Math 6 · 6.EE.B.8

Writing an inequality for a condition

An airline says a checked bag may weigh at most 23 kg. That one short rule allows endlessly many weights. How do you write it in symbols, and know whether a bag of exactly 23 kg passes?

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Inequalities on the number lineMath 6 Translating between words and expressionsMath 6

Where you are 7 in Inequalities

Algebra9

Inequalities7

Inequalities on the number line Interval and set-builder notation

Writing an inequality for a condition

Solving linear inequalities3 Compound inequalities: and, or ▾ Show all 7 (2 more)

When you reach for it

Many rules do not name one number. An airline says a checked bag may weigh at most 23 kg. A festival sells child tickets to anyone under 12. A dog walker takes no more than 5 dogs on a walk. Each rule draws a line at one number and lets through everything on one side of it. An equation cannot say that: w = 23 allows a single weight, while the rule allows 23 kg, 22.5 kg, 18 kg and every weight below.

An inequality can. The four symbols <, >, \le, \ge and their graphs belong to Inequalities on the number line; this lesson is about getting from the words to the symbols. Reach for it whenever a rule sets a limit, a minimum, a maximum, a floor or a ceiling, and you want to say exactly which values pass. Once the rule is in symbols, testing a value means putting it in and seeing whether the claim is true (What a solution is, and checking one).

The procedure

Four decisions turn a condition in words into an inequality. Each is settled by asking what the situation allows, not by matching a key word.

Name the quantity. Pick a letter and say in words what it stands for, units included: “let w be the weight of the bag in kilograms”. The rule is a claim about one quantity, so the letter must stand for exactly that quantity, and its units must match the number in the rule. This is the same habit as naming the unknown in Translating between words and expressions.

Find the boundary. The number the rule names, 23 in the bag rule, is the boundary. On one side of it values pass; on the other they fail.

Find the side. Ask whether the values that pass are larger or smaller than the boundary, and answer from the situation. A weight limit is there to stop heavy bags, so the weights that pass are the smaller ones. A minimum age keeps out the young, so the ages that pass are the larger ones. When a rule describes what is not allowed (“no walks below -10 °C”), the values that pass lie on the other side.

Decide whether the boundary itself passes. Put the boundary into the rule as a story. A bag of exactly 23 kg under “at most 23 kg”: the most it may weigh is 23 kg, and it weighs 23 kg, so it passes and the symbol is inclusive, \le. The same bag under “under 23 kg”: 23 kg is not under 23 kg, so it fails and the symbol is strict, <.

Then write the inequality with the letter first, w \le 23, and read it back as a sentence: “the weight is less than or equal to 23 kilograms”. If the sentence says what the rule says, the inequality is right.

What the everyday phrases already mean. None of these needs memorizing; each is the plain meaning of the words, tested at the boundary.

“At most 23”, “no more than 23”, “maximum 23” name a ceiling the value may touch, so the value is 23 or less: \le. “No more than” is worth reading slowly. A weight that is not more than 23 is either equal to 23 or less than it.
“At least 12”, “no fewer than 12”, “minimum 12”, “12 and over” name a floor the value may stand on: 12 or more, \ge.
“More than”, “over”, “above” put the value strictly past the boundary on the high side, >, because a value equal to 12 is not more than 12.
“Less than”, “fewer than”, “under”, “below” put it strictly on the low side, <, because a value equal to the boundary is not under it.

How many values pass. The inequality w \le 23 is true for 23, 22.5, 18, 0.3 and endlessly many more, since between any two allowed weights sits another, such as the number halfway between them. A condition on a measured quantity has infinitely many solutions, and its graph is a shaded stretch. When the letter counts things (dogs, people, nights), only whole numbers make sense, and the graph is dots. A ceiling on a count then allows a short list: d \le 2 allows 0, 1 and 2. A floor on a count still allows infinitely many: d \ge 2 allows 2, 3, 4 and so on, with no largest.

Check against the story. Try three values: the boundary, one just inside, one just outside. Each should pass or fail exactly as the rule in words says.

Worked examples

Example 1

A weight limit and a refusal

An airline’s rules say: “A checked bag may weigh at most 23 kg. A bag weighing more than 32 kg is refused.” Write an inequality for the weights within the allowance and one for the weights that are refused, and graph both. Eliska’s bag weighs 23 kg and Fenn’s weighs 32 kg. What does each rule say about each bag?

Solution

Let w be the weight of a bag in kilograms.

The allowance is a limit on heavy bags, so the weights within it are at or below 23. A bag of exactly 23 kg weighs the most the rule permits, so it is within: the inequality is w \le 23. On the graph, a filled dot at 23, then shade to the left.

The refusal is about the heavy end, so refused weights lie above 32. A bag of exactly 32 kg is not more than 32 kg, so it is not refused: the inequality is w > 32. On the graph, an open circle at 32, then shade to the right.

Figure 1Example 1. The allowance lets 23 in; the refusal leaves 32 out.

Eliska: 23 \le 23 is true, so her bag is within the allowance. Fenn: 32 \le 23 is false, so his bag is over the allowance, but 32 > 32 is also false, so it is not refused. It flies, outside the allowance. Both strict and inclusive symbols appear in one set of rules, and only testing the boundary in each tells them apart.

Example 2

Ticket bands at a festival

A festival sells child tickets to people under 12 and senior tickets to people aged 65 and over. Everyone else buys an adult ticket. Ages are in whole years, the way people state them. Write an inequality for each band. Which ticket do Idun, aged 12, Jiri, aged 65, and Fenn, aged 64, buy?

Solution

Let a be a person’s age in whole years.

Child band: a 12-year-old is not under 12, so 12 is left out, and the ages that pass are below it: a < 12. Senior band: “65 and over” names 65 as the first age in, so 65 passes and so does every age above: a \ge 65.

Idun: 12 < 12 is false and 12 \ge 65 is false, so Idun buys an adult ticket, by one birthday. Jiri: 65 \ge 65 is true, so a senior ticket. Fenn: 64 < 12 and 64 \ge 65 are both false, so an adult ticket.

Because a is a whole number, the child band is exactly the twelve ages 0 to 11. The senior band a \ge 65 has no last age written into it; the inequality allows 65, 66, 67 and so on without end, and real lifespans cut the list short, not the mathematics.

Example 3

A ceiling and a floor on a count

Kalle walks dogs. She takes no more than 5 dogs on one walk, and she drives out to the big park only when she has no fewer than 3 dogs. Write each condition as an inequality, list its solutions and graph them.

Solution

Let d be the number of dogs on a walk. Dogs are counted, so d is a whole number.

“No more than 5”: the number of dogs is not more than 5, so it is 5 or fewer, and 5 itself is allowed: d \le 5. The whole numbers that pass are 0, 1, 2, 3, 4 and 5, which is six values. The value 0 makes the inequality true; a walk with no dogs is Kalle out alone, and the rule does not forbid it.

“No fewer than 3”: the number is not fewer than 3, so it is 3 or more: d \ge 3. The solutions are 3, 4, 5, 6 and so on. Even though only whole numbers count, the inequality d \ge 3 has infinitely many solutions, because the floor has no ceiling to stop it.

Figure 2Example 3. Dogs are counted, so each graph is dots. A ceiling gives a short list; a floor never ends.

Kalle’s real walks do have a ceiling: her first rule stops her at 5 dogs. The inequality d \ge 3 on its own does not know that; the story does. Put the two rules together and only 3, 4 and 5 pass both; two conditions at once are the subject of Compound inequalities: and, or.

Example 4

Saying when not

Kalle’s winter rule is: “No walks when the temperature is below -10 °C.” Write an inequality for the temperatures at which walks go ahead, graph it, and decide for -10 °C, -12 °C and -9.5 °C.

Solution

Let t be the temperature in degrees Celsius. The rule names the temperatures with no walk: those below -10. Walks go ahead at every other temperature, which is every temperature not below -10. At exactly -10 °C it is not below -10 °C, so the no-walk rule does not apply and the walk goes ahead. The walking temperatures are -10 and above: t \ge -10. On the graph, a filled dot at -10, then shade to the right.

Figure 3Example 4. The temperatures at which walks go ahead.

Check the three temperatures. -10 \ge -10 is true, so a walk. -12 lies to the left of -10 on the number line, so -12 \ge -10 is false: no walk. -9.5 lies to the right of -10, so it is greater and the walk goes ahead.

The rule used the strict word “below”, yet the answer is inclusive. The boundary belongs to exactly one of the two sides; when the rule describes the other side, the boundary changes sides with it.

Example 5

Different words, the same children

One festival poster says “Children under 10 go free”. Another says “Free entry for children aged 9 and under”. With ages in whole years, write both as inequalities and show they let in the same children. Would two such rules still agree for a quantity that need not be whole?

Solution

Let a be the age in whole years. “Under 10” leaves 10 out: a < 10. “Aged 9 and under” lets 9 in: a \le 9.

The whole numbers less than 10 are 0 to 9; the whole numbers less than or equal to 9 are 0 to 9. The two lists are the same, and at the age 10 both fail (10 < 10 and 10 \le 9 are false). So the posters agree.

For a quantity that can take any value, the two forms differ. The number 9.5 makes x < 10 true and x \le 9 false, since 9.5 lies to the right of 9. The posters agree only because ages are counted in whole years.

Example 6

From symbols back to a rule

A festival campsite prints its rules in symbols: n \ge 2, where n is the number of nights booked, and p \le 8, where p is the number of people on one pitch. Write each as a sign a visitor would read, in two different ways, and list the values that make sense.

Solution

n \ge 2 allows 2 and everything above, so it is a floor that may be met: “Minimum stay 2 nights”, or “Book at least 2 nights”. Read back: a stay of 2 nights passes because 2 \ge 2. Nights are counted, so the solutions are 2, 3, 4 and so on. The inequality has no largest solution; the length of the festival limits stays, and that would be a second condition.

p \le 8 allows 8 and everything below, so it is a ceiling that may be reached: “At most 8 people per pitch”, or “No more than 8 people on a pitch”. People are counted, so the whole numbers 0 to 8 make the inequality true, which is nine values. A booked pitch has at least one person, so the story uses only 1 to 8, which is eight. The inequality states the limit; the story can rule out values the inequality allows.

Boundaries

This move writes a condition that compares one quantity with one fixed number, giving x > c, x < c, x \ge c or x \le c. A condition with a calculation inside it, such as a total price that may not go over a budget, gives an inequality like 4d \le 30 that has to be solved.
One inequality holds one limit. A rule with a floor and a ceiling together (at least 3 and at most 5 dogs) is two inequalities joined by “and”: Compound inequalities: and, or.
The inequality allows every number on its side, including values the situation rules out: negative weights, an age of 500. The story decides which solutions can happen.
Some everyday wording leaves the boundary unclear: “up to 5” usually includes 5, while “between 3 and 5” does not say. Writing the inequality forces the choice, and the choice comes from what the rule’s maker meant, not from the mathematics.
When the letter counts things, two inequalities with different boundaries can have the same solutions, as a < 10 and a \le 9 do. For a measured quantity they never do.

Where this goes next

Solving comes next: One-step inequalities and the sign flip handles conditions with a calculation inside, such as 4d \le 30, by undoing the operation exactly as for an equation. A pair of limits on one quantity becomes a single stretch of the line in Compound inequalities: and, or, and the stretch gets a short written name in Interval and set-builder notation.

Exercises

Writing the inequality

For each rule, choose a letter, say what it stands for with its units, and write an inequality for the values described.

A cabin bag may weigh at most 8 kg.
Tickets for the late-night stage are sold to people aged 16 and over, ages in whole years.
Each of Kalle’s walks lasts at least 40 minutes.
The festival shuttle bus carries no more than 52 passengers.
A bag longer than 45 cm cannot go in the cabin. Write the inequality for the lengths turned away.
The toddler tent is for children under 3.
A dog must be no younger than 6 months to join a group walk.
Kalle carries extra water whenever the temperature is above 20 °C. Write the inequality for the temperatures at which she carries it.

Is the boundary in?

Bags weighing more than 20 kg get a heavy-bag tag. Write the inequality for tagged weights. Is a bag of (a) 20 kg (b) 20.5 kg tagged?
The kids’ zone has a maximum age of 11. Write the inequality. May a child aged (a) 11 (b) 12 go in?
There are no walks when the temperature is -15 °C or lower. Write the inequality for the temperatures at which walks go ahead, and decide for (a) -15 °C (b) -14.5 °C (c) -20 °C.
A festival stall takes card payments of at least 5 dollars. Write the inequality for accepted payments. Is a payment of (a) 5 dollars (b) 4.99 dollars accepted?

How many solutions?

A dance tent holds at most 9 dancers. Write the inequality, and say how many different numbers of dancers it allows, counting an empty tent.
A second airline’s cabin limit is w < 8.5, with w in kilograms. Give three allowed weights, one of them not a whole number, and say how many allowed weights there are.
The festival sends an extra shuttle bus whenever more than 70 people are waiting. Write the inequality, find the smallest number of people waiting that brings the extra bus, and write a different inequality with the same whole-number solutions.
Kalle’s plan asks for at least 4 dogs a day. Write the inequality, list its four smallest solutions, and say how many solutions it has.

Writing and graphing

Write the inequality and describe its graph.

A carry-on case must be no longer than 55 cm.
The outdoor stage stays open while the temperature is above -2 °C.
Kalle’s car takes at most 4 dogs.

Explain

Explain why “no more than 30” and “at most 30” give the same inequality.
A sign reads “Under-8s go free”. With a the age in whole years, explain why a < 8 describes it and a \le 8 does not.
A small bag has the limit w \le 3, in kilograms, and a small car takes d \le 3 dogs. Explain why the first has infinitely many solutions and the second has only four.
Can a condition about a count of things have infinitely many solutions? Explain with an example.

Answers

w \le 8, with w the bag’s weight in kilograms. A bag of exactly 8 kg weighs the most it may, so it passes.
a \ge 16, with a the age in years. “16 and over” lets 16 in.
m \ge 40, with m the length of a walk in minutes. A walk of exactly 40 minutes lasts at least 40 minutes.
p \le 52, with p the number of passengers, a whole number. Not more than 52 means 52 or fewer.
l > 45, with l the bag’s length in centimeters. A bag exactly 45 cm long is not longer than 45 cm, so it is not turned away.
a < 3, with a the age in years. A child aged 3 is not under 3.
g \ge 6, with g the dog’s age in months. Not younger than 6 months means 6 months or older.
t > 20, with t the temperature in °C. At exactly 20 °C it is not above 20 °C.
w > 20. (a) no, 20 > 20 is false. (b) yes, 20.5 lies to the right of 20.
a \le 11. (a) yes, 11 \le 11 is true. (b) no, 12 \le 11 is false.
t > -15. The no-walk temperatures are t \le -15, so walks go ahead at all the others. (a) no walk, since -15 is itself a no-walk temperature. (b) walk, since -14.5 lies to the right of -15. (c) no walk, since -20 lies to the left of -15.
c \ge 5, with c the payment in dollars. (a) yes, 5 \ge 5 is true. (b) no, 4.99 is less than 5.
d \le 9, with d the number of dancers. The whole numbers 0 to 9 pass: 10 different numbers.
For example 2, 5 and 7.25, each less than 8.5. There are infinitely many: between any two allowed weights sits another.
n > 70, with n the number of people waiting, a whole number. The smallest is 71, since 70 > 70 is false. The inequality n \ge 71 has the same whole-number solutions: 71, 72, 73 and so on.
d \ge 4. The smallest solutions are 4, 5, 6 and 7. The inequality has infinitely many: every whole number from 4 upward. Her real day caps the list; d \ge 4 does not.
l \le 55, with l the length in centimeters: a filled dot at 55, then shade to the left.
t > -2, with t the temperature in °C: an open circle at -2, then shade to the right.
d \le 4, with d the number of dogs: dots at 0, 1, 2, 3 and 4, nothing shaded between them.
Both say the value may reach 30 but not pass it. “No more than 30” rules out every value above 30, leaving 30 and everything below; “at most 30” names 30 as the largest value allowed. Both are x \le 30.
A child aged 8 is not under 8, so pays. 8 < 8 is false, so a < 8 leaves 8 out, as the sign does. 8 \le 8 is true, so a \le 8 would let an 8-year-old in free.
A weight can be any number: 3, 2.5, 2.95, and between any two allowed weights another. Dogs are counted, so only the whole numbers 0, 1, 2 and 3 fit: four.
Yes, when the condition is a floor. “At least 10 tickets sold” allows 10, 11, 12 and so on: all whole numbers, but with no largest. Only a ceiling on a count gives a list that ends. A real ticket office runs out, but that limit comes from the story, not the inequality.