How to Solve Inequalities and When to Flip the Sign

Every equation so far had one exact answer, the number that makes both sides match. Many questions are not like that. A recipe wants the oven at least $350$ degrees, so $351$ and $400$ work too, and a bridge bars trucks above a posted height, so many heights pass. A statement that compares two amounts instead of setting them equal is an inequality, and its solution is a stretch of the number line. Every balance move from 6.1 through 6.4 works here. Only one rule is new. Multiplying or dividing by a negative flips the sign.

Problem

A solo ride only lets you board if your ticket count is strictly more than

5

, written

t > 5

. The word strictly matters, because holding exactly

5

is not enough, you have to clear the line, not land on it. Five friends hold

3

5

7

12

, and

5.5

tickets. How many of those counts satisfy

t > 5

? Type the single number.

Show a hint

The rule is $t > 5$ . Go through the friends one at a time and ask whether each ticket count is bigger than $5$ . Watch the friend holding exactly $5$ , since $> 5$ does not include $5$ itself.
Mark each value yes or no for $t > 5$ . $3$ is no, $5$ is no because it only ties the line, $7$ is yes, $12$ is yes, and $5.5$ is yes. Count the yeses.

Show the full solution

Check each count against

t > 5

3

is no,

5

is no because it only ties the line,

7

is yes,

12

is yes, and

5.5

is yes. That is

3

friends. Strict means the boundary is out, so holding exactly

5

tickets fails even though it reaches the number.

A hollow circle and a filled circle say two different things at the endpoint. The top line graphs

x > 3

, so the circle at

3

is hollow because the sign is strict and

3

itself is left out, while the ray sweeps right over every value above

3

at once. The bottom line graphs

x \leq 7

, so the circle at

7

is filled because the sign allows equality and

7

is kept in, while the ray sweeps left through

0

and on over every value below

7

. Same idea both times, a hollow ring drops the boundary and a filled dot holds onto it, and the arrow shades in the whole direction in one stroke.

Problem

A delivery cart already holds

8

crates, and the running total must stay under

20

, so with

x

more crates the rule is

x + 8 < 20

. Solve it the same way as

x + 8 = 20

, by subtracting

8

from both sides, with no sign flip for subtraction. The boundary is strict, so a value sitting exactly on it does not count. What is the largest whole number

x

can be?

Show a hint

Make the first move you would make for an equation. Subtract $8$ from both sides to get $x$ alone. Adding or subtracting the same amount never flips an inequality, so the $<$ sign stays pointing the same way. Then notice the bound is strict, which matters for a whole number answer.
Subtracting $8$ from both sides gives $x < 12$ . Since $x$ must be strictly less than $12$ , the value $12$ itself is out. The largest whole number that still fits is the integer just below $12$ .

Show the full solution

Subtract

8

from both sides of

x + 8 < 20

to get

x < 12

. Subtracting never flips the sign. The bound is strict, so

12

is out and the largest whole number that fits is

11

. Check it,

11 + 8 = 19

which is under

20

, while

12 + 8 = 20

is not.

Problem

A diver's gauge reads negatively below the surface. Her reading is

x

, she rises

5

meters to

x + 5

, and the safety check needs the new reading at least

2

, so

x + 5 \geq 2

. Subtract

5

from both sides, and subtracting never flips the sign. Because the sign is

\geq

, the boundary itself is allowed. What is the smallest value

x

can be?

Show a hint

Treat the inequality just like an equation for the first move. To get $x$ by itself, undo the $+ 5$ by subtracting 5 from both sides. You only flip the inequality sign when you multiply or divide both sides by a negative, and subtracting does not do that, so the $\geq$ stays pointing the same way.
Subtract 5 from both sides of $x + 5 \geq 2$ and you get $x \geq - 3$ . The sign is $\geq$ , which means $- 3$ itself is allowed, so the gate is closed. The smallest reading that still works is the boundary number $- 3$ .

Show the full solution

Subtract

5

from both sides of

x + 5 \geq 2

to get

x \geq - 3

. Subtracting does not flip the sign. Because the sign is

\geq

, the boundary is allowed, so the smallest reading she can start from is

- 3

. The sign only reverses when you multiply or divide both sides by a negative, which never happens here.

Problem

Four identical sandbags sit on a shelf rated for at most

26

kilograms total, so with one bag weighing

x

, the rule is

4 x \leq 26

. Solve it like an equation, keeping the inequality sign. Divide both sides by the positive number

4

, which does not flip the sign. The word most means the boundary counts. What is the largest value

x

can be? Give a fraction in lowest terms.

Show a hint

The bags share the load, so undo the $4$ by dividing both sides by $4$ . You are dividing by a positive number here, so watch what happens to the inequality sign. When you divide by a positive, the sign keeps pointing the same way, it does not flip.
Divide both sides of $4 x \leq 26$ by $4$ . The sign stays as $\leq$ , giving $x \leq \frac{26}{4}$ . Simplify $\frac{26}{4}$ to lowest terms, and since the bound is "at most" the boundary itself is allowed, so the largest value is that boundary.

Show the full solution

Divide both sides of

4 x \leq 26

4

, a positive number, so the sign stays put. That gives

x \leq \frac{26}{4}

, which reduces to

\frac{13}{2}

. At most means the boundary counts, so the heaviest one bag can be is

13 / 2

kilograms. If the shelf were rated for under

26

kilograms instead,

\frac{13}{2}

would be shut out and there would be no largest allowed weight.

Problem

Priya pays a flat

$ 45

booth fee and clears

$ 5

on each eraser she sells, so selling

n

erasers gives a profit of

5 n - 45

dollars. She wants that profit to be at least

$ 200

, so the goal is

5 n - 45 \geq 200

. Solve it, dividing by the positive

5

without flipping the sign. What is the fewest whole erasers she must sell to reach her goal?

Show a hint

Each eraser adds $5$ to her profit, but the $45$ booth fee gets subtracted once. Start by undoing that subtraction. Add $45$ to both sides so the $n$ term sits alone, and notice you are only adding and dividing by positive numbers here, so the $\geq$ sign stays pointing the same way the whole time.
After adding $45$ you get $5 n \geq 245$ . Divide both sides by $5$ , a positive number, so no flip, and you land on $n \geq 49$ . Since she cannot sell part of an eraser and $49$ is already a whole number that is allowed, the fewest is $49$ itself.

Show the full solution

Add

45

to both sides of

5 n - 45 \geq 200

to get

5 n \geq 245

, then divide by

5

, a positive, so no flip. That gives

n \geq 49

, and

49

is a whole number that is allowed, so she needs

49

erasers. At

49

the profit is

245 - 45 = 200

dollars, which counts because at least includes hitting the goal exactly.

Problem

Start from a statement nobody would argue with,

3 < 8

. Now multiply both sides by

- 1

, so the

3

becomes

- 3

and the

8

becomes

- 8

. On a number line, the one farther left is the smaller, so

- 8

is now the smaller value. Of the two results

- 3

and

- 8

, which one is the larger? Type that value.

Show a hint

Put $- 3$ and $- 8$ on a number line and remember the rule from earlier chapters. Farther to the left means smaller, farther to the right means larger. Watch what happened to the order. The statement started as $3 < 8$ , with the left number smaller, but after the move the smaller one is no longer on the left.
Mark both numbers. $- 8$ is eight steps left of zero, and $- 3$ is only three steps left of zero, so $- 3$ sits to the RIGHT of $- 8$ . The number on the right is the larger one. That is $- 3$ .

Show the full solution

Multiplying

3 < 8

- 1

turns the sides into

- 3

and

- 8

. On the number line

- 8

sits farther left, so it is the smaller one, and the larger of the two is

- 3

. Copying the old direction would give

- 3 < - 8

, which is false. That is why multiplying or dividing both sides by a negative flips the sign.

Start with something plainly true,

3 < 8

. Multiplying both sides by

- 1

sends every number to its opposite, which on the line is a reflection straight through zero. The point at

3

lands on

- 3

and the point at

8

lands on

- 8

. But a reflection turns left into right, so

8

, once the rightmost and largest, becomes

- 8

, now the leftmost and smallest. To keep the statement true the relationship has to read

- 3 > - 8

. That single forced swap, from

<

>

, is the whole reason the inequality sign flips whenever you multiply or divide both sides by a negative.

Problem

A freezer alarm trips whenever

- 5 x

climbs above

17

, so the danger zone is

- 5 x > 17

. To free

x

, divide both sides by

- 5

, and dividing by a negative flips the

>

<

. What is the largest integer

x

can be?

Show a hint

Get $x$ alone by dividing both sides by $- 5$ . The moment you divide by a negative number, something has to happen to the inequality sign. Decide what it becomes before you read off the answer.
Dividing by $- 5$ flips $>$ into $<$ , so $x < - \frac{17}{5}$ , which is $x < - 3.4$ . The gate at $- 3.4$ is open, so $- 3.4$ itself is not allowed. Now name the largest integer that sits strictly to the left of $- 3.4$ on the number line.

Show the full solution

Divide both sides of

- 5 x > 17

- 5

. The divisor is negative, so the sign flips and

x < - \frac{17}{5} = - 3.4

. The bound is strict, and the largest integer sitting left of

- 3.4

- 4

. Check it,

- 5 \cdot (- 4) = 20 > 17

is true, while

- 5 \cdot (- 3) = 15 > 17

is false.

Problem

This one pulls everything together. The variable appears on both sides and there is a group to clear. Solve

3 (2 - x) > 2 x + 1

by distributing, then gathering the variable, and stay alert because one step divides by a negative and forces the sign to flip. Once you have the range, what is the largest integer

x

can be?

Show a hint

Start exactly the way you would for an equation. Distribute the $3$ across the parentheses to get $6 - 3 x$ on the left, then move the variable terms to one side and the numbers to the other. Keep one eye out for a moment where you divide by a negative, since that is the only thing here that behaves differently from an equation.
After distributing you have $6 - 3 x > 2 x + 1$ . Subtract $2 x$ from both sides to reach $6 - 5 x > 1$ , then subtract $6$ to get $- 5 x > - 5$ . Now divide both sides by $- 5$ , and because that divisor is negative, FLIP the sign. That gives $x < 1$ . The bound is strict, so $1$ itself is not allowed. Name the largest whole number that still sits below $1$ .

Show the full solution

Distribute the

3

, so

3 (2 - x) > 2 x + 1

becomes

6 - 3 x > 2 x + 1

. Subtract

2 x

to get

6 - 5 x > 1

, then subtract

6

to get

- 5 x > - 5

. Dividing both sides by

- 5

flips the sign, leaving

x < 1

. The bound is strict, so the largest integer that fits is

0

. At

x = 1

both sides equal

3

, and

3 > 3

is false, which is exactly why

1

is shut out.

Problem

Three runners are scored so a bigger number means faster. The sheet gives two facts. Priya beat Quinn,

P > Q

, and Quinn was at least as fast as Rafe,

Q \geq R

. A teammate scribbles five claims,

P > R

P \geq R

P > Q

Q > R

, and

R > P

. Using only the two facts and the way inequalities chain when they point the same way, how many of the five claims are forced to be true? Type the single count.

Show a hint

Line the two facts up so the arrows point the same way, $P > Q \geq R$ . Anything you can read straight off this chain by following greater-than signs from left to right is forced. Watch the spot where $Q$ meets $R$ , because that link is $\geq$ and not a strict $>$ .
Forced because they read straight off the chain, $P > Q$ is given outright, $P > R$ jumps across the whole chain, and $P \geq R$ is a weaker true statement that $P > R$ already guarantees. Not forced, $Q > R$ fails whenever Quinn and Rafe tie since that link is only $\geq$ , and $R > P$ is actually backwards. Count the forced ones.

Show the full solution

Stack the two facts into one chain,

P > Q \geq R

. Reading across it,

P > Q

is given outright,

P > R

follows because the trip across the chain crosses a strict

>

, and

P \geq R

follows since it is a weaker claim than

P > R

. The other two are not forced.

Q > R

fails if Quinn and Rafe tie, since that link is only

\geq

, and

R > P

points backwards. That is

3

forced claims. One strict link anywhere makes the whole chain strict, but a chain of all

\geq

leaves ties possible.

A compound inequality traps the variable between two bounds at the same moment, so the picture is a finite segment instead of an endless ray. The left gate at

- 3

is a closed dot because

- 3 \leq x

lets

x

actually be

- 3

, while the right gate at

2

is an open dot because

x < 2

never lets

x

reach

2

. Together they say

- 3 \leq x < 2

, the gold bar between the gates. Counting the integer dots caught inside gives a finite answer, here the five values

- 3, - 2, - 1, 0, 1

Problem

A compound inequality traps the variable in the middle. Solve

1 < 5 - 2 x \leq 11

. Whatever you do, do it to all three parts at once so the middle stays fenced in on both sides, and remember that dividing by the negative coefficient flips both signs together. Once you have

x

cornered, how many integers make the whole thing true? Type the single count.

Show a hint

Peel away the constant from all three parts first. Subtract $5$ from the left, the middle, and the right, and you should land on $- 4 < - 2 x \leq 6$ . Now you have to divide every part by $- 2$ , and that negative is about to do something to the signs, so stay alert.
Dividing all three parts by $- 2$ flips BOTH inequality signs, which gives $2 > x \geq - 3$ . Read that the tidy way around and it says $- 3 \leq x < 2$ , closed at $- 3$ and open at $2$ . Now just list the integers from $- 3$ up to but not including $2$ and count them.

Show the full solution

Subtract

5

from all three parts of

1 < 5 - 2 x \leq 11

to get

- 4 < - 2 x \leq 6

. Divide every part by

- 2

, and since

- 2

is negative both signs flip, giving

2 > x \geq - 3

. Written small to large that is

- 3 \leq x < 2

, so the integers are

- 3, - 2, - 1, 0, 1

, which is

5

of them. Dividing by a negative also swaps which end is the floor and which is the ceiling, so rewrite the range in increasing order before you count.

Key idea

Two bounds at once. A compound inequality like

4 < x \leq 9

sets two conditions at the same time, a lower bound and an upper bound. Its solution is a segment with both ends marked, not a ray that runs on forever. Solve it with the same balance work, except that every move now applies to the left, middle, and right together. Multiplying all three by a negative flips both signs. Chaining is a related move. When several comparisons all point the same direction, combine them into one statement and read across it. If a first amount is greater than a second and the second is at least as large as a third, the first is greater than the third. Watch the symbols. One strict

>

anywhere keeps the two ends strictly apart, but a chain of

\geq

does not force the ends to differ, since they might be equal.

Problem

The trickiest one, folding everything together. Solve

5 - 2 x - 8 > 4 - x

. The left side has two number pieces sitting apart, so tidy it first, then gather the variables. One move near the end forces the sign to turn around. Once you have the range, what is the largest integer

x

can be?

Show a hint

Start by combining the two number terms on the left, since $5 - 8$ is a single value. That turns the left side into $- 3 - 2 x$ . Then gather the variables on one side the same way you did when solving equations. Keep an eye out for the step that might flip the inequality.
After tidying you have $- 3 - 2 x > 4 - x$ . Add $x$ to both sides to get $- 3 - x > 4$ , then add $3$ to both sides to get $- x > 7$ . Now multiply both sides by $- 1$ , and because the multiplier is negative you must reverse the sign, giving $x < - 7$ . The bound is strict, so $- 7$ itself is not allowed. The largest integer that fits is the whole number just below $- 7$ .

Show the full solution

Combine the numbers on the left,

5 - 8 = - 3

, so the inequality reads

- 3 - 2 x > 4 - x

. Add

x

to both sides to get

- 3 - x > 4

, then add

3

to get

- x > 7

. Multiplying both sides by

- 1

flips the sign, leaving

x < - 7

. The bound is strict, so the largest integer is

- 8

. A bare

- x

carries a negative coefficient, so clearing it flips the sign just like dividing by

- 5

would.

Problem

A metal sample starts at

22

degrees Celsius and the cooler pulls it down

1.5

degrees per minute, so after

m

minutes its temperature is

22 - 1.5 m

. It freezes only once that drops strictly below

- 8

, so the condition is

22 - 1.5 m < - 8

. Solve it, remembering that dividing by the negative coefficient flips the sign. What is the fewest whole minutes the lab must wait until it freezes?

Show a hint

Solve $22 - 1.5 m < - 8$ just like a normal equation. Subtract $22$ from both sides to get $- 1.5 m < - 30$ . Now watch closely, because your next move divides both sides by a negative number, and that does something to the inequality sign.
Divide both sides by $- 1.5$ . Dividing by a negative flips the sign, so $- 1.5 m < - 30$ turns into $m > 20$ . The bound is strict, so $m = 20$ does not count. Check it. At $20$ minutes the temperature is exactly $- 8$ , which is not below the line yet, and at $21$ minutes it is $- 9.5$ , which finally is. So the answer is the first whole number bigger than $20$ .

Show the full solution

Subtract

22

from both sides of

22 - 1.5 m < - 8

to get

- 1.5 m < - 30

. Divide by

- 1.5

, and the negative divisor flips the sign, giving

m > 20

. The bound is strict, so the lab waits

21

minutes. At

20

minutes the sample is at exactly

- 8

, which is not yet below the line, and at

21

minutes it is

- 9.5

Practice these ideas

Practice

A rewards account stays open as long as its balance

w

satisfies

w \geq - 4

, since a small overdraft is forgiven. Five accounts have balances

- 6

- 4

- 3

0

, and

5

. How many satisfy

w \geq - 4

Show the solution

Check each balance against

w \geq - 4

- 6

fails,

- 4

passes, and

- 3

0

, and

5

all pass. That is

4

accounts. The equals half of

\geq

is what saves the account sitting exactly at

- 4

Practice

A scanner keeps a reading

m

only when

m < 7

and tosses anything that hits or passes the threshold, so the strict sign leaves the boundary

7

out. One shift gives

2

7

6.9

8

, and

- 1

. How many readings does the scanner keep?

Show the solution

Test each reading against

m < 7

2

is kept,

7

is not since

7 < 7

is false,

6.9

is kept,

8

is not, and

- 1

is kept. That is

3

readings. A strict less-than never lets a value equal the boundary, so the reading of exactly

7

goes out.

Practice

A ferry deck already holds

13

crates, and loading

x

more must keep the total under

25

, so

x + 13 < 25

. Solve it. What is the largest integer number of extra crates

x

can be?

Show the solution

Subtract

13

from both sides of

x + 13 < 25

to get

x < 12

. Subtracting does not flip the sign. The bound is strict, so the largest whole number of extra crates is

11

. Eleven brings the deck to

24

, while

12

would hit

25

exactly, which the rule does not allow.

Practice

A pilot's dial setting

x

must satisfy

x - 9 \geq - 2

. Add

9

to both sides, with no flip for addition. Because the sign is

\geq

, the boundary is allowed. What is the smallest value

x

can be?

Show the solution

Add

9

to both sides of

x - 9 \geq - 2

to get

x \geq 7

. Adding never flips the sign. The

\geq

lets

x

equal the boundary, so the smallest setting is

7

Practice

A garden permit allows at most

28

liters, and each visit takes

8

liters, so over

x

visits the rule is

8 x \leq 28

. Divide both sides by the positive

8

, no flip. The word most means the boundary counts. What is the largest value

x

can be? Give a fraction in lowest terms.

Show the solution

Divide both sides of

8 x \leq 28

8

, a positive, so no flip. That gives

x \leq \frac{28}{8} = \frac{7}{2}

, and at most means the boundary counts, so the largest is

7 / 2

. At

x = \frac{7}{2}

the water used is exactly

28

liters, right on the permit limit.

Practice

Cookie packs cost

$ 7

each, and Priya wants to spend strictly more than

$ 50

to clear a free-shipping bar, so

7 x > 50

. Solve it. The strict sign means

50

itself does not count. What is the smallest integer number of packs she can buy?

Show the solution

Divide both sides of

7 x > 50

7

, a positive, so the sign stays put. That gives

x > \frac{50}{7}

, about

7.14

, and the first whole number above it is

8

. Seven packs cost

$ 49

, which does not clear

$ 50

, so rounding down would miss the bar.

Practice

A deep-sea sensor's reading

x

must satisfy

- 6 x < 30

. Divide both sides by

- 6

and flip the sign because the divisor is negative. What is the smallest integer

x

can be?

Show the solution

Divide both sides of

- 6 x < 30

- 6

, and the negative divisor flips the sign, giving

x > - 5

. The bound is strict, so

- 5

is out and the smallest integer allowed is

- 4

. Check it,

- 6 \cdot (- 4) = 24 < 30

is true, while

- 6 \cdot (- 5) = 30 < 30

is false.

Practice

A frost alarm trips whenever

- 5 x > 12

, where

x

is a temperature that runs colder as it drops. Solve it, dividing by

- 5

and flipping the sign. What is the largest integer value

x

can be?

Show the solution

Divide both sides of

- 5 x > 12

- 5

, and the negative divisor flips the sign, giving

x < - \frac{12}{5}

, which is

x < - 2.4

. Walking left from

- 2.4

, the first integer is

- 3

. It is tempting to grab

- 2

, but

- 2

is greater than

- 2.4

, so it misses the bound.

Practice

A vending machine credits Mara once it counts more than

19

tokens past a deposit, and

x

cups give

4 x - 5

such tokens, so she needs

4 x - 5 > 19

. Add

5

, then divide by the positive

4

, no flip. What is the smallest integer number of cups

x

she can buy?

Show the solution

Add

5

to both sides of

4 x - 5 > 19

to get

4 x > 24

, then divide by

4

, a positive, so no flip. That gives

x > 6

, and since the bound is strict the smallest whole number is

7

. Six cups land on exactly

19

tokens, which is not more than

19

Practice

A drone starts at a battery reading of

10

and the lamp drains

4

per minute, so after

x

minutes the reading is

10 - 4 x

, and the pilot keeps it on while

10 - 4 x \geq - 6

. Subtract

10

, then divide by

- 4

and flip the sign. What is the largest value

x

can be?

Show the solution

Subtract

10

from both sides of

10 - 4 x \geq - 6

to get

- 4 x \geq - 16

. Divide by

- 4

, and the negative divisor flips

\geq

into

\leq

, giving

x \leq 4

. The boundary is included, so the largest value is

4

. At

x = 4

the reading is exactly

- 6

, which the rule still allows, and at

x = 5

it drops to

- 10

Practice

One battery mode reads

7 - x

and another reads

2 x + 1

after

x

minutes, and the first beats the second when

7 - x > 2 x + 1

. Gather the variable, then divide by a negative and flip the sign. What is the largest integer

x

can be?

Show the solution

Subtract

2 x

from both sides of

7 - x > 2 x + 1

to get

7 - 3 x > 1

, then subtract

7

to get

- 3 x > - 6

. Divide by

- 3

, which flips the sign, leaving

x < 2

. The bound is strict, so the largest integer is

1

. At

x = 2

the two readings tie at

5

, and a tie is not a win.

Practice

A batch of

x

panels uses

3 (x - 2)

ounces of grout because the first

2

panels are scrapped, and the bucket holds

18

ounces, so

3 (x - 2) \leq 18

. Distribute, then solve, dividing by the positive

3

with no flip. What is the largest value

x

can be?

Show the solution

Distribute to get

3 x - 6 \leq 18

, add

6

for

3 x \leq 24

, then divide by

3

, a positive, so no flip. That gives

x \leq 8

, and the bound is closed, so the largest is

8

. At

x = 8

the grout used is

3 (6) = 18

ounces, exactly the bucket, and

x = 9

needs

21

Practice

A ferry logs every stop

x

that is at least

- 5

and strictly west of

4

, so

- 5 \leq x < 4

. The left end is closed, so

- 5

counts, and the right end is open, so

4

does not. How many integer positions get logged?

Show the solution

Since

- 5

is included and

4

is not, the logged integers run

- 5, - 4, - 3, - 2, - 1, 0, 1, 2, 3

. That is

9

positions. Counting is faster as

3 - (- 5) = 8

steps plus one for the endpoint you started on.

Practice

A candle stands

30

cm tall and burns down

2.5

cm each hour, so after

h

hours its height is

30 - 2.5 h

, and it must stay at least

5

cm tall to sit in its holder, giving

30 - 2.5 h \geq 5

. Dividing by

- 2.5

flips the sign. What is the most whole hours the candle can burn?

Show the solution

Subtract

30

from both sides of

30 - 2.5 h \geq 5

to get

- 2.5 h \geq - 25

. Divide by

- 2.5

, and the negative divisor flips

\geq

into

\leq

, giving

h \leq 10

. The boundary counts, so the candle can burn

10

whole hours. At

10

hours it stands exactly

5

cm tall, which still sits in the holder, and at

11

it is down to

2.5

cm.