Inequalities

An inequality compares: bigger or smaller, not equal

How to solve them

1. 1

   Read the four signs

   An inequality uses one of four signs. The open side of the sign faces the bigger number, and the point faces the smaller one. A line under the sign means “or equal to”: the number itself is allowed.

   is less than<3 < 8

   is greater than>9 > 5

   is less than or equal to≤x ≤ 6: 6 is allowed

   is greater than or equal to≥x ≥ 6: 6 is allowed

   With negative numbers, think of the number line. A number to the right is always bigger, even when it looks smaller without its minus sign.

   Two negatives

   −2 > −5

   −2 is to the right of −5 on the number line, so it is bigger: true.

   Equal is allowed

   7 ≤ 7

   7 is equal to 7, and ≤ allows equal: true.

   Watch the sign

   −8 > −1

   Without the minus signs 8 is bigger, but −8 is farther left, below −1: false.

   Next step: Number line ↓
2. 2

   Graph it: open dot or closed dot

   The answer to an inequality is not one number but a whole stretch of numbers. On a number line you draw it with a dot and shading.

   The dot sits on the number. It is open (empty, ○) when the number is not included: < and >. It is closed (filled in, ●) when the number is included: ≤ and ≥. Then shade toward the numbers that work: to the right for greater, to the left for less.

   How to draw it

   Inequality	Dot on the number	Shade
   x > 3	open ○ at 3	to the right →
   x ≥ 3	closed ● at 3	to the right →
   x < −1	open ○ at −1	to the left ←
   x ≤ −1	closed ● at −1	to the left ←

   Shortcut

   Put x first

   If the number comes first, as in 4 > x, turn the whole thing around so the sign still points at x: x < 4. With x on the left, the sign points the way you shade: < points left, > points right.

   Next step: Solve it ↓
3. 3

   Solve it like an equation

   You solve an inequality with the same moves as an equation. Undo the adding or subtracting first, then the multiplying or dividing, and always do the same thing to both sides. The sign stays where it is.

   Then check with one number from your answer. If it makes the first inequality true, and a number outside your answer makes it false, you are done.

   1. 3x − 7 < 14
   2. 3x − 7 + 7 < 14 + 7
   3. 3x < 21
   4. 3x ÷ 3 < 21 ÷ 3
   5. x < 7

   Example

   Check: try 6, which is less than 7

   1. 3 × 6 − 7 = 18 − 7 = 11.
   2. 11 < 14 is true, so the numbers below 7 work.
   3. Now try 8, which is not in the answer: 3 × 8 − 7 = 17, and 17 < 14 is false. x < 7 is right.

   Answer Checked

   Next step: Flip the sign ↓
4. 4

   Multiply or divide by a negative: flip the sign

   There is one new rule. When you multiply or divide both sides by a negative number, the sign turns around: < becomes >, and ≤ becomes ≥.

   Why? Multiplying by −1 moves every number to the other side of zero, like a mirror, so the order reverses. 2 is less than 5, but −2 is greater than −5.

   Adding or subtracting never flips the sign, even with negative numbers. And a negative answer is not a reason to flip: only multiplying or dividing by a negative number is.

   1. 2 < 5
   2. 2 × (−1) ? 5 × (−1)
   3. −2 ? −5
   4. −2 > −5

   1. −4x ≥ 28
   2. −4x ÷ (−4) ≤ 28 ÷ (−4)
   3. x ≤ −7

   Next step: Compound ↓
5. 5

   Compound inequalities: two conditions at once

   −3 < x ≤ 4 says two things at once: x is greater than −3 and x is at most 4. On the number line it is the piece in between: an open dot at −3, a closed dot at 4, and the shading between them.

   To solve one, do the same move to all three parts. The middle part has the x; the two outside parts are plain numbers.

   With the word or, as in x < −2 or x > 3, the answer is two pieces that go away from each other: one arrow to the left and one to the right.

   1. 5 ≤ 2x + 1 < 13
   2. 5 − 1 ≤ 2x + 1 − 1 < 13 − 1
   3. 4 ≤ 2x < 12
   4. 4 ÷ 2 ≤ 2x ÷ 2 < 12 ÷ 2
   5. 2 ≤ x < 6

   Next step: Word problems ↓
6. 6

   Word problems: turn the words into a sign

   Most inequalities on the test come as a story. Find the phrase that sets the limit and write it as a sign.

   at least, no less than, minimum≥at least 8 → ≥ 8

   at most, no more than, maximum≤no more than 6 → ≤ 6

   more than, over>more than 3 → > 3

   less than, fewer than, under<fewer than 9 → < 9

   Then think about the answer. You cannot buy part of a ticket. For “the greatest number,” round down, because going up breaks the limit. For “the least number,” round up, because going down falls short.

   Example

   A van can carry no more than 1,200 pounds. Each box weighs 45 pounds. What is the greatest number of boxes it can carry?

   1. “No more than” means ≤. With b for the number of boxes: 45b ≤ 1,200.
   2. Divide both sides by 45: b ≤ 1,200 ÷ 45, which is about 26.7.
   3. You cannot load part of a box. 27 boxes would weigh 27 × 45 = 1,215 pounds, too much. 26 boxes weigh 26 × 45 = 1,170 pounds, which fits. Round down: 26.

   Answer 26 boxes

   Next step: Three examples ↓
7. 7

   Three worked examples, from easy to test level

   Each one starts with the first try many people make, crossed out, and then the way that works. Read the wrong try too: it is the kind of answer the test puts in the choices.

   In the test-level one the hard part is not the algebra. It is reading every cost in the story, and then rounding the right way at the end.

   Example · Easy

   Solve x + 9 > 4.

   First tryx > 4 + 9, so x > 13

   1. The + 9 is undone by subtracting 9, not by adding it. Adding moves you away from x alone.
   2. Subtract 9 from both sides: x + 9 − 9 > 4 − 9.
   3. 4 − 9 is −5, so x > −5. Subtracting does not flip the sign.
   4. Check x = 0, which is greater than −5: 0 + 9 = 9, and 9 > 4 is true.

   Answer x > −5

   Example · Medium

   Solve 11 − 3x ≤ 26.

   First try−3x ≤ 15, so x ≤ −5

   1. The first try divides by −3 and keeps ≤. Test it with −6, which is in “x ≤ −5”: 11 − 3 × (−6) is 11 + 18, which is 29, and 29 ≤ 26 is false.
   2. Subtract 11 from both sides: −3x ≤ 15.
   3. Divide by −3 and flip the sign: x ≥ 15 ÷ (−3), so x ≥ −5.
   4. Check x = 0: 11 − 0 = 11, and 11 ≤ 26 is true.

   Answer x ≥ −5

   Example · Test level

   A parking garage charges $6 to enter plus $4 an hour. Jordan wants to spend no more than $40. What is the greatest number of hours Jordan can park?

   First try40 ÷ 4 = 10, so 10 hours

   1. The first try forgets the $6 to enter. The total is the entry fee plus the hours: 6 + 4h, where h is the number of hours.
   2. “No more than $40” means ≤: 6 + 4h ≤ 40.
   3. Subtract 6 from both sides: 4h ≤ 34. Divide by 4: h ≤ 34 ÷ 4, which is 8.5.
   4. Round down. With 9 hours Jordan would pay 6 + 4 × 9, which is 6 + 36 = 42 dollars, over the limit. With 8 hours: 4 × 8 = 32, and 6 + 32 = 38, which fits. The answer is 8.

   Answer 8 hours

4 traps in the answer choices

The test writes its wrong answers on purpose: each one is what you get if you make one of these mistakes.

15 questions in three levels

Warm-up first, then core, then test level. Pick an answer, then read why.

12 flashcards

Tap a card to flip it. Keep the ones you miss until they stick.