Properties of Inequalities

What happens to an inequality when the same number is added to both sides, subtracted from them, or when both sides are multiplied or divided by the same number; when the sign must be reversed and when it must not; and how to solve one-step inequalities.

Before you start

To solve an equation, we do the same thing to both sides until the unknown is alone: $x + 5 = 12$ gives $x = 12 - 5 = 7$ .

An inequality is solved in the same way, with one difference, which this lesson is about.

On a number line, an open circle shows $<$ or $>$ and a filled circle shows $\le$ or $\ge$ . With the unknown on the left, the arrow goes right for $>$ and $\ge$ , and left for $<$ and $\le$ .

Adding and Subtracting

Start with a true inequality, $3 < 5$ . Add 2 to both sides: $5 < 7$ , still true. Subtract 4 from both sides of $3 < 5$ : $-1 < 1$ , still true. Adding or subtracting moves both numbers the same distance along the number line, so the one on the left stays on the left.

Rule 1: adding or subtracting

If $a < b$ , then $a + c < b + c$ and $a - c < b - c$ , for any number $c$ .

Adding or subtracting the same number on both sides of an inequality does not change the sign. This is true for $<$ , $>$ , $\le$ and $\ge$ , and for any number $c$ , positive or negative.

Example 1: Solving by adding or subtracting

Solve each inequality and show the solution on a number line.
(a) $x + 5 < 12$
(b) $x - 3 \ge 8$
(c) $x + 4 > 1$

Solution

(a) Subtract 5 from both sides: $x + 5 - 5 < 12 - 5$ , so $x < 7$ .

(b) Add 3 to both sides: $x - 3 + 3 \ge 8 + 3$ , so $x \ge 11$ .

(c) Subtract 4 from both sides: $x + 4 - 4 > 1 - 4$ , so $x > -3$ . The answer is negative, but nothing was multiplied or divided, so the sign stays as it was.

(a) $x < 7$ (b) $x \ge 11$ (c) $x > -3$

Multiplying and Dividing by a Positive Number

Start with $2 < 5$ . Multiply both sides by 3: $6 < 15$ , still true. Divide both sides of $2 < 5$ by 2: $1 < 2.5$ , still true. Multiplying or dividing by a positive number stretches or shrinks the distances from zero, but keeps the order.

Rule 2: multiplying or dividing by a positive number

If $a < b$ and $c$ is positive, then $ac < bc$ and $\frac{a}{c} < \frac{b}{c}$ .

Multiplying or dividing both sides by the same positive number does not change the sign.

Example 2: Solving by multiplying or dividing by a positive number

Solve each inequality and show the solution on a number line.
(a) $3x \le 15$
(b) $\frac{x}{2} > 4$
(c) $3x < -6$

Solution

(a) Divide both sides by 3, a positive number: $\frac{3x}{3} \le \frac{15}{3}$ , so $x \le 5$ .

(b) Multiply both sides by 2, a positive number: $\frac{x}{2} \times 2 > 4 \times 2$ , so $x > 8$ .

(c) Divide both sides by 3: $\frac{3x}{3} < \frac{-6}{3}$ , so $x < -2$ . The number on the right is negative, but we divided by positive 3, so the sign does not reverse.

(a) $x \le 5$ (b) $x > 8$ (c) $x < -2$

Multiplying and Dividing by a Negative Number

Start with $2 < 5$ and multiply both sides by $-1$ . The two sides become $-2$ and $-5$ . On the number line $-2$ is to the right of $-5$ , so $-2 > -5$ : the order has been reversed. Multiplying $2 < 5$ by $-3$ gives $-6$ and $-15$ , and again $-6 > -15$ .

Multiplying by a negative number moves every number other than 0 to the other side of zero, like a reflection in a mirror at 0, and that turns the order round. So the sign must be reversed too.

Rule 3: multiplying or dividing by a negative number

If $a < b$ and $c$ is negative, then $ac > bc$ and $\frac{a}{c} > \frac{b}{c}$ .

When both sides are multiplied or divided by the same negative number, the inequality sign must be reversed: $<$ becomes $>$ , $>$ becomes $<$ , $\le$ becomes $\ge$ and $\ge$ becomes $\le$ .

Write the reversed sign in the same line as the multiplication or division, for example $\frac{-2x}{-2} < \frac{10}{-2}$ from $-2x > 10$ . Every line you write must be true.

Example 3: When does the sign reverse?

It is true that $8 > 6$ . Do each of these to both sides, and say whether the inequality sign stays the same or is reversed.
(a) Multiply by 3.
(b) Multiply by $-5$ .
(c) Divide by 2.
(d) Divide by $-4$ .

Solution

(a) $8 \times 3 = 24$ and $6 \times 3 = 18$ . $24 > 18$ , so the sign stays the same.

(b) $8 \times (-5) = -40$ and $6 \times (-5) = -30$ . $-40$ is to the left of $-30$ , so $-40 < -30$ : the sign is reversed.

(d) $8 \div (-4) = -2$ and $6 \div (-4) = -1.5$ . $-2$ is to the left of $-1.5$ , so $-2 < -1.5$ : the sign is reversed.

The sign is reversed in (b) and (d), where a negative number is used, and stays the same in (a) and (c).

Example 4: Solving by multiplying or dividing by a negative number

Solve each inequality and show the solution on a number line.
(a) $-2x > 10$
(b) $-\frac{x}{4} \le 3$
(c) $-5x \ge -15$

Solution

(a) Divide both sides by $-2$ . Dividing by a negative number reverses the sign, so $>$ becomes $<$ :

$\frac{-2x}{-2} < \frac{10}{-2}\text{,}$ so $x < -5$ .

Check with a number from the answer, $x = -6$ : $-2(-6) = 12$ , and $12 > 10$ is true.

(b) Multiply both sides by $-4$ . Multiplying by a negative number reverses the sign, so $\le$ becomes $\ge$ :

$-\frac{x}{4} \times (-4) \ge 3 \times (-4)\text{,}$ so $x \ge -12$ .

Check with $x = 0$ : $-\frac{0}{4} = 0$ , and $0 \le 3$ is true.

Check with $x = 0$ : $-5(0) = 0$ , and $0 \ge -15$ is true.

(a) $x < -5$ (b) $x \ge -12$ (c) $x \le 3$

Does the Sign Reverse?

Only one step reverses the sign while the sides stay in place: multiplying or dividing both sides by a negative number. (Swapping the sides, Rule 4, also turns the sign round, but does not change what the inequality says.) A negative number somewhere else in the inequality does not matter.

Step	Inequality	Solution	Sign reversed?
Adding or subtracting a positive number	${x - 6 > -2}$	$x > 4$	No
Adding or subtracting a negative number	${x + (-3) < 4}$	$x < 7$	No
Dividing by a positive number, with a negative number on the other side	${4x \ge -20}$	$x \ge -5$	No
Multiplying or dividing by a negative number	${-4x \ge 20}$	$x \le -5$	Yes

Example 5: Reverse or not?

Solve each inequality, saying each time whether the sign is reversed. Show each solution on a number line.
(a) $x - 8 > -3$
(b) $6x \ge -18$
(c) $-6x \ge 18$
(d) $-x < 7$

Solution

(a) Add 8 to both sides: $x > -3 + 8$ , so $x > 5$ . Adding does not reverse the sign.

(b) Divide both sides by 6, a positive number: $x \ge \frac{-18}{6}$ , so $x \ge -3$ . Not reversed: the $-18$ does not matter.

(d) $-x$ is $-1 \times x$ . Multiply both sides by $-1$ , which reverses the sign: $x > -7$ .

(a) $x > 5$ , not reversed (b) $x \ge -3$ , not reversed (c) $x \le -3$ , reversed (d) $x > -7$ , reversed

Swapping the Two Sides

Rule 4: swapping the sides

An inequality may be written the other way round, with its two sides swapped and its sign turned round with them: $12 > x$ is the same as $x < 12$ , and $-6 \le 2x$ is the same as $2x \ge -6$ .

This does not change what the inequality says. It is used to put the unknown on the left before you solve or draw.

Example 6: The unknown on the right

Solve each inequality and show the solution on a number line.
(a) $12 > x + 4$
(b) $-6 \le 2x$
(c) $10 < -5x$

Solution

(a) Swap the sides: $x + 4 < 12$ . Subtract 4 from both sides: $x < 8$ .

(b) Swap the sides: $2x \ge -6$ . Divide both sides by 2: $x \ge -3$ .

(c) Swap the sides: $-5x > 10$ . Divide both sides by $-5$ , which reverses the sign: $x < \frac{10}{-5}$ , so $x < -2$ . The sign changed twice: once because the sides were swapped, and once because we divided by $-5$ .

(a) $x < 8$ (b) $x \ge -3$ (c) $x < -2$

Using the Rules on Letters

The rules hold for every number, so they can be used when the numbers are letters. Choosing numbers that fit, such as $a = 1$ and $b = 4$ , is a good check.

Example 7: Two letters in order

It is given that $a < b$ . Write $<$ or $>$ between each pair.
(a) $a + 3$ and $b + 3$
(b) $a - 2$ and $b - 2$
(c) $3a$ and $3b$
(d) $\frac{a}{2}$ and $\frac{b}{2}$
(e) $-a$ and $-b$
(f) $-2a$ and $-2b$

Solution

(a), (b) Adding or subtracting keeps the sign (Rule 1).

(c), (d) Multiplying by 3 and dividing by 2 keep the sign, because 3 and 2 are positive (Rule 2).

(e), (f) Multiplying by $-1$ and by $-2$ reverses the sign (Rule 3).

Check with $a = 1$ and $b = 4$ : (e) $-1 > -4$ and (f) $-2 > -8$ are both true.

(a) $a + 3 < b + 3$ (b) $a - 2 < b - 2$ (c) $3a < 3b$ (d) $\frac{a}{2} < \frac{b}{2}$ (e) $-a > -b$ (f) $-2a > -2b$

Example 8: True or false

Is each statement true or false?
(a) If $a > b$ , then $-3a > -3b$ .
(b) If $a < b$ , then $\frac{a}{4} < \frac{b}{4}$ .
(c) If $a < b$ , then $a - 5 < b - 5$ .
(d) If $p \ge q$ , then $2 - p \le 2 - q$ .

Solution

(a) Multiplying by $-3$ reverses the sign, so $-3a < -3b$ . Check with $a = 2$ and $b = 1$ : $-6 > -3$ is false.

(b) Dividing by 4, a positive number, keeps the sign.

(d) Multiply $p \ge q$ by $-1$ : $-p \le -q$ . Then add 2 to both sides: $2 - p \le 2 - q$ . Check with $p = 5$ and $q = 1$ : $-3 \le 1$ is true.

(a) False (b) True (c) True (d) True

Common Mistakes

Forgetting to reverse the sign when multiplying or dividing by a negative number. From $-2x > 10$ , the answer is $x < -5$ , not $x > -5$ . Check: $x = 0$ satisfies $x > -5$ , but $-2(0) = 0$ and $0 > 10$ is false, so $x > -5$ cannot be the answer.
Reversing the sign one line too late. Writing $\frac{-2x}{-2} > \frac{10}{-2}$ and then reversing on the next line makes the first line false. Reverse the sign in the same line as the division.
Reversing the sign because a number is negative. $3x < -6$ gives $x < -2$ : we divide by positive 3, so the sign stays. $x + (-3) < 4$ gives $x < 7$ : adding never reverses.
Mixing up swapping and reversing. $12 > x$ and $x < 12$ say the same thing. Dividing by a negative number changes what the inequality says, and that is why the sign must be reversed.

Practice

Work each question on paper before you open its solution.

Question 1

It is true that $4 < 9$ . Do each of these to both sides, write the new inequality, and say whether the sign stays the same or is reversed.
(a) Add 5.
(b) Subtract 10.
(c) Multiply by 2.
(d) Multiply by $-2$ .
(e) Divide by $-1$ .

ShowHide solution

(a) $9$ and $14$ : $9 < 14$ .

(b) $-6$ and $-1$ : $-6$ is to the left of $-1$ , so $-6 < -1$ .

(d) $-8$ and $-18$ : $-8$ is to the right of $-18$ , so $-8 > -18$ .

(e) $-4$ and $-9$ : $-4 > -9$ .

(a) $9 < 14$ , same (b) $-6 < -1$ , same (c) $8 < 18$ , same (d) $-8 > -18$ , reversed (e) $-4 > -9$ , reversed

Question 2

It is given that $p < q$ . Write $<$ or $>$ between each pair.
(a) $p + 7$ and $q + 7$
(b) $5p$ and $5q$
(c) $-p$ and $-q$
(d) $\frac{p}{3}$ and $\frac{q}{3}$
(e) $-4p$ and $-4q$

ShowHide solution

(a) Adding keeps the sign. (b), (d) 5 and 3 are positive, so the sign stays.

(c), (e) Multiplying by $-1$ or $-4$ reverses the sign. Check with $p = 1$ , $q = 2$ : $-1 > -2$ and $-4 > -8$ .

(a) $p + 7 < q + 7$ (b) $5p < 5q$ (c) $-p > -q$ (d) $\frac{p}{3} < \frac{q}{3}$ (e) $-4p > -4q$

Question 3

It is given that $x > 4$ . Write an inequality for
(a) $x + 2$ ,
(b) $x - 9$ ,
(c) $3x$ ,
(d) $-x$ ,
(e) $-\frac{x}{2}$ .

ShowHide solution

(a) Add 2 to both sides: $x + 2 > 6$ .

(b) Subtract 9 from both sides: $x - 9 > -5$ .

(d) Multiply both sides by $-1$ , which reverses the sign: $-x < -4$ .

(e) Divide both sides by $-2$ , which reverses the sign: $-\frac{x}{2} < -2$ .

Check with $x = 6$ : $8 > 6$ , $-3 > -5$ , $18 > 12$ , $-6 < -4$ and $-3 < -2$ are all true.

(a) $x + 2 > 6$ (b) $x - 9 > -5$ (c) $3x > 12$ (d) $-x < -4$ (e) $-\frac{x}{2} < -2$

Question 4

Is each statement true or false?
(a) If $a > b$ , then $a + 10 > b + 10$ .
(b) If $a < b$ , then $-5a < -5b$ .
(c) If $a \ge b$ , then $\frac{a}{3} \ge \frac{b}{3}$ .
(d) If $a < b$ , then $-\frac{a}{2} > -\frac{b}{2}$ .
(e) If $a > b$ , then $a - 8 < b - 8$ .

ShowHide solution

(a) Adding keeps the sign.

(b) Multiplying by $-5$ reverses the sign, so $-5a > -5b$ . With $a = 1$ and $b = 2$ : $-5 < -10$ is false.

(d) Dividing by $-2$ reverses the sign. With $a = 2$ and $b = 4$ : $-1 > -2$ is true.

(e) Subtracting keeps the sign, so $a - 8 > b - 8$ .

(a) True (b) False (c) True (d) True (e) False

Question 5

Solve each inequality and show the solution on a number line.
(a) $x + 7 > 3$
(b) $x - 2 \le -5$
(c) $6 + x \ge 6$

ShowHide solution

(a) Subtract 7 from both sides: $x > 3 - 7$ , so $x > -4$ .

(b) Add 2 to both sides: $x \le -5 + 2$ , so $x \le -3$ .

(a) $x > -4$ (b) $x \le -3$ (c) $x \ge 0$

Question 6

Solve each inequality and show the solution on a number line.
(a) $6x < 42$
(b) $\frac{x}{3} \ge -2$
(c) $5x > -10$

ShowHide solution

(a) Divide both sides by 6: $x < 7$ .

(b) Multiply both sides by 3, a positive number: $x \ge -6$ .

(a) $x < 7$ (b) $x \ge -6$ (c) $x > -2$

Question 7

Solve each inequality and show the solution on a number line.
(a) $-4x \le 20$
(b) $-x > 2$
(c) $-\frac{x}{3} < 1$
(d) $-6x \ge -9$

ShowHide solution

(a) Divide both sides by $-4$ , which reverses the sign: $x \ge \frac{20}{-4}$ , so $x \ge -5$ .

(b) Multiply both sides by $-1$ , which reverses the sign: $x < -2$ .

(d) Divide both sides by $-6$ , which reverses the sign: $x \le \frac{-9}{-6}$ , so $x \le 1.5$ .

(a) $x \ge -5$ (b) $x < -2$ (c) $x > -3$ (d) $x \le 1.5$

Question 8

Solve each inequality, saying whether the sign is reversed. Show each solution on a number line.
(a) $x + (-6) < 1$
(b) $2x > -8$
(c) $-2x > 8$
(d) $-2x > -8$

ShowHide solution

(a) Add 6 to both sides: $x < 1 + 6$ , so $x < 7$ . Not reversed.

(b) Divide both sides by 2: $x > -4$ . Not reversed: 2 is positive.

(d) Divide both sides by $-2$ , which reverses the sign: $x < 4$ .

(a) $x < 7$ (b) $x > -4$ (c) $x < -4$ (d) $x < 4$ . Only (c) and (d) reverse the sign.

Question 9

Write each inequality with the unknown on the left, then solve it and show the solution on a number line.
(a) $15 \ge x + 9$
(b) $-12 < 4x$
(c) $6 > -3x$

ShowHide solution

(a) $x + 9 \le 15$ . Subtract 9 from both sides: $x \le 6$ .

(b) $4x > -12$ . Divide both sides by 4: $x > -3$ .

(a) $x \le 6$ (b) $x > -3$ (c) $x > -2$

Question 10

Wei Jie solved $-3x < 12$ and wrote the answer $x < -4$ .
(a) Test his answer with $x = -5$ , which satisfies $x < -4$ .
(b) Solve the inequality correctly.

ShowHide solution

(a) When $x = -5$ : $-3(-5) = 15$ , and $15 < 12$ is false. So $-5$ is not a solution of $-3x < 12$ , and $x < -4$ cannot be the answer.

(b) Divide both sides by $-3$ , which reverses the sign: $x > \frac{12}{-3}$ , so $x > -4$ . Wei Jie divided by $-3$ but kept the sign $<$ .

Check with $x = 0$ : $-3(0) = 0$ , and $0 < 12$ is true.

(a) $x = -5$ does not satisfy $-3x < 12$ . (b) $x > -4$

Question 11

During a winter evening in a town in Hokkaido, Japan, the temperature fell by 5 °C, and after that it was below $-2$ °C. Let the temperature before the fall be $t$ °C.
(a) Write an inequality in $t$ .
(b) Solve it, and show the solution on a number line.
(c) Could the temperature before the fall have been 4 °C?

ShowHide solution

(a) After the fall the temperature was $t - 5$ °C, and "below $-2$ " means less than $-2$ .

(b) Add 5 to both sides: $t < -2 + 5$ , so $t < 3$ .

(a) $t - 5 < -2$ (b) $t < 3$ (c) No: the temperature before the fall was below 3 °C.

Question 12

Mei Ling saves $6 every week. She wants to save at least $45. Let $w$ be the number of weeks she saves for.
(a) Write an inequality in $w$ , and solve it.
(b) Is 7 weeks enough? Is 8 weeks enough?

ShowHide solution

(a) In $w$ weeks she saves $6w$ dollars, and this must be at least 45. Divide both sides of $6w \ge 45$ by 6: $w \ge 7.5$ .

(b) 7 is less than 7.5: in 7 weeks she saves $42. 8 is greater than 7.5: in 8 weeks she saves $48.

$w$ is a whole number of weeks, and the first whole number to the right of 7.5 is 8. So she needs at least 8 weeks.

(a) $6w \ge 45$ , so $w \ge 7.5$ . $w$ is a whole number, so she needs at least 8 weeks. (b) 7 weeks is not enough ($42); 8 weeks is enough ($48).

Key Points

Adding or subtracting the same number, positive or negative, on both sides does not change the sign.
Multiplying or dividing both sides by the same positive number does not change the sign, even when a side is negative: $3x < -6$ gives $x < -2$ .
Multiplying or dividing both sides by the same negative number reverses the sign. Write the reversed sign in the same line: $-2x > 10$ gives $x < -5$ .
Swapping the two sides turns the sign round without changing the meaning: $12 > x$ is $x < 12$ .
Check an answer by substituting a number from it into the inequality you started with.