broken both promises.

Sports

The qualifying time

In a made-up track meet, a runner reaches the final if the 400-meter time is at most 52 seconds. Let t be the time in seconds. The rule is t ≤ 52. A runner at exactly 52 seconds reaches the final. A runner at 52.4 seconds does not. A runner at 49 seconds does, because 49 is less than 52.

How it connects: “At most” means ≤: a closed circle at 52 and shading to the left, toward smaller numbers. It can feel backwards that smaller times are better, but the inequality only compares numbers: 49 is less than 52, so 49 is in the shaded part. The limit: a time cannot be 0 or negative, so only part of the shaded ray makes sense in this story.

Gaming

Coins for the shield

In a made-up game, a shield costs 100 coins. You can buy it when your number of coins c satisfies c ≥ 100. With exactly 100 coins you can buy it. With 99 you cannot. With 250 you can, and you have coins left over.

How it connects: c ≥ 100 is a closed circle at 100 with shading to the right: 100 counts, and so does everything above it. The limit: coins are whole numbers, so the real picture is a row of separate dots (100, 101, 102, and so on), not a continuous shaded line. On a number line for real numbers, we shade the whole ray.

Nature

The lake mirror

A calm lake reflects a mountain upside down. Take the waterline as 0 and measure heights in units above it. A ridge 2 units high and a peak 5 units high satisfy 2 < 5. In the reflection, their tips are 2 and 5 units below the waterline, at −2 and −5. Now −2 > −5: the taller peak has the lower tip.

How it connects: A reflection across the waterline is multiplication by −1, and it swaps the order of any two heights. That is why the inequality sign reverses when you multiply or divide both sides by a negative number. The limit: a real reflection only multiplies by −1. Multiplying by −3 also stretches distances, but the order swaps just the same. Adding or subtracting only slides everything up or down, so the order and the sign stay.

Idiom

The sky’s the limit

The idiom “the sky’s the limit” means there is no limit to what is possible.

How it connects: The shaded part of x > 3 has no end: 4, 400 and 4 million all work, so the shading runs off the number line with an arrow. The limit of the idiom: it fits one direction only. x > 3 is unlimited upward, but it has a boundary at 3 on the other side, and 3 itself is not a solution.

Proverb

Measure twice, cut once

The proverb “Measure twice, cut once” says to check carefully before you do something you cannot undo, like cutting a board.

How it connects: Check your solution before you trust it: put one number from the shaded part, one from the unshaded part, and the boundary into the original inequality. The limit: a board has one cut, but an inequality has infinitely many solutions. A few tests can catch a mistake like a forgotten sign reversal, but they cannot prove the whole answer is right. Correct steps are what guarantee that.

Everyday joke

The dizzy sign

Why did the less-than sign get dizzy? Someone divided both sides by −2 and spun it around, and it landed facing the other way: >.

How it connects: Multiplying or dividing both sides by a negative number reverses the sign: < becomes >. The spinning happens only for that move. Adding, subtracting, or multiplying or dividing by a positive number leaves the sign alone.

Memory help

Acronym

FLIP

When you multiply or divide both sides by a number, ask: Find the number you multiply or divide by. Look at its sign. If it is negative, Put in the reversed sign (< becomes >, ≤ becomes ≥). If it is positive, keep the sign. Adding and subtracting never reverse anything.

Rhyme

Turn it around

“Negative times, negative divide: turn the sign around with pride.” Only multiplying or dividing both sides by a negative number does that. For adding, subtracting, or multiplying or dividing by a positive number, leave the sign alone.

Picture it

The hungry alligator

Picture an alligator whose open mouth faces the bigger number. In 5 > 2 and in 2 < 5, the mouth opens toward the 5. The line under ≤ and ≥ is half of an equals sign, which means “or equal”. Limit: the alligator tells you which number is bigger. It does not tell you which way to shade unless x is on the left.

Mnemonic

A line under it means a filled circle

The symbols ≤ and ≥ have a line under them, like half of an equals sign. “Or equal” means the boundary counts, so the circle is filled. The symbols < and > have no line, so the boundary does not count and the circle is open (empty).

Trap alert

A negative in the problem is not a flip signal

Only multiplying or dividing both sides by a negative number reverses the sign. In x − 3 > −8 there are two negatives in sight, but you add 3 to both sides and get x > −5 with the sign unchanged.

Story

In, out and edge

To check an answer, send three scouts into the original inequality: one from inside the shaded part (it should be true), one from outside (it should be false), and one from the edge, the boundary (both sides should be equal, and it counts only if the sign has a line under it).

Study notes

Key ideas

An inequality compares two expressions with <, >, ≤ or ≥. Its solution set is usually infinite, so we show it on a number line.
Open circle: the boundary is not included (<, >). Closed circle: the boundary is included (≤, ≥).
Adding or subtracting the same number on both sides, or multiplying or dividing both sides by a positive number, keeps the sign. Multiplying or dividing both sides by a negative number reverses it.
If the variable vanishes, a true leftover (like 1 > 0) means every real number is a solution, and a false leftover (like 1 < 0) means no solution.
“At least” is ≥, “at most” is ≤, “more than” is > and “fewer than” is <. Define the variable first, and say what the answer means in the story.

Steps

Define the variable (in word problems) and write the inequality.
Simplify each side: distribute and combine like terms.
Gather the variable terms on one side and the plain numbers on the other. The sign stays.
Multiply or divide both sides to free the variable. If the number is negative, reverse the sign.
Graph it: an open or closed circle at the boundary, and shade the true side.
Check with a number from the shaded part, a number from outside, and the boundary.

Formulas

Add, or multiply by a positive: sign stays: If a < b, then a + c < b + c for any number c. If also c > 0, then ac < bc.
Negative multiplier: sign reverses: If a < b and c < 0, then ac > bc and a/c > b/c
Words to symbols: at least: ≥. at most: ≤. more than: >. fewer than: <

Watch out

Reverse the sign only when you multiply or divide both sides by a negative number. A negative number somewhere else in the problem does not reverse it.
Adding or subtracting never reverses the sign: x − 3 > −8 becomes x > −5.
< and > get an open circle. ≤ and ≥ get a closed circle.
Put x on the left before you shade: 3 < x means x > 3, so shade to the right.
When the variable vanishes, read the leftover statement. 1 > 1 is false (no solution), but 1 ≥ 1 is true (every real number).
In a word problem, interpret the answer. When only whole numbers make sense, read the solved inequality: for m ≤ 5.4 pick the greatest whole number that works (5), and for m ≥ 5.4 pick the least (6). Then test it in the original story.

Why it works

Why does the sign reverse? If a < b, then b − a is positive. A negative number c times a positive number is negative, so c(b − a) < 0. That means cb − ca < 0, so cb < ca, which is the same as ca > cb.
Why can you add the same number to both sides? Adding slides both numbers the same distance along the number line, so their order does not change.
Why does a true leftover mean every real number works? When the variable terms cancel, the comparison no longer depends on x. It is the same true (or false) statement for every x.

Common mistakes

✗ Mistake−4x < 20 becomes x < −5.

✓ InsteadDivide both sides by −4 and reverse the sign: x > −5. Test x = 0: 0 < 20 is true, and 0 > −5 is true, but 0 < −5 is false.

✗ MistakeAny negative number in the problem means I flip the sign.

✓ InsteadReverse the sign only when you multiply or divide both sides by a negative number. In x − 3 > −8 you add 3 to both sides, and the sign stays: x > −5.

✗ Mistakex < 5 and x ≤ 5 look the same on a number line.

✓ InsteadBoth shade to the left of 5, but x < 5 has an open circle (5 is not a solution) and x ≤ 5 has a closed circle (5 is a solution).

✗ MistakeThe symbol in 3 < x points left, so I shade to the left.

✓ Instead3 < x means x > 3. Read it with x first, then shade the numbers greater than 3: to the right.

✗ MistakeIf the variable terms cancel, the answer is x = 0.

✓ InsteadRead what is left. A true statement (like 3 < 8) means every real number is a solution. A false statement (like 4 < 4) means there is no solution.

✗ MistakeIf the number of months m satisfies m ≤ 65/12, then the answer to the word problem is 65/12 months.

✓ InsteadThe solved inequality m ≤ 65/12 allows whole numbers up to 65/12, which is a little more than 5. The greatest whole number that works is 5, so the answer is 5 months.

Teach it back

Explain to a friend (or a pet!) how to solve −3x + 7 ≥ 25, how to graph the answer, and why the sign reverses.

Explaining it out loud shows what you really know. Did you…

I subtracted 7 from both sides to get −3x ≥ 18.
I divided both sides by −3 and reversed the sign to get x ≤ −6.
I drew a closed circle at −6 with shading to the left, and tested x = −10 (true) and x = 0 (false).
I explained why the sign reverses: multiplying or dividing by a negative number swaps the order of numbers, as in 1 < 4 but −3 > −12 after multiplying by −3.

Your 3-week memory plan

TodayRead the lesson, then try the flashcards.
TomorrowRecall it from memory first, then check.
In 3 daysTake the true/false and practice questions.
In 1 weekMix it with older topics and teach it back.
In 3 weeksA last quick review before it fades.

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