Solving Two-Step Inequalities: When the Symbol Reverses | Grade 7 Math | LevelKite

You already know how to solve a two-step equation such as $3x + 5 = 20$ , where $x$ stands for an unknown number. You undo the addition, then undo the multiplication, and you get one number: $x = 5$ .

An inequality compares two quantities with $<$ (less than), $>$ (greater than), $\le$ (less than or equal to) or $\ge$ (greater than or equal to). A two-step inequality, such as $3x + 5 < 20$ , takes two operations to solve. In this lesson, each inequality multiplies or divides the variable by a number other than zero and then adds or subtracts a number. The number that multiplies the variable is its coefficient. In $3x + 5 < 20$ , the coefficient of $x$ is $3$ .

To solve an inequality is to find every value of the variable that makes it true. Those values form the solution set. In this lesson a solution set holds many numbers, not just one, so you describe it with a short inequality such as $x < 5$ .

The number in that short inequality is the boundary. It is the value of the variable that makes the two sides of the original inequality equal. For $3x + 5 < 20$ the boundary is $5$ , because $3(5) + 5 = 15 + 5 = 20$ . The symbol tells you whether the boundary itself is a solution: $<$ and $>$ leave it out, while $\le$ and $\ge$ include it. When an inequality comes from a real situation, the situation may limit the solutions further, for example to whole numbers or to numbers that are not negative.

With an equation, you do the same operation to both sides (never dividing by zero) and the two sides stay equal. An inequality says that one side is less, so after each operation you ask a new question: is the same side still less?

Try it on a true statement, $2 < 6$ . The right side is $4$ more than the left side. Call that difference the gap.

Add $3$ to both sides: $2 + 3 = 5$ and $6 + 3 = 9$ , so $5 < 9$ . Both sides moved by the same amount, so the gap is still $4$ . Subtracting the same number works the same way.
Multiply both sides by $3$ : $2 \cdot 3 = 6$ and $6 \cdot 3 = 18$ , so $6 < 18$ . This multiplies the gap by $3$ as well: $4 \cdot 3 = 12$ . The gap is still positive, so the right side is still greater. Dividing by a positive number works the same way.

So the symbol stays when you add or subtract the same number, and when you multiply or divide by a positive number. This reasoning works for any two different numbers. Multiplying or dividing by a negative number is different: it reverses the symbol, for example from $<$ to $>$ . You will see why after the first example.

Worked example

A positive coefficient

Solve $3x + 5 < 20$ , where $x$ can be any number.

The left side multiplies x by 3 and then adds 5. Undo the addition first: subtract 5 from both sides. Both sides go down by the same amount, so the side that was less is still less and the symbol does not change.
$3x + 5 - 5 < 20 - 5$
Do the subtraction on each side. On the left, 5 minus 5 is 0, which leaves 3x. On the right, 20 minus 5 is 15.
$3x < 15$
Undo the multiplication: divide both sides by 3. Because 3 is positive, dividing by it keeps the order of the two sides, so the symbol does not change.
$\frac{3x}{3} < \frac{15}{3}$
Do the division on each side. On the left, 3x divided by 3 is x. On the right, 15 divided by 3 is 5. The symbol is 'less than', so the boundary 5 is not part of the solution set.
$x < 5$
Check the boundary in the original inequality. With x equal to 5, the two sides are equal, so 5 is the correct boundary. It is left out because 20 is not less than 20.
$3(5) + 5 = 15 + 5 = 20$
Check one other number from the solution set, such as 4. It makes the original inequality true, so the symbol points the right way.
$3(4) + 5 = 12 + 5 = 17 \quad \text{and} \quad 17 < 20$

Answer

The solution is $x < 5$ . Every number less than $5$ makes $3x + 5 < 20$ true, and $5$ itself does not. No context limits $x$ here, so the solutions include negative numbers and fractions.

Now see what a negative number does. Start again with $2 < 6$ , where the gap is $4$ , and multiply both sides by $-3$ . The left side becomes $2 \cdot (-3) = -6$ and the right side becomes $6 \cdot (-3) = -18$ . This multiplies the gap by $-3$ as well: $4 \cdot (-3) = -12$ . A negative gap means the right side is now $12$ less than the left side. The side that was greater is now the side that is less, so the true statement is $-6 > -18$ . The symbol has reversed.

Dividing by a negative number changes the sign of the gap in the same way, so it reverses the symbol too. The same is true for any two different numbers. Two equal sides stay equal under all of these operations, which is why $\le$ and $\ge$ keep their or equal to part when they reverse.

The table compares positive and negative numbers. Notice the last row: adding a negative number does not reverse the symbol. Only multiplying or dividing by a negative number does.

The same operation done to both sides of a true inequality

Operation on both sides	Left side	Right side	True statement	What the symbol does
Multiply by $3$	$2 \cdot 3 = 6$	$6 \cdot 3 = 18$	$6 < 18$	Stays $<$
Multiply by $-3$	$2 \cdot (-3) = -6$	$6 \cdot (-3) = -18$	$-6 > -18$	Reverses to $>$
Divide by $2$	$2 \div 2 = 1$	$6 \div 2 = 3$	$1 < 3$	Stays $<$
Divide by $-2$	$2 \div (-2) = -1$	$6 \div (-2) = -3$	$-1 > -3$	Reverses to $>$
Add $-3$	$2 + (-3) = -1$	$6 + (-3) = 3$	$-1 < 3$	Stays $<$

Key concept

Same steps as an equation, plus one sign check

For the two-step inequalities in this lesson, do the same operation to both sides, as with a two-step equation: first undo the addition or subtraction, then undo the multiplication or division. At each step, decide what the symbol does.

Add or subtract the same number: the symbol stays.
Multiply or divide by a positive number: the symbol stays.
Multiply or divide by a negative number: the symbol reverses. $<$ becomes $>$ and $>$ becomes $<$ ; $\le$ becomes $\ge$ and $\ge$ becomes $\le$ .

$<$ and $>$ leave the boundary out of the solution set; $\le$ and $\ge$ include it. A context may limit the solutions further, for example to whole numbers or to numbers that are not negative.

Worked example

A negative coefficient

Solve $-4x + 7 \le 31$ , where $x$ can be any number.

Undo the addition first: subtract 7 from both sides. Subtracting the same number from both sides keeps their order, so the symbol does not change. On the right, 31 minus 7 is 24.
$-4x + 7 - 7 \le 31 - 7, \quad \text{so} \quad -4x \le 24$
Undo the multiplication: divide both sides by negative 4. Dividing by a negative number reverses the order of the two sides, so the symbol reverses from 'less than or equal to' to 'greater than or equal to'.
$\frac{-4x}{-4} \ge \frac{24}{-4}$
Do the division on each side. On the left, negative 4x divided by negative 4 is x. On the right, 24 divided by negative 4 is negative 6. The symbol includes 'or equal to', so the boundary negative 6 is part of the solution set.
$x \ge -6$
Check the boundary in the original inequality. With x equal to negative 6, the two sides are equal, so negative 6 is the correct boundary. It is included because 31 is less than or equal to 31.
$-4(-6) + 7 = 24 + 7 = 31$
Check one other number from the solution set, such as 0. It makes the original inequality true, so the reversed symbol points the right way.
$-4(0) + 7 = 0 + 7 = 7 \quad \text{and} \quad 7 \le 31$

Answer

The solution is $x \ge -6$ . The number $-6$ and every number greater than $-6$ make $-4x + 7 \le 31$ true. No context limits $x$ here, so the solutions include numbers that are not whole, such as $-5.5$ .

Before the second step, check the sign of the number you multiply or divide by (x is any number)

After the first step	Second step on both sides	Sign of that number	Arithmetic	Solution
$5x \ge -20$	Divide by $5$	Positive, so the symbol stays	$-20 \div 5 = -4$	$x \ge -4$
$-5x \ge 20$	Divide by $-5$	Negative, so the symbol reverses	$20 \div (-5) = -4$	$x \le -4$
$\frac{x}{2} < -3$	Multiply by $2$	Positive, so the symbol stays	$-3 \cdot 2 = -6$	$x < -6$
$\frac{x}{-2} < 3$	Multiply by $-2$	Negative, so the symbol reverses	$3 \cdot (-2) = -6$	$x > -6$

Worked example

A context that limits the solutions

You have $60$ arcade tokens, and each game uses $4$ tokens. You want to keep at least $12$ tokens, meaning $12$ or more, for a prize. Let $g$ be the number of games you play. Then $60 - 4g$ is the number of tokens left, so $60 - 4g \ge 12$ . How many games can you play?

The coefficient of g is negative 4, because 60 minus 4g is the same as 60 plus negative 4 times g. Undo the 60 first: subtract 60 from both sides. Subtracting the same number keeps the order, so the symbol does not change. On the right, 12 minus 60 is negative 48.
$60 - 4g - 60 \ge 12 - 60, \quad \text{so} \quad -4g \ge -48$
Divide both sides by negative 4. Dividing by a negative number reverses the order, so the symbol reverses from 'greater than or equal to' to 'less than or equal to'. On the right, negative 48 divided by negative 4 is positive 12.
$\frac{-4g}{-4} \le \frac{-48}{-4}, \quad \text{so} \quad g \le 12$
Apply the context. You cannot play a negative number of games or part of a game, so g must be a whole number: 0, 1, 2 and so on. Keep only the whole numbers that are 12 or less. The symbol includes 'or equal to', so 12 itself is a solution.
$g = 0, 1, 2, \dots, 12$
Check the boundary in the original inequality. Playing 12 games leaves exactly 12 tokens, so the two sides are equal and 12 is the correct boundary.
$60 - 4(12) = 60 - 48 = 12$
Check one other number from the solution set, such as 10. Playing 10 games leaves 20 tokens, and 20 is at least 12, so the reversed symbol points the right way.
$60 - 4(10) = 60 - 40 = 20 \quad \text{and} \quad 20 \ge 12$

Answer

You can play any whole number of games from $0$ to $12$ , so at most $12$ games. Playing $12$ games leaves exactly $12$ tokens, and playing fewer games leaves more.

Takeaway. You solve a two-step inequality with the same two undo steps as a two-step equation, and you watch the symbol at each step. The symbol stays when you add or subtract, and when you multiply or divide by a positive number. It reverses when you multiply or divide by a negative number, because that swaps which side is greater. Then read your answer carefully: $<$ and $>$ leave the boundary out, $\le$ and $\ge$ include it, and a context may limit the solutions to whole numbers or to numbers that are not negative.

Next, you will graph the solution set of a two-step inequality.