Math 6 · 6.EE.B.8 · 7.EE.B.4B

Inequalities on the number line

A ride sign says riders must be at least 120 cm tall. That allows not one height but endlessly many. How can one picture show every one of them?

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Expression, equation and inequalityMath 6 VariablesGrade 6 The number lineGrades 1 to 2

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Interval and set-builder notationAlgebra 1 Writing an inequality for a conditionMath 6 One-step inequalities and the sign flipMath 7 Solving linear inequalitiesGrades 7 to 8 Graphing a linear inequality in two variablesAlgebra 1 +2 more

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

An equation such as x = 3 is true for exactly one number. An inequality is a claim that one amount is larger or smaller than another (Expression, equation and inequality), and it is usually true for a great many numbers. The claim x > 3 is true when x is 4, when x is 20, when x is 3.5 and when x is 3.001. It is false when x is 3 itself, and false for every number below 3.

A number that makes an inequality true when it is put in for the variable is a solution of the inequality. The solutions of x > 3 cannot be listed, not even with “and so on”: between 3 and 4 alone sit 3.1, 3.01, 3.001 and endlessly many more. The number line can hold them. Every number has a position on the line (The number line), so a whole stretch of the line can stand for a whole stretch of numbers, and one picture shows every solution at once.

Reach for this graph whenever a condition allows a range of values rather than a single one: a minimum height for a ride, a weight limit for a lift, a temperature below freezing.

The procedure

The four symbols. Two of them are the comparison signs already used to order numbers (Comparing numbers and the symbols >, < and =, Comparing and ordering rational numbers); the other two add “or equal to”.

Symbol	Read as	Example	Is 3 a solution?
<	is less than	x < 3	no
>	is greater than	x > 3	no
\le	is less than or equal to	x \le 3	yes
\ge	is greater than or equal to	x \ge 3	yes

The open side of each symbol faces the larger amount and the point faces the smaller. The bar under \le and \ge is half of an equals sign, so 3 \le 3 is true (the two sides are equal) while 3 < 3 is false. The symbols < and > are called strict: they leave the boundary number out. The symbols \le and \ge are inclusive: they let it in.

Either side may come first. “x is greater than 3” and “3 is less than x” describe the same two positions on the line, read from opposite ends, so x > 3 and 3 < x make the same claim. When the two sides swap, the symbol turns round with them, and its point still faces the smaller amount: 5 \ge y says the same as y \le 5. Before graphing, it helps to write the variable first, because then the symbol names the direction: > or \ge for greater, < or \le for less.

The graph comes from three decisions, each read straight off the inequality.

The boundary is the number in the inequality, 3 in every line of the figure. It is where the claim switches between true and false, so the graph is built around it.

The circle comes from testing the boundary itself. In x > 3, putting in 3 gives 3 > 3, which is false, so 3 is not a solution: an open circle marks the edge of the solutions without including it. In x \ge 3, 3 \ge 3 is true, so a filled dot includes it. A strict symbol always gives an open circle and an inclusive one always gives a filled dot.

The side comes from the direction of the line. Numbers grow to the right, so the numbers greater than 3 are exactly the ones to the right of 3, and the numbers less than 3 are exactly the ones to its left. Shade that whole side with a heavy line, and end it in an arrow, because it never stops: x > 3 is true for 1000 and for a million. One number on the shaded side, put back into the inequality, confirms the choice; 0 is usually the easiest, as long as it is not the boundary.

Figure 1The circle says whether 3 itself is a solution; the shading says which side the solutions are on.

The heavy line covers every point, not only the marked whole numbers, so 3.5 and 4\tfrac{1}{3} are shaded along with 4 and 5.

Reading a graph back runs the same decisions in reverse. The circle’s position is the boundary. An open circle means a strict symbol and a filled dot an inclusive one. Shading to the right means greater; shading to the left means less.

When only whole numbers make sense. If the variable counts things (people, tickets, buses), then 2.5 is not a possible value even when it makes the inequality true. The graph then marks only the whole numbers that work, one filled dot each, with nothing shaded between them. If the whole numbers that work go on forever, draw the first few dots and write “and so on” beside them.

Worked examples

Example 1

Two graphs

Graph (a) x < 4 and (b) x \ge -2.

Solution

(a) The boundary is 4. Testing it gives 4 < 4, which is false, so draw an open circle at 4. “Less than” means to the left of 4 on the line, so shade to the left. Check with 0: 0 < 4 is true, and 0 is on the shaded side.

(b) The boundary is -2. Testing it gives -2 \ge -2, which is true because the two sides are equal, so draw a filled dot at -2. “Greater than” means to the right, so shade to the right. Check with 0: it lies to the right of -2 on the line, so 0 \ge -2 is true, and 0 is shaded.

Figure 2Example 1.

Example 2

The variable on the right

Graph (a) -3 < x and (b) 6 \ge y.

Solution

(a) Read from the variable: “x is greater than -3”. Swapping the sides turns the symbol round, so -3 < x is the same claim as x > -3. Testing the boundary gives -3 < -3, false, so draw an open circle at -3 and shade to the right. Check with 0: -3 < 0 is true, and 0 is to the right of -3.

(b) 6 \ge y is the same claim as y \le 6. Testing the boundary gives 6 \ge 6, true, so draw a filled dot at 6 and shade to the left. Check with 0: 6 \ge 0 is true, and 0 is to the left of 6.

Figure 3Example 2. With the variable on the right, the symbol points one way and the shading runs the other.

In (a) the symbol < has its point to the left, yet the shading runs right. The shape of the symbol is not an instruction to shade; the direction comes from what the variable is said to be. Written variable first, as x > -3, symbol and shading agree.

Example 3

Solutions that are not whole numbers

Graph x \le 2.5, then decide which of -4, 0, 2.5, 2.49 and 2.6 are solutions.

Solution

The boundary 2.5 sits halfway between 2 and 3. Testing it gives 2.5 \le 2.5, true, so draw a filled dot at 2.5, between the marks, and shade to the left.

Figure 4Example 3. The dot sits between two marked numbers, and the shading covers every number to its left, whole or not.

Now each number in turn. -4 \le 2.5 is true: -4 lies to the left of 2.5. 0 \le 2.5 is true. 2.5 \le 2.5 is true, since the two sides are equal. 2.49 \le 2.5 is true, because 2.49 is 0.01 less than 2.5. 2.6 \le 2.5 is false, because 2.6 lies to the right of 2.5. So the solutions among them are -4, 0, 2.5 and 2.49.

The shading covers 2.49, 2.499, 2.4999 and endlessly many other numbers crowded just below 2.5. That is why the graph is a solid stretch and not a row of dots.

Example 4

Reading an inequality from a graph

Write an inequality in x for each graph.

Figure 5Example 4. Two graphs to read back as inequalities.

Solution

(a) The circle at -1 is open, so -1 is not a solution and the symbol is strict. The shading runs to the left, so the solutions are less than -1: the inequality is x < -1, which may also be written -1 > x. Check with -3, which is shaded: -3 < -1 is true.

(b) The dot at 7 is filled, so 7 is a solution and the symbol is inclusive. The shading runs to the right, so the inequality is x \ge 7, or 7 \le x. Check with 8, which is shaded: 8 \ge 7 is true.

Example 5

Ride heights

At a theme park, the first ride takes only riders who are at least 120 cm tall. The second ride takes only riders shorter than 195 cm. Write each rule as an inequality in h, a rider’s height in centimeters, and graph it. Yosef is 118 cm tall, Aiko 120 cm and Benno 197 cm. Who may go on each ride?

Solution

“At least 120” means 120 or more, and 120 itself counts, so the first rule is h \ge 120: a filled dot at 120, then shade to the right. “Shorter than 195” leaves 195 out, since 195 cm is not shorter than 195 cm, so the second rule is h < 195: an open circle at 195, then shade to the left. Turning a sentence into an inequality has its own lesson, Writing an inequality for a condition.

Figure 6Example 5. A minimum height and a maximum height.

First ride: 118 \ge 120 is false, 120 \ge 120 is true and 197 \ge 120 is true, so Aiko and Benno may ride and Yosef may not. Second ride: 118 < 195 and 120 < 195 are true and 197 < 195 is false, so Yosef and Aiko may ride and Benno may not. Only Aiko may go on both.

The arrow on the second graph runs on past 0 and into the negative numbers, which are not heights at all. The graph shows the inequality; the situation decides which part of it can actually happen.

Example 6

A lift: a measure and a count

A lift carries a plate reading “450 kg, 6 persons”. Let w be the total load in kilograms and p the number of passengers. Graph the allowed loads and the allowed numbers of passengers. Cira counts 6 people already inside when Darius arrives. May he get in?

Solution

A load can be any number of kilograms, such as 312.5 or 449.9, so the allowed loads w \le 450 get the full graph: 450 \le 450 is true, so a filled dot at 450, then shade to the left.

The number of passengers satisfies p \le 6, but p counts people, so it must be a whole number. The solutions are 0, 1, 2, 3, 4, 5 and 6: seven values, with 0 standing for an empty lift. The graph marks each with a dot and shades nothing between them, since 2.5 passengers is not a thing a lift can carry.

Figure 7Example 6. A load can be any number, so it is shaded; passengers are counted, so they get dots.

With Darius aboard there would be 7 passengers, and 7 \le 6 is false, so he waits.

Example 7

Is there a smallest solution?

Find the smallest solution, if there is one, of (a) x \ge 8, (b) x > 8, and (c) x > 8 when x must be a whole number.

Solution

(a) 8 \ge 8 is true, and every number below 8 makes the claim false. The smallest solution is 8, the filled dot.

(b) Now 8 itself fails. Take any solution, say 8.1. It is not the smallest: halfway between 8 and 8.1 is 8.05, which is still greater than 8 and less than 8.1. The same move works on any solution, since the number halfway between 8 and it is always a smaller solution. So x > 8 has no smallest solution. The open circle shows this: the shading runs right up to 8 with no first point.

(c) The whole numbers greater than 8 are 9, 10, 11 and so on, so the smallest solution is 9. Its graph is dots at 9, 10, 11 and onward, with “and so on” beside them.

Boundaries

A number line graphs an inequality with one variable compared with a number. An inequality with two variables needs the coordinate plane instead: Graphing a linear inequality in two variables.
Strict and inclusive differ at exactly one number. The graphs of x > 3 and x \ge 3 are the same except for the circle at 3.
Shading assumes every number is allowed. When the variable counts things, only whole numbers are solutions and the graph is dots. Then two different inequalities can share their solutions: for whole numbers, x > 3 and x \ge 4 both give 4, 5, 6 and so on, although for all numbers 3.5 solves the first and not the second.
A graph shows every solution of the inequality, including any the situation rules out, such as negative heights. Showing two limits at once, such as at least 120 and under 195, takes two inequalities joined by “and”: Compound inequalities: and, or.
The circle and the side are read straight off the inequality only when the variable stands alone on one side. An inequality such as x + 4 > 9 must first be solved.

Where this goes next

Solving comes next: One-step inequalities and the sign flip undoes an addition or a multiplication to get the variable alone, exactly as for an equation, with one surprise when multiplying by a negative number. Words become symbols in Writing an inequality for a condition. The shaded stretch gets a short written name in Interval and set-builder notation, where x \ge 3 becomes [3, \infty). Two conditions at once shade the overlap of two graphs in Compound inequalities: and, or. And on the coordinate plane, the same choice of open or filled edge returns as a dashed or solid boundary line in Graphing a linear inequality in two variables.

Exercises

Symbols and solutions

Is each number a solution of x \le 11? (a) 11 (b) 11.1 (c) -20 (d) 10\tfrac{3}{4}
Is each number a solution of x > -7? (a) -7 (b) -6.9 (c) 0 (d) -8
True or false: (a) 3 \ge 3 (b) 3 > 3 (c) -2 \le -5 (d) 0 < \tfrac{1}{2}
Rewrite each with the variable first: (a) 12 > x (b) -6 \le x (c) 0 < x
Write in symbols, variable first: (a) x is less than or equal to 15. (b) x is greater than -9.

Graphing

Graph each inequality on a number line.

x > 1
x \le -5
2 \ge x
-10 < x
x \ge 1.5
The whole-number solutions of x < 3.

Reading a graph

Write an inequality in x for each graph.

Figure 8Graphs for exercises 12 to 15. The number at the left of each line is the exercise number.

The graph labeled 12.
The graph labeled 13.
The graph labeled 14.
The graph labeled 15.

In context

A ride’s sign reads h \ge 107, where h is height in centimeters. Which of these riders may ride: 106 cm, 107 cm, 115.5 cm, 98 cm?
A children’s ride allows riders with h < 137, in centimeters. Describe the graph, and say whether a child exactly 137 cm tall may ride.
A small lift allows p \le 13 passengers, where p is a whole number. How many different numbers of passengers are allowed, counting an empty lift?
A ride’s graph of allowed heights h, in centimeters, has a filled dot at 132 and shading to the right. Write the inequality, and decide whether a rider 131.5 cm tall may ride.

Explain

Explain why the graph of x \ge 3 has a filled dot at 3 but the graph of x > 3 has an open circle there.
Explain why x > 11 and 11 < x have the same graph.
Does x < 10 have a largest solution? Explain.
Do x \ge 15 and x > 14 have the same solutions (a) when x must be a whole number, (b) when x may be any number?
A lift’s load w must satisfy w \le 630 (kilograms) and its number of passengers p must satisfy p \le 10. Explain why the graph for w is shaded but the graph for p is drawn as dots.

Answers

Figure 9Answers to exercises 6 to 11, each line labeled with its exercise number.

(a) yes, 11 \le 11 because the sides are equal. (b) no, 11.1 lies to the right of 11. (c) yes. (d) yes, 10\tfrac{3}{4} is less than 11.
(a) no, -7 > -7 is false. (b) yes, -6.9 lies to the right of -7. (c) yes. (d) no, -8 lies to the left of -7.
(a) true, the sides are equal. (b) false, 3 is not greater than itself. (c) false, -2 lies to the right of -5, so it is greater. (d) true.
(a) x < 12 (b) x \ge -6 (c) x > 0. Each time the sides swap and the symbol turns round.
(a) x \le 15 (b) x > -9
1 > 1 is false, so an open circle at 1; greater means shade to the right.
-5 \le -5 is true, so a filled dot at -5; less means shade to the left.
2 \ge x is x \le 2: a filled dot at 2, then shade to the left.
-10 < x is x > -10: an open circle at -10, then shade to the right.
1.5 \ge 1.5 is true, so a filled dot at 1.5, halfway between 1 and 2; shade to the right.
Dots at 0, 1 and 2 only, nothing shaded; 3 < 3 is false, so 3 gets no dot.
x < 5: open circle, so strict; shading left, so less.
x \ge -4: filled dot, so inclusive; shading right, so greater.
x \le 0: filled dot at 0, shading left.
Whole numbers x with x \le 5, or equally x < 6 (the dots are 0 to 5, with no shading and no dot at -1).
107 cm and 115.5 cm. 106 \ge 107 and 98 \ge 107 are false.
An open circle at 137, then shade to the left. The child may not ride, since 137 < 137 is false.
14: the whole numbers 0, 1, 2, \ldots, 13.
h \ge 132. The rider may not ride: 131.5 lies to the left of 132.
Put 3 in. 3 \ge 3 is true, so 3 is a solution and the dot is filled to include it. 3 > 3 is false, so 3 is not a solution, and the open circle marks the edge while leaving it out.
“x is greater than 11” and “11 is less than x” describe the same positions: x to the right of 11. Both have an open circle at 11 and shading to the right.
No. Any solution is below 10, and the number halfway between it and 10 is larger but still below 10. For example 9.9 is beaten by 9.95.
(a) Yes: both are 15, 16, 17 and so on. (b) No: 14.5 solves x > 14 but not x \ge 15.
A load can be any number of kilograms, such as 500.5, so every point up to 630 is a solution and the stretch is shaded. Passengers are counted, so only whole numbers from 0 to 10 make sense, and each gets its own dot.