Unit 2 · Lesson 5 of 6

Linear and compound inequalities, interval notation

About 25 minutes · 5 steps · one idea: solve like an equation, flip for negatives

You'll be able to: solve inequalities such as \(5 - 3x \ge 17\) and \(-3 \lt 2x + 1 \le 7\), and write the answer in interval notation.

  1. 1 Warm-up
  2. 2 The idea
  3. 3 Examples
  4. 4 Practice
  5. 5 Wrap-up

Step 1 of 5Warm-up

Two quick questions from earlier lessons. The second one is about signs, which matter a lot today.

The sum of two consecutive integers is 45. What is the smaller one?

Let \(x\) = the smaller integer. The next one is \(x + 1\).

\(x + (x + 1) = 45\), so \(2x + 1 = 45\), \(2x = 44\), \(x = 22\).

Check: \(22 + 23 = 45\).

Compute \((-12) \div 3 - (-5)\).

Divide first: \((-12) \div 3 = -4\) (different signs give a negative).

Then \(-4 - (-5) = -4 + 5 = 1\).

Step 2 of 5The idea

An inequality like \(x \lt 3\) has many solutions: every number less than 3. You solve it almost exactly like an equation, with one new rule.

The new rule: when you multiply or divide both sides by a negative number, flip the sign. For example, \(-2x \lt 6\) becomes \(x \gt -3\).

Interval notation: write the two endpoints of the answer: where it starts on the left and where it ends on the right. A round bracket ( ) means the endpoint is not included (open circle). A square bracket [ ] means it is included (filled dot). Infinity \(\infty\) always gets a round bracket.

\(x \gt 3\) is \((3, \infty)\). \(\;x \le -2\) is \((-\infty, -2]\). \(\;-2 \lt x \le 3\) is \((-2, 3]\).

Compound inequalities: a three-part one like \(-3 \lt 2x + 1 \le 7\) is an "and": do the same thing to all three parts. For an "or", solve each piece and join them with \(\cup\) (union).

Why this shows up on the placement test: inequality questions usually want the answer in interval notation, and the flip rule is the classic trap.

Step 3 of 5Worked examples

Example 1: \(5 - 3x \ge 17\)

  1. Subtract 5 from both sides: \(-3x \ge 12\).
  2. Divide both sides by \(-3\) and flip the sign: \(x \le -4\). Dividing by a negative flips \(\ge\) into \(\le\).
  3. Test \(x = -5\): \(5 - 3(-5) = 20\), and \(20 \ge 17\) is true.
  4. Interval notation: \((-\infty, -4]\). The bracket is square because \(-4\) is included.
−6−5−4−3−2−10123456
\(x \le -4\): a filled dot at \(-4\), shaded to the left.

Example 2: \(-3 \lt 2x + 1 \le 7\), then an "or"

  1. Subtract 1 from all three parts: \(-4 \lt 2x \le 6\).
  2. Divide all three parts by 2 (positive, so no flip): \(-2 \lt x \le 3\). In interval notation: \((-2, 3]\).
−6−5−4−3−2−10123456
\(-2 \lt x \le 3\): an open circle at \(-2\), a filled dot at 3, shaded between.
  1. Now: \(x + 2 \lt -1\) or \(3x \gt 12\). Solve each piece: \(x \lt -3\) or \(x \gt 4\).
  2. Join with a union: \((-\infty, -3) \cup (4, \infty)\).
−6−5−4−3−2−10123456
\(x \lt -3\) or \(x \gt 4\): two separate pieces going outward.

Step 4 of 5Practice

Give your answers in interval notation if you can, like (-inf, 4]. Type inf for \(\infty\) and U for \(\cup\). Then press Check.

Which interval notation means \(x \ge -2\)?

  1. \((-2, \infty)\)
  2. \((-\infty, -2]\)
  3. \([-2, \infty)\)
  4. \([-2, \infty]\)

The numbers start at \(-2\) (included, so a square bracket) and go up forever (infinity always gets a round bracket): \([-2, \infty)\). Answer (c).

(a) leaves out \(-2\), (b) points the wrong way, and (d) puts a square bracket on \(\infty\).

Solve \(3x + 2 \lt 14\).

Subtract 2: \(3x \lt 12\). Divide by 3 (positive, no flip): \(x \lt 4\).

Answer: \((-\infty, 4)\).

Solve \(-4x \le 20\).

Divide both sides by \(-4\) and flip the sign: \(x \ge -5\).

Test \(x = 0\): \(0 \le 20\) is true. Answer: \([-5, \infty)\).

Solve \(7 - 2x \gt 1\).

Subtract 7: \(-2x \gt -6\).

Divide by \(-2\) and flip: \(x \lt 3\).

Test \(x = 0\): \(7 \gt 1\) is true. Answer: \((-\infty, 3)\).

Solve \(2(x - 3) \ge 5x + 3\).

Distribute: \(2x - 6 \ge 5x + 3\).

Subtract \(5x\): \(-3x - 6 \ge 3\). Add 6: \(-3x \ge 9\).

Divide by \(-3\) and flip: \(x \le -3\). Answer: \((-\infty, -3]\).

Solve \(-5 \le 3x + 1 \lt 10\).

Subtract 1 from all three parts: \(-6 \le 3x \lt 9\).

Divide all three parts by 3: \(-2 \le x \lt 3\).

Answer: \([-2, 3)\).

Solve \(x - 1 \lt -4\) or \(2x + 3 \ge 9\).

First piece: add 1, so \(x \lt -3\).

Second piece: subtract 3, \(2x \ge 6\), so \(x \ge 3\).

"Or" means a union. Answer: \((-\infty, -3) \cup [3, \infty)\).

Solve \(-1 \le \dfrac{3 - x}{2} \lt 4\).

Multiply all three parts by 2 to clear the fraction: \(-2 \le 3 - x \lt 8\).

Subtract 3 from all three parts: \(-5 \le -x \lt 5\).

Multiply all three parts by \(-1\) and flip both signs: \(5 \ge x \gt -5\), which reads \(-5 \lt x \le 5\).

Answer: \((-5, 5]\).

Step 5 of 5Wrap-up

You can now solve linear and compound inequalities, flip the sign when you multiply or divide by a negative, and write the answer in interval notation.

Read or watch next: OpenStax Intermediate Algebra 2e, 2.5 Solve Linear Inequalities (free textbook, with more practice at the end of the section).

Also good: Khan Academy: Algebra 1: Multi-step inequalities (short videos). Quick rules: Equations and inequalities cheat sheet.

Something unclear? Ask your teacher bot a follow-up question, for example: "Why does dividing by a negative number flip the inequality sign?" It knows this course and can explain things another way.