Solving Inequalities and Reversing the Symbol | Grade 8 Math | LevelKite

An inequality is a statement that compares two quantities. In this lesson it uses one of four symbols. For real numbers $a$ and $b$ :

$a < b$ means $a$ is less than $b$ .
$a > b$ means $a$ is greater than $b$ .
$a \le b$ means $a$ is less than or equal to $b$ .
$a \ge b$ means $a$ is greater than or equal to $b$ .

You have already written inequalities such as $x > 3$ to describe sets of numbers. Now you will solve inequalities such as $3x + 4 \le 25$ .

A variable here stands for any real number unless a problem says otherwise. A solution of an inequality is a value of the variable that makes the statement true. The solution set is the collection of all the solutions. To solve an inequality is to describe that whole set.

Start by substituting a few numbers for $x$ in $x + 3 > 7$ . To substitute is to replace the variable with a number, so that you can see whether the statement is true.

Which values of x make the statement true?

Value of x	Left side: x + 3	Is it greater than 7?	Solution?
$3$	$3 + 3 = 6$	$6 > 7$ is false	No
$4$	$4 + 3 = 7$	$7 > 7$ is false	No
$4.5$	$4.5 + 3 = 7.5$	$7.5 > 7$ is true	Yes
$5$	$5 + 3 = 8$	$8 > 7$ is true	Yes
$20$	$20 + 3 = 23$	$23 > 7$ is true	Yes

To solve an inequality, you rewrite it in simpler forms that have exactly the same solutions, until the variable stands alone. Which steps are safe?

Think about the gap: the right side minus the left side. The statement $a < b$ is true exactly when the gap $b - a$ is positive.

Add or subtract the same number on both sides. Call the number $c$ . Both sides shift by the same amount, so the gap does not change. For adding: $(b + c) - (a + c) = b - a$ . The symbol keeps its direction.
Multiply or divide both sides by the same positive number. Call it $c$ . This multiplies or divides the gap by $c$ , so the gap keeps its sign. For multiplying: $cb - ca = c(b - a)$ . The symbol keeps its direction.

The other symbols work the same way: $a > b$ means the gap is negative, and $\le$ and $\ge$ also allow a gap of zero. Multiplying or dividing by a negative number behaves differently, as you will see soon.

A one-step inequality needs one step to leave the variable alone.

To solve $x + 3 > 7$ , subtract $3$ from both sides: $x + 3 - 3 > 7 - 3$ , so $x > 4$ . This confirms the pattern from the table.
To solve $\frac{x}{2} \le 6$ , multiply both sides by $2$ . Because $2$ is positive, the symbol keeps its direction: $x \le 12$ . The symbol is non-strict, so the boundary $12$ is a solution. Check it: $\frac{12}{2} = 6$ , and $6 \le 6$ is true.

A two-step inequality such as $3x + 4 \le 25$ needs two steps. It is simplest to undo the addition first and the multiplication second.

Worked example

A two-step inequality

Solve $3x + 4 \le 25$ , where $x$ is a real number. Check the answer by substitution.

Subtract 4 from both sides to undo the addition of 4. Subtracting the same number from both sides does not change the gap, so the symbol keeps its direction.
$3x + 4 - 4 \le 25 - 4$
Simplify each side: 4 - 4 is 0 on the left, and 25 - 4 is 21 on the right.
$3x \le 21$
Divide both sides by 3 to undo the multiplication by 3. Because 3 is positive, the gap keeps its sign and the symbol keeps its direction.
$\frac{3x}{3} \le \frac{21}{3}$
Simplify: 21 divided by 3 is 7. The symbol is non-strict, so the boundary 7 belongs to the solution set.
$x \le 7$
Check the boundary by substituting 7 for x in the original inequality. The left side equals 25, and 25 is less than or equal to 25, so 7 is a solution.
$3(7) + 4 = 21 + 4 = 25 \le 25$
Check one value on each side of the boundary. For x = 6 the left side is 22, which is less than 25, so 6 is a solution. For x = 8 the left side is 28, which is greater than 25, so 8 is not a solution. Both results match the answer.
$3(6) + 4 = 22 \le 25, \qquad 3(8) + 4 = 28 > 25$

Answer

The solution set is every real number less than or equal to $7$ , written $x \le 7$ .

Every step so far kept the symbol's direction. Multiplying or dividing both sides by a negative number does not. To see what happens, start with a true statement about two plain numbers, $2 < 6$ . Do the same thing to both sides, then compare the two results.

Starting from the true statement that 2 is less than 6

Step on both sides	New left side	New right side	True comparison	Direction
Add $5$	$2 + 5 = 7$	$6 + 5 = 11$	$7 < 11$	Same
Subtract $9$	$2 - 9 = -7$	$6 - 9 = -3$	$-7 < -3$	Same
Multiply by $3$	$2 \cdot 3 = 6$	$6 \cdot 3 = 18$	$6 < 18$	Same
Divide by $2$	$2 \div 2 = 1$	$6 \div 2 = 3$	$1 < 3$	Same
Multiply by $-3$	$2 \cdot (-3) = -6$	$6 \cdot (-3) = -18$	$-6 > -18$	Reversed
Divide by $-2$	$2 \div (-2) = -1$	$6 \div (-2) = -3$	$-1 > -3$	Reversed

The direction changed only when you multiplied or divided by a negative number. Subtracting $9$ gave negative results, yet the direction stayed the same.

The gap explains why. For $2 < 6$ the gap is $6 - 2 = 4$ , which is positive. Multiplying both sides by $-3$ multiplies the gap by $-3$ too: the new gap is $-18 - (-6) = -12$ . A negative number times a positive gap is negative, so the side that was greater is now the smaller one.

In general, multiplying both sides by any number $c$ turns the gap $b - a$ into $cb - ca = c(b - a)$ . When $c$ is negative, a positive gap becomes negative and a negative gap becomes positive, so the symbol must reverse to keep the statement true. A gap of zero stays zero, and a gap that is not zero stays not zero. So a non-strict symbol stays non-strict, and a strict symbol stays strict.

Dividing by a negative number $c$ does the same, because it equals multiplying by $\frac{1}{c}$ , which is also negative. Zero is different: you cannot divide by $0$ , and multiplying both sides by $0$ erases the variable, so neither is a solving step.

Key concept

When the inequality symbol reverses

You can change both sides of an inequality with the same number by adding, subtracting, multiplying or dividing. The symbol reverses in exactly one case: you multiply or divide by a negative number.

Add or subtract any number: the direction stays.
Multiply or divide by a positive number: the direction stays.
Multiply or divide by a negative number: the direction reverses, so $<$ and $>$ trade places, and so do $\le$ and $\ge$ .

For example, dividing both sides of $-5x \ge 30$ by $-5$ gives $x \le -6$ .

A strict symbol stays strict and a non-strict symbol stays non-strict, so the boundary stays out or stays in. Never multiply or divide by $0$ . A context may also rule out some solutions.

\text{If } a < b \text{ and } c < 0, \text{ then } ac > bc \text{ and } \frac{a}{c} > \frac{b}{c}

Worked example

Dividing by a negative number

Solve $-2x + 7 < 15$ , where $x$ is a real number. Check the answer by substitution.

Subtract 7 from both sides to undo the addition of 7. Subtracting does not change the gap, so the symbol keeps its direction.
$-2x + 7 - 7 < 15 - 7$
Simplify each side: 7 - 7 is 0 on the left, and 15 - 7 is 8 on the right.
$-2x < 8$
Divide both sides by -2 to undo the multiplication by -2. The number you divide by is negative, so the gap changes sign and the symbol reverses from less than to greater than.
$\frac{-2x}{-2} > \frac{8}{-2}$
Simplify: 8 divided by -2 is -4. The symbol is still strict, so the boundary -4 is not in the solution set.
$x > -4$
Check the boundary by substituting -4 for x in the original inequality. The left side equals 15, and 15 is not less than 15, so -4 is not a solution. That fits a strict symbol.
$-2(-4) + 7 = 8 + 7 = 15$
Check one value on each side of the boundary. For x = 0 the left side is 7, which is less than 15, so 0 is a solution. For x = -5 the left side is 17, which is not less than 15, so -5 is not a solution. Both results match the answer.
$-2(0) + 7 = 7 < 15, \qquad -2(-5) + 7 = 17 > 15$

Answer

The solution set is every real number greater than $-4$ , written $x > -4$ .

A multi-step inequality needs more than two steps. You may first have to multiply out parentheses, or move every term that contains the variable to one side. The direction rules stay the same at every step, so keep asking one question: did I just multiply or divide both sides by a negative number?

Worked example

A multi-step inequality

Solve $3(x - 2) \ge 5x + 4$ , where $x$ is a real number. Check the answer by substitution.

Multiply 3 by each term inside the parentheses: 3 times x is 3x, and 3 times 2 is 6, so the left side is 3x - 6. This rewrites the left side without changing its value, so the symbol stays.
$3x - 6 \ge 5x + 4$
Subtract 5x from both sides to bring the terms with x together on the left. For any value of x, 5x is one number, so subtracting it from both sides keeps the direction.
$3x - 6 - 5x \ge 5x + 4 - 5x \quad\Rightarrow\quad -2x - 6 \ge 4$
Add 6 to both sides to leave only the term with x on the left. Adding the same number to both sides keeps the direction.
$-2x - 6 + 6 \ge 4 + 6 \quad\Rightarrow\quad -2x \ge 10$
Divide both sides by -2. The number you divide by is negative, so the symbol reverses from greater than or equal to into less than or equal to. It stays non-strict, so the solution set includes the boundary.
$\frac{-2x}{-2} \le \frac{10}{-2} \quad\Rightarrow\quad x \le -5$
Check the boundary by substituting -5 for x in the original inequality. Both sides equal -21, and -21 is greater than or equal to -21, so -5 is a solution.
$3(-5 - 2) = 3(-7) = -21, \qquad 5(-5) + 4 = -25 + 4 = -21$
Check one value on each side of the boundary. For x = -6 the sides are -24 and -26, and -24 is greater than -26, so -6 is a solution. For x = 0 the sides are -6 and 4, and -6 is less than 4, so 0 is not a solution. Both results match the answer.
$3(-6 - 2) = -24 \ge -26 = 5(-6) + 4, \qquad 3(0 - 2) = -6 < 4 = 5(0) + 4$

Answer

The solution set is every real number less than or equal to $-5$ , written $x \le -5$ .

You can now solve these inequalities. The last skill is to explain what happens to the symbol. For any step, say what you did to both sides. If you multiplied or divided, also say whether the number you used is positive or negative.

"I subtracted $7$ from both sides. Both sides shifted by the same amount, so the symbol stays."
"I divided both sides by $-2$ . Dividing by a negative number changes the sign of the gap, so the symbol reverses."

In a real situation, say one more thing: which solutions make sense. A number of people must be a whole number, and a time measured from a starting moment cannot be negative. The context can remove solutions that the algebra allows.

A mistake to avoid: reversing the symbol because you see a negative number. Suppose you solve $4x < -12$ and reverse the symbol because of the $-12$ . You get $x > -3$ . Test $x = 0$ , which that answer allows: $4(0) = 0$ , and $0 < -12$ is false. The answer is wrong.

It fails because you divided both sides by $4$ , a positive number. That keeps the sign of the gap, so the direction stays. The correct solution set is $x < -3$ . Test $x = -4$ : $4(-4) = -16$ , and $-16 < -12$ is true.

A negative number that you add or subtract, or one that already sits on one side, does not reverse the symbol. Among the solving steps in this lesson, only multiplying or dividing both sides by a negative number does.

Worked example

A context that limits the solutions

A tank holds $120$ liters of water. A valve drains $8$ liters each minute until the tank is empty. Let $t$ be the number of minutes since the valve opened, so $t \ge 0$ . For which times does the tank hold more than $40$ liters? Explain what happens to the symbol at each step.

Write the inequality. After t minutes, 8t liters have drained, so the tank holds 120 - 8t liters until it is empty. More than 40 liters means strictly greater than 40, so use a strict symbol.
$120 - 8t > 40$
Subtract 120 from both sides to leave only the term with t on the left. Subtracting shifts both sides by the same amount, so the symbol stays, even though the right side is now negative: 40 - 120 is -80.
$-8t > -80$
Divide both sides by -8. The number you divide by is negative, so the symbol reverses from greater than to less than, and -80 divided by -8 is 10. This fits the situation: more water left means less time has passed.
$t < 10$
Apply the context. A time since the valve opened cannot be negative, so keep only the solutions with t at least 0. Here t may be any real number in that range, not only a whole number of minutes.
$t \ge 0 \quad\text{and}\quad t < 10$
Check by substitution. At t = 5 the tank holds 80 liters, which is more than 40. At t = 10 it holds exactly 40 liters, which is not more than 40, so 10 is not a solution. At t = 12 it holds 24 liters, which is not more than 40.
$120 - 8(5) = 80, \qquad 120 - 8(10) = 40, \qquad 120 - 8(12) = 24$

Answer

The tank holds more than $40$ liters at every time $t$ with $t \ge 0$ and $t < 10$ , that is, from the moment the valve opens until just before $10$ minutes have passed.

Takeaway. To solve an inequality is to find every value that makes it true. Undo the operations one at a time. Adding, subtracting, and multiplying or dividing by a positive number keep the symbol's direction. Multiplying or dividing both sides by a negative number reverses it, because that changes the sign of the gap between the two sides. A strict symbol leaves the boundary out, and a non-strict symbol keeps it in. Check your answer by substituting the boundary and one value on each side of it. When a problem has a context, keep only the solutions that make sense there.