Unit 2, lesson 13

Solving Inequalities on a Number Line

How to solve linear and compound inequalities step by step, why multiplying by a negative number flips the sign, and how to write the exact solution set as an interval.

  • Theory part 25
  • Practice part 26
  • Solving Equations and Inequalities

A phone plan costs 20 dollars a month plus 5 dollars per gigabyte, and you want the bill to stay under 45 dollars. With gg gigabytes the bill is 20+5g20 + 5g, and 20+5g<4520 + 5g < 45 holds exactly when 5g<255g < 25, that is when g<5g < 5. So the answer is not one number but a whole stretch, every amount of data from 00 up to but not including 55 gigabytes. Inequalities describe stretches like that, and calculus uses them constantly, from saying how close xx is to a point to saying where a function is increasing.

Key terms

Inequality

Definition An inequality is a statement that compares two real expressions with one of the order symbols <<, ≤\le, >> or ≥\ge. For real numbers, a<ba < b means that b−ab - a is positive, and a>ba > b means the same as b<ab < a. An inequality in a variable xx may be true for some real values of xx and false for others, so it makes sense to ask exactly which values make it true.

In plain words An inequality says which of two quantities is bigger. On a number line, a<ba < b means aa sits to the left of bb.

Example The inequality 3x−5<73x - 5 < 7 is true at x=2x = 2, since 6−5=1<76 - 5 = 1 < 7. It is false at x=5x = 5, since 15−5=1015 - 5 = 10 is not less than 77.

Strict and weak inequalities

Definition The inequalities a<ba < b and a>ba > b are strict, and they exclude equality. The inequalities a≤ba \le b and a≥ba \ge b are weak (also called non-strict). The statement a≤ba \le b means a<ba < b or a=ba = b.

In plain words A strict inequality leaves the boundary number out. A weak one lets it in. On a number line the boundary gets an open circle when it is left out and a filled dot when it is included.

Example The solutions of x<4x < 4 are every number to the left of 44, drawn with an open circle at 44. The solutions of x≤4x \le 4 are the same numbers together with 44 itself, drawn with a filled dot at 44. The only difference between the two sets is the single number 44.

Solution set

Definition The solution set of an inequality in xx is the set of all real numbers xx in the domain of both sides that make the inequality true. Two inequalities are equivalent when they have the same solution set. Solving an inequality means replacing it by simpler equivalent inequalities until the solution set can be read off.

In plain words The solution set is the complete list of numbers that work, nothing missing and nothing extra. Each step of a solution must keep exactly the same set.

Example Solve 2x+1≥72x + 1 \ge 7. Subtracting 11 from both sides gives 2x≥62x \ge 6, and dividing by the positive number 22 gives x≥3x \ge 3. A quick test agrees. At x=3x = 3 the left side is 77, which satisfies 7≥77 \ge 7, and at x=2x = 2 it is 55, which does not.

Operations that keep the order

Definition For all real numbers aa, bb and cc,

a<bimpliesa+c<b+c,a < b \quad \text{implies} \quad a + c < b + c,

and if c>0c > 0,

a<bimpliesac<bcandac<bc.\begin{gathered} a < b \quad \text{implies} \\[0.4em] ac < bc \quad \text{and} \quad \frac{a}{c} < \frac{b}{c}. \end{gathered}

The same holds with ≤\le in place of <<. Each of these steps can be undone by subtracting cc or by multiplying by 1c>0\frac{1}{c} > 0, so they turn an inequality into an equivalent one.

In plain words Sliding both numbers the same distance along the line does not change which is on the left. Stretching the line by a positive factor does not change it either.

Example Solve 3x−5<73x - 5 < 7. Adding 55 to both sides gives 3x<123x < 12, and dividing by 33 gives x<4x < 4. The solution set is every real number less than 44.

Order reversal

Definition If a<ba < b and c<0c < 0, then

ac>bcandac>bc.ac > bc \quad \text{and} \quad \frac{a}{c} > \frac{b}{c}.

So multiplying or dividing both sides of an inequality by a negative number reverses the direction of the inequality sign, and ≤\le becomes ≥\ge in the same way. The reason is that b−a>0b - a > 0 and c<0c < 0 give (b−a)c<0(b - a)c < 0, which is bc<acbc < ac. Because of this, never multiply an inequality by an expression such as xx or x−1x - 1 unless you know its sign. If the sign can change, split into cases where it is positive and where it is negative, and leave out the values where it is zero.

In plain words Multiplying by a negative number mirrors the number line around 00, so left and right trade places. Since 2<52 < 5 but −2>−5-2 > -5, the sign has to flip.

Example Solve −3x+4>10-3x + 4 > 10. Subtracting 44 gives −3x>6-3x > 6. Dividing by −3-3 reverses the sign,

x<6−3=−2.x < \frac{6}{-3} = -2.

Check a number on each side. At x=−3x = -3 the left side is 13>1013 > 10, true. At x=0x = 0 it is 4>104 > 10, false. The solution set is all x<−2x < -2.

Now the warning about variable factors. In 1x<2\frac{1}{x} < 2 you may not just multiply by xx. For x>0x > 0 the sign stays and you get x>12x > \frac{1}{2}. For x<0x < 0 it flips, giving 1>2x1 > 2x, which every negative xx satisfies. The solution set is (−∞,0)∪(12,∞)(-\infty, 0) \cup \left(\frac{1}{2}, \infty\right), and multiplying blindly would have lost every negative solution.

Compound inequality

Definition A compound inequality such as a<x<ba < x < b is shorthand for two comparisons joined by “and”, here a<xa < x and x<bx < b. A number is a solution exactly when it satisfies both, so the solution set is the intersection of the two separate solution sets. The same goes for mixed forms like a≤x<ba \le x < b. Such a chain is only written when a<ba < b, since a chain like 5<x<25 < x < 2 has no solutions at all.

In plain words A compound inequality traps xx between two walls. You can apply the same legal step to all three parts at once, as long as you flip both signs when you multiply or divide by a negative number.

Example Solve −1≤2x+3<9-1 \le 2x + 3 < 9. Subtract 33 from all three parts and then divide all three by 22.

−4≤2x<6so−2≤x<3.-4 \le 2x < 6 \quad \text{so} \quad -2 \le x < 3.

The left end −2-2 is included and the right end 33 is not. At x=−2x = -2 the middle is −1-1, which satisfies −1≤−1<9-1 \le -1 < 9, and at x=3x = 3 it is 99, which fails 9<99 < 9.

Interval notation

Definition For real numbers a<ba < b, interval notation names these sets.

IntervalMeaningEndpoints
(a,b)(a, b)a<x<ba < x < bboth excluded
[a,b][a, b]a≤x≤ba \le x \le bboth included
[a,b)[a, b)a≤x<ba \le x < baa included
(a,b](a, b]a<x≤ba < x \le bbb included
(a,∞)(a, \infty)x>ax > aaa excluded
[a,∞)[a, \infty)x≥ax \ge aaa included
(−∞,b)(-\infty, b)x<bx < bbb excluded
(−∞,b](-\infty, b]x≤bx \le bbb included

A parenthesis marks an open endpoint that is left out, and a square bracket marks a closed endpoint that is included. The symbols ∞\infty and −∞-\infty are not real numbers, so they always get a parenthesis. The whole real line is (−∞,∞)(-\infty, \infty), and a union of separate pieces is written with ∪\cup.

In plain words An interval is a shorthand for a stretch of the number line. Round brackets match open circles in the picture and square brackets match filled dots.

Example The solution of −1≤2x+3<9-1 \le 2x + 3 < 9 was −2≤x<3-2 \le x < 3, which is the interval [−2,3)[-2, 3). The solution of −3x+4>10-3x + 4 > 10 was x<−2x < -2, which is (−∞,−2)(-\infty, -2).

Check yourself

  1. Solve 4−5x≥194 - 5x \ge 19 and write the answer in interval notation.
Show the answer

Subtracting 44 gives −5x≥15-5x \ge 15. Dividing by −5-5 reverses the sign, so x≤−3x \le -3. The solution set is (−∞,−3](-\infty, -3]. As a check, at x=−3x = -3 the left side is exactly 1919.

  1. Solve the compound inequality 1<5−2x≤91 < 5 - 2x \le 9 and draw the solution on a number line.
Show the answer

Subtract 55 from all parts to get −4<−2x≤4-4 < -2x \le 4. Dividing by −2-2 flips both signs,

2>x≥−2,2 > x \ge -2,

which reads −2≤x<2-2 \le x < 2. The solution set is [−2,2)[-2, 2), with a filled dot at −2-2 and an open circle at 22. At x=2x = 2 the middle equals 11, and 1<11 < 1 is false.

  1. Solve 2(x−1)<5x+72(x - 1) < 5x + 7.
Show the answer

Expanding gives 2x−2<5x+72x - 2 < 5x + 7. Subtracting 5x5x and adding 22 gives −3x<9-3x < 9, and dividing by −3-3 reverses the sign, so x>−3x > -3. The solution set is (−3,∞)(-3, \infty).

How this lesson connects