One-step inequalities and the sign flip
An equation is true for one number. When the condition says “at most” or “more than”, which numbers pass, and which moves keep the condition true?
Where you are 3 in Solving linear inequalities
When you reach for it
An equation such as x + 3 = 8 is true for exactly one number, x = 5. Many conditions are not like that. A ride lets on anyone at least 120 cm tall; a ferry ramp takes any vehicle up to its weight limit; a driver may go at any speed up to the limit. Each of these is an inequality, and an inequality is usually true for a whole stretch of numbers rather than one.
Reach for this lesson when the unknown appears once, on one side, and a single operation stands between it and being alone: x + 3 < 8, y - 7 \le -10, 6x > 27, -4x \ge 28. Solving means finding every number that makes the inequality true. The moves are the ones that solve one-step equations, doing the same thing to both sides to undo the operation, with one exception: multiplying or dividing both sides by a negative number reverses the inequality sign. Drawing the answer uses inequalities on the number line.
This lesson is taught over two days. Part 1 is what a solution set is, how to graph it, and the moves that never change the sign: adding or subtracting, and multiplying or dividing by a positive number. Part 2 is multiplying or dividing by a negative number, and why that reverses the sign.
The procedure
Part 1: solution sets, and moves that keep the order
A solution is any number that makes the inequality true. Take x + 3 < 8 and try some numbers.
| x | x + 3 | Is x + 3 < 8? |
|---|---|---|
| -10 | -7 | yes |
| 4 | 7 | yes |
| 4.9 | 7.9 | yes |
| 5 | 8 | no, 8 is not less than 8 |
| 6 | 9 | no |
Every number less than 5 works, fractions, decimals and negatives included, and every number from 5 upward fails. The solution set is all the numbers that work, taken together. There are infinitely many, so they cannot be listed; instead the set is written as an inequality with the variable alone, x < 5, and drawn on a number line, as in the upper line of Figure 1. The circle at 5 is left open because 5 itself fails, and the heavy line runs left forever, which the arrow shows. For \le or \ge the endpoint is a solution, and its dot is filled in.
| Solution set | Mark at a | Heavy line runs |
|---|---|---|
| x < a | open circle | left |
| x \le a | filled dot | left |
| x > a | open circle | right |
| x \ge a | filled dot | right |
Solving means rewriting into that form without changing the set. Each move below turns an inequality into one that is true for exactly the same numbers, as each move on an equation does. What has to be asked of a move is whether it keeps the order of two numbers: if one number is less than another before the move, is it still less after?
Adding or subtracting the same number keeps the order. a < b means a sits to the left of b on the number line. Adding c to both slides every number the same distance the same way, to the right if c is positive and to the left if it is negative, so a + c is still to the left of b + c. For example 3 < 7, and subtracting 10 from both gives -7 < -3, still true. Subtracting c is adding -c, so it keeps the order too. For any numbers a, b and c:
The same holds with >, \le or \ge in place of <. This is the addition property of inequality. So x + 3 < 8 becomes x < 5 by subtracting 3 from both sides, the same move that solves x + 3 = 8.
Multiplying or dividing by a positive number keeps the order. Multiplying every number by 3 stretches the number line away from 0: each number lands three times as far from 0, on the same side it started. Nothing overtakes anything else, so -1 < 2 becomes -3 < 6 and 2 < 5 becomes 6 < 15. The same thing in symbols: if a < b then b - a is positive, and 3b - 3a = 3(b - a) is a positive number times a positive number, so it is positive too, which says 3a < 3b. Dividing by a positive number c is multiplying by \displaystyle \frac{1}{c}, which is also positive. For any c > 0:
This is the multiplication property of inequality for a positive multiplier.
An answer is checked with two numbers. The endpoint should make the two sides equal, which shows the boundary is in the right place. A number from the shaded side should make the inequality true, which shows the sign points the right way. For x < 5: 5 + 3 = 8, equal, and 0 + 3 = 3 < 8, true.
Part 2: multiplying or dividing by a negative number
Multiplying by -1 reverses the order. Start from 2 < 5 and multiply both numbers by -1. The results are -2 and -5, and -2 is the larger: -2 > -5. Figure 2 shows why. Multiplying by -1 sends every number to its opposite, the same distance from 0 on the other side, so it reflects the whole number line in 0. A mirror swaps left and right, so the number that was on the left ends up on the right.
The difference argument agrees: b - a is positive, and (-a) - (-b) = b - a, so -a is the larger of the two.
Every negative multiplier does the same. Multiplying by -3 is multiplying by 3, which keeps the order, and then by -1, which reverses it, so the order ends reversed. That uses (-3)a = -(3a), one of the sign rules for products. Dividing by a negative number is multiplying by its reciprocal, which is negative too. For any c < 0:
So < turns into >, > into <, \le into \ge, and \ge into \le. Whether the sign includes “or equal” never changes: a strict sign stays strict.
The flip goes with the multiplier and nothing else. To solve -3x < 12, divide both sides by -3, a negative number, and reverse the sign:
Check: at x = -4 the sides are equal, 12 = 12, and x = 0 gives 0 < 12, true. A negative number anywhere else does not flip anything. To solve 3x < -12, divide by 3, which is positive, and the sign stays: x < -4. Adding or subtracting a negative number keeps the order too, since the addition property holds whatever the sign of c.
The flip can be avoided, which confirms it. Solve -3x < 12 again using only the moves from Part 1. Adding 3x to both sides gives 0 < 12 + 3x. Subtracting 12 gives -12 < 3x. Dividing by 3, a positive number, gives -4 < x. An inequality can be read from either end, since p < q and q > p say the same thing, so -4 < x is x > -4: the answer the flip gave.
Worked examples
Examples 1 to 4 belong to Part 1 and Examples 5 to 7 to Part 2.
Adding or subtracting to leave the variable alone
Solve each inequality and graph its solution set. (a) n - 6 \ge -2 \qquad (b) 9 > y + 13
Solution(a) The 6 is subtracted from n, so add 6 to both sides. Adding keeps the order.
The graph is the lower line of Figure 1: a filled dot at 4, because \ge includes 4, and a heavy line running right. Check: n = 4 gives 4 - 6 = -2, equal to the right side, and n = 10 gives 4 \ge -2, true.
(b) The variable is on the right, which changes nothing. Subtract 13 from both sides:
Read from the other end, -4 > y is y < -4. Graph: an open circle at -4, since the sign is strict, and a heavy line running left. Check: y = -4 gives 9 and 9, equal; y = -5 gives 9 > 8, true.
Fractions and decimals
Solve (a) \displaystyle a + \frac{3}{4} < \frac{1}{2} \qquad (b) -1.6 \le k - 2.5
Solution(a) Subtract \displaystyle \frac{3}{4} from both sides, over the common denominator 4:
So \displaystyle a < -\frac{1}{4}: an open circle a quarter of the way from 0 to -1, and a heavy line running left. Check: \displaystyle a = -\frac{1}{4} gives \displaystyle -\frac{1}{4} + \frac{3}{4} = \frac{1}{2}, equal; a = -1 gives \displaystyle -\frac{1}{4} < \frac{1}{2}, true.
(b) Add 2.5 to both sides: -1.6 + 2.5 \le k, so 0.9 \le k, which is k \ge 0.9. Check: k = 1 gives -1.6 \le -1.5, true, since -1.6 lies to the left of -1.5.
Multiplying or dividing by a positive number
Solve (a) 6x > 27 \qquad (b) \displaystyle \frac{k}{5} \le -2
Solution(a) x is multiplied by 6, so divide both sides by 6. Six is positive, so the sign stays.
Check: x = 4.5 gives 27, equal to the right side, and x = 5 gives 30 > 27, true.
(b) k is divided by 5, so multiply both sides by 5, a positive number: k \le -2 \cdot 5, so k \le -10. The right side is negative, but the number multiplying is 5, so the sign stays. Check: k = -15 gives -3 \le -2, true, and k = 0, outside the set, gives 0 \le -2, false.
Two limits on the road
(a) Eskil drives a truck 270 km along a highway with a speed limit of 90 km/h. How long can the trip take if he never goes over the limit? (b) A ferry ramp has a weight limit of 3500 kg. Bettina’s van weighs 2140 kg empty. How much cargo can it carry onto the ferry?
Solution(a) Let the trip take t hours. Never going over 90 km/h, the truck covers at most 90t km in that time, and it has to cover 270 km, so
Dividing both sides by 90, a positive number, gives t \ge 3. The trip takes at least 3 hours. Check: in exactly 3 hours at 90 km/h the truck covers 270 km, and a shorter time, say 2.5 hours, would need 270 \div 2.5 = 108 km/h.
(b) Let the cargo weigh c kg. The loaded van weighs 2140 + c kg, which must be at most the limit:
Subtracting 2140 from both sides gives c \le 1360. The algebra allows every number up to 1360, but a load cannot weigh less than nothing, so the cargo is anywhere from 0 to 1360 kg.
Dividing by a negative number
Solve and graph (a) -4x \ge 28 \qquad (b) 4x \ge -28
Solution(a) x is multiplied by -4, so divide both sides by -4. That is a negative number, so the sign reverses, \ge to \le:
The graph is the upper line of Figure 3, a filled dot at -7 and a heavy line running left. Check: x = -7 gives 28 \ge 28, equal, and x = -10 gives 40 \ge 28, true.
(b) Now divide by 4, which is positive, so the sign stays: \displaystyle x \ge \frac{-28}{4} = -7. The graph is the lower line of Figure 3. Check: x = 0 gives 0 \ge -28, true.
The two inequalities share the endpoint -7 and point opposite ways. The -28 in (b) is negative, but it is the number being divided, not the divisor, so it flips nothing.
Negative fractions and a lone minus sign
Solve (a) \displaystyle \frac{m}{-3} < 2 \qquad (b) \displaystyle -\frac{2}{3}p \le 8 \qquad (c) -y > 5
Solution(a) m is divided by -3, so multiply both sides by -3 and reverse the sign: m > 2 \cdot (-3), so m > -6. Check: m = -6 gives 2 < 2, the boundary case of equality; m = 0 gives 0 < 2, true; m = -9, outside the set, gives 3 < 2, false.
(b) Multiply both sides by \displaystyle -\frac{3}{2}, the reciprocal of \displaystyle -\frac{2}{3}. It is negative, so \le becomes \ge:
Check: p = -12 gives \displaystyle -\frac{2}{3} \cdot (-12) = 8, equal, and p = 0 gives 0 \le 8, true.
(c) -y means -1 \cdot y. Multiply both sides by -1 and reverse the sign: y < -5. This is the reflection of Figure 2 exactly: -y lies to the right of 5 just when y lies to the left of -5. Check: y = -6 gives 6 > 5, true.
What each move does to any pair
Suppose a < b, without knowing what the numbers are. Decide which is larger in each pair. (a) a - 9 and b - 9 \qquad (b) -2a and -2b \qquad (c) \displaystyle \frac{a}{-5} and \displaystyle \frac{b}{-5} \qquad (d) 10 - a and 10 - b
Solution(a) Subtracting 9 keeps the order, so a - 9 < b - 9.
(b) Multiplying by -2, a negative number, reverses it: -2a > -2b.
(c) Dividing by -5 reverses it too: \displaystyle \frac{a}{-5} > \frac{b}{-5}.
(d) This takes two facts. Multiplying a < b by -1 gives -a > -b, and adding 10 to both sides keeps that order, so 10 - a > 10 - b. Taking a number away from 10 reverses order, because it reflects first and then slides.
Testing with a = 1 and b = 4: (a) -8 < -5, (b) -2 > -8, (c) -0.2 > -0.8, (d) 9 > 6. All four agree, but the test only illustrates; the reasons above are what make the answers hold for every pair.
Boundaries
Never multiply or divide both sides by zero. Dividing by zero is undefined, and multiplying by zero destroys the order: 2 < 5 times 0 is 0 < 0, which is false. Zero is neither positive nor negative, so neither rule covers it.
The sign of the multiplier must be known. The whole rule turns on whether the number is positive or negative. To solve kx < 6 for x when k is a letter, there are three cases: k positive keeps the sign, k negative reverses it, and k = 0 leaves 0 < 6, true for every x. The answers differ, so a single answer cannot be given until the sign of k is known.
Only numbers on the number line can be compared. The order these rules preserve or reverse is left and right on the real number line; the rules mean nothing for objects with no such order.
A setting can trim the set. The algebra returns every number on one side of the endpoint. A weight cannot be negative, a count of jars is a whole number, and a time is positive, so in a story the answer is the part of the solution set the setting allows, as in Example 4.
The unknown appears once, on one side. With more than one operation on the unknown, or the unknown on both sides, the same properties apply several times in a row; that is the work of the next two lessons.
Where this goes next
Two-step inequalities undo two operations in turn, which means watching for the flip at the step that multiplies or divides. Multi-step inequalities add distributing and collecting the unknown from both sides, and there the alternative from Part 2, moving the term across so its coefficient stays positive, is often the neater path. Together these make up solving linear inequalities.
The solution sets here are rays on the number line, and interval notation writes them compactly: x < 5 is (-\infty, 5) and n \ge 4 is [4, \infty). Turning a sentence such as “at most” or “no fewer than” into a symbol is writing an inequality for a condition. Much later, quadratic inequalities meet the same question in a new form: before multiplying by an expression, you must know its sign, and when it can be either, the work splits into cases.
Exercises
Part 1: solution sets and graphs
- Which of -3, 0, 2, 2.5 and 4 are solutions of x + 1.5 < 4?
- Write the inequality in x that each graph shows. (a) A filled dot at -2, heavy line running right. (b) An open circle at 3, heavy line running left.
Part 1: adding and subtracting
Solve each inequality and describe its graph.
- x + 9 > 4
- y - 7 \le -10
- -6 \ge k - 1.5
- \displaystyle p - \frac{2}{3} < \frac{1}{6}
Part 1: multiplying and dividing by a positive number
- 8x \le 36
- \displaystyle \frac{w}{6} > -3
- 0.4n \ge 6
- \displaystyle \frac{3}{5}t < 9
Part 1: in context
- A ride lets on riders at least 120 cm tall. Aurelio is 104 cm tall. How much must he grow before he can ride?
- Fiorella packs jars weighing 0.75 kg each into a shipping box rated for contents of up to 18 kg. How many jars can she pack?
- A bridge has a weight limit of 7.5 tonnes. A truck weighs 4.8 tonnes empty. What loads can it carry across?
Part 2: multiplying and dividing by a negative number
- -5x > 35
- -2y \le -14
- \displaystyle \frac{z}{-4} \ge 3
- -q < 6
- \displaystyle -\frac{3}{4}m > -6
- 1.2 > -0.3r
- Solve 6x < -42 and -6x < 42, and say why only one of them reverses the sign.
- When a number is multiplied by -4, the result is at least 20. Which numbers could it be?
Reasoning
- Using the number line, explain why a < b means -a > -b.
- Does solving \displaystyle \frac{x}{3} < -2 reverse the sign? Solve it, and justify with a test number.
- Explain why both sides of an inequality may not be multiplied by 0.
- Solve -7x \ge 21 without dividing by a negative number, and show the answer agrees with the one the flip gives.
Answers
- -3, 0 and 2. Subtracting 1.5 gives x < 2.5. The number 2.5 is not a solution, since 2.5 + 1.5 = 4 and 4 is not less than 4.
- (a) x \ge -2: the filled dot includes -2, and the line running right means greater. (b) x < 3: the open circle leaves 3 out, and the line running left means less.
- x > -5. Subtract 9: x > 4 - 9. Open circle at -5, heavy line running right.
- y \le -3. Add 7: y \le -10 + 7. Filled dot at -3, heavy line running left.
- k \le -4.5. Add 1.5: -4.5 \ge k, which is read from the other end. Filled dot at -4.5, heavy line running left.
- \displaystyle p < \frac{5}{6}. Add \displaystyle \frac{2}{3} = \frac{4}{6}: \displaystyle p < \frac{1}{6} + \frac{4}{6}. Open circle at \displaystyle \frac{5}{6}, heavy line running left.
- \displaystyle x \le \frac{9}{2}, that is x \le 4.5. Divide by 8, positive, so the sign stays: \displaystyle \frac{36}{8} = \frac{9}{2}.
- w > -18. Multiply by 6, positive: w > -3 \cdot 6.
- n \ge 15. Divide by 0.4: 6 \div 0.4 = 15.
- t < 15. Multiply by \displaystyle \frac{5}{3}, positive: \displaystyle 9 \cdot \frac{5}{3} = 15.
- At least 16 cm. With growth g cm, 104 + g \ge 120, so g \ge 16.
- At most 24 jars. With j jars, 0.75j \le 18, so j \le 24; a number of jars is a whole number, so she can pack anywhere from 0 to 24.
- Any load from 0 up to 2.7 tonnes. With load \ell tonnes, 4.8 + \ell \le 7.5, so \ell \le 2.7, and a load cannot be negative.
- x < -7. Divide by -5 and reverse: \displaystyle x < \frac{35}{-5}.
- y \ge 7. Divide by -2 and reverse: \displaystyle y \ge \frac{-14}{-2}.
- z \le -12. Multiply by -4 and reverse: z \le 3 \cdot (-4).
- q > -6. Multiply by -1 and reverse.
- m < 8. Multiply by \displaystyle -\frac{4}{3} and reverse: \displaystyle -6 \cdot \left(-\frac{4}{3}\right) = 8.
- r > -4. Divide by -0.3 and reverse: \displaystyle \frac{1.2}{-0.3} < r, that is -4 < r.
- 6x < -42 gives x < -7; -6x < 42 gives x > -7. The first is divided by 6, positive, so its sign stays even though -42 is negative. The second is divided by -6, so its sign reverses.
- Any number n \le -5. The condition is -4n \ge 20; divide by -4 and reverse. Check: n = -6 gives 24 \ge 20.
- Multiplying by -1 sends each number to the point the same distance from 0 on the other side, which reflects the number line in 0. a < b says a is to the left of b; a reflection swaps left and right, so -a is to the right of -b, which is -a > -b.
- No. x < -6: multiply by 3, which is positive, so the sign stays, and the negative number on the right changes nothing. Test: x = -9 gives -3 < -2, true, while x = 0 gives 0 < -2, false, so the set does lie to the left of -6.
- Multiplying by 0 makes both sides 0, which wipes out the difference the inequality was about: 2 < 5 would become 0 < 0, which is false. A move that turns a true statement false cannot be used to solve. Zero is neither positive nor negative, so neither property covers it.
- x \le -3. Add 7x: 0 \ge 21 + 7x. Subtract 21: -21 \ge 7x. Divide by 7, positive: -3 \ge x, which is x \le -3. Dividing -7x \ge 21 by -7 and reversing gives the same x \le -3.