Two-Step Inequality Word Problems: Solve and Interpret | Grade 7 Math | LevelKite

Imagine you can spend at most 50 dollars at a fair. Getting in costs 14 dollars, and each ride ticket costs 4 dollars. How many tickets can you buy?

There is not just one answer. You could buy 2 tickets, or 6, or none at all, and your total would stay within 50 dollars each time.

To describe a situation with a limit like this, you use an inequality: a statement that compares two quantities with one of the symbols $\lt$ , $\gt$ , $\le$ or $\ge$ . All the values that make an inequality true form its solution set.

In this lesson you will turn a word problem into a two-step inequality (one that takes two operations to solve), solve it, and then say exactly which values the situation allows.

Words that choose the symbol

Words in the problem	Symbol	Can the two sides be equal?	Example
at most, no more than, a maximum of	$\le$	Yes	Spend at most 50 dollars: total $\le 50$
at least, no less than, a minimum of	$\ge$	Yes	Score at least 80 points: score $\ge 80$
less than, fewer than, under	$\lt$	No	Weigh under 20 kilograms: weight $\lt 20$
more than, greater than, over	$\gt$	No	Raise more than 200 dollars: total $\gt 200$

Many word problems have two parts: a fixed amount that does not change, and an amount that goes up or down by the same number each time. The unknown is usually how many times that change happens, such as how many tickets you buy. That is where you put your variable, a letter that stands for the unknown number.

Use the same plan each time:

Say what the variable stands for and name its unit.
Write the total using the variable.
Let the words of the problem choose the symbol, and write the inequality.
Solve in two steps: undo the addition or subtraction, then undo the multiplication.
Describe the solution set in the situation.

As you solve, watch the symbol. Adding or subtracting the same number on both sides keeps the symbol the same. So does multiplying or dividing both sides by a positive number. Multiplying or dividing both sides by a negative number reverses the symbol.

Worked example

Ride tickets at the fair

You can spend at most 50 dollars at a fair. Admission costs 14 dollars and each ride ticket costs 4 dollars. How many ride tickets can you buy?

Name the variable and its unit. Let t be the number of ride tickets you buy. Each ticket costs 4 dollars, so t tickets cost 4t dollars. Add the 14 dollars for admission to get your total cost in dollars.
$14 + 4t$
Choose the symbol. The words "at most 50 dollars" mean your total may be less than 50 or exactly 50, so use the less-than-or-equal-to symbol, which lets the two sides be equal.
$14 + 4t \le 50$
Undo the addition by subtracting 14 from both sides. Subtracting the same number from both sides keeps the symbol the same. The right side becomes 50 minus 14, which is 36.
$14 + 4t - 14 \le 50 - 14 \quad\text{so}\quad 4t \le 36$
Undo the multiplication by dividing both sides by 4. Because 4 is positive, the symbol stays the same. The right side becomes 36 divided by 4, which is 9.
$\frac{4t}{4} \le \frac{36}{4} \quad\text{so}\quad t \le 9$
Describe the result in the situation. The symbol lets t equal 9. With 9 tickets, your total is 14 plus 36, which is exactly 50 dollars, and "at most 50" allows exactly 50.
$14 + 4(9) = 14 + 36 = 50$

Answer

You can buy at most $9$ ride tickets. Buying $9$ tickets brings your total to exactly 50 dollars.

Look again at $t \le 9$ . As a statement about numbers alone, it is true for $9$ , for $8.5$ and even for $-2$ . But the fair does not sell half a ticket, and you cannot buy a negative number of tickets. In this situation, the solution set is only the whole numbers (0, 1, 2, 3 and so on) up to 9: $0, 1, 2, \dots, 9$ .

To interpret a solution set is to say which values it allows in the situation. First find the boundary, the value of the variable that makes the two sides exactly equal. Here the boundary is $9$ . Then ask three questions:

Which direction? Are the solutions less than the boundary or greater than it?
Does the boundary count? The symbols $\le$ and $\ge$ include the boundary. The symbols $\lt$ and $\gt$ exclude it.
Which numbers make sense? A count of objects must be a whole number. A length or an amount of time can include fractions and decimals, but it must be nonnegative, which means zero or greater. A few quantities, such as a temperature, can even be negative.

The table below shows how these answers change with the situation.

The situation changes what a solved inequality allows

Solved inequality	What x stands for	Numbers that make sense	Solution set in words
$x \le 6$	Number of pizzas you order	Whole numbers: 0, 1, 2, 3 and so on	0, 1, 2, 3, 4, 5 or 6 pizzas
$x \lt 6$	Number of pizzas you order	Whole numbers: 0, 1, 2, 3 and so on	0, 1, 2, 3, 4 or 5 pizzas, because 6 is excluded
$x \le 6$	Kilograms of apples you buy	Zero or greater, with fractions and decimals allowed	Any amount from 0 kilograms up to and including 6 kilograms
$x \le 6$	Temperature in degrees Celsius	Negative numbers, zero and positive numbers, with decimals allowed	6 degrees or colder, such as 2.5 degrees or $-4$ degrees

Key concept

The situation decides which solutions count

Solving a two-step inequality from a word problem gives a solution set: usually a whole range of values rather than one answer. To state it correctly, put three facts together:

Direction. The solved inequality tells you whether the solutions are less than or greater than the boundary. Multiplying or dividing both sides by a negative number reverses the symbol.
Boundary. The symbols $\le$ and $\ge$ include the boundary value. The symbols $\lt$ and $\gt$ exclude it.
Context. The situation may limit the solutions to whole numbers or to nonnegative numbers.

Worked example

Washing cars to pass a goal

A school club wants to raise more than 200 dollars. It already has 65 dollars and earns 12 dollars for each car it washes. How many cars must the club wash?

Let c be the number of cars the club washes. Washing c cars earns 12c dollars, and adding the 65 dollars the club already has gives its total in dollars. The words "more than 200" do not allow a total of exactly 200, so use the greater-than symbol.
$65 + 12c \gt 200$
Subtract 65 from both sides. Subtracting the same number from both sides keeps the symbol the same. The right side becomes 200 minus 65, which is 135.
$65 + 12c - 65 \gt 200 - 65 \quad\text{so}\quad 12c \gt 135$
Divide both sides by 12. Because 12 is positive, the symbol stays the same. The right side becomes 135 divided by 12, which is 11.25.
$\frac{12c}{12} \gt \frac{135}{12} \quad\text{so}\quad c \gt 11.25$
Interpret the result. The solutions are greater than the boundary 11.25, and the symbol excludes 11.25 itself. A number of cars must be a whole number, so the solutions are the whole numbers greater than 11.25. The smallest is 12.
$c = 12, 13, 14, \dots$
Check the whole numbers just below and just above the boundary in the original inequality. With 11 cars the club has 197 dollars, which is not more than 200. With 12 cars it has 209 dollars, which is more than 200.
$65 + 12(11) = 65 + 132 = 197 \qquad 65 + 12(12) = 65 + 144 = 209$

Answer

The club must wash at least $12$ cars. Any whole number of cars that is $12$ or greater raises more than 200 dollars.

Worked example

A draining water tank

A tank starts with 120 liters of water. A valve opens, and the water drains at a steady 8 liters per minute. For which times does the tank have more than 40 liters left?

Let m be the number of minutes since the valve opened. After m minutes, 8m liters have drained, so 120 minus 8m liters are left. The words "more than 40 liters" do not allow exactly 40, so use the greater-than symbol.
$120 - 8m \gt 40$
Subtract 120 from both sides. Subtracting the same number from both sides keeps the symbol the same. The right side becomes 40 minus 120, which is negative 80.
$120 - 8m - 120 \gt 40 - 120 \quad\text{so}\quad -8m \gt -80$
Divide both sides by negative 8. Dividing both sides by a negative number reverses the symbol, so greater than becomes less than. This fits the situation: more water left goes with less draining time. Negative 80 divided by negative 8 is 10.
$\frac{-8m}{-8} \lt \frac{-80}{-8} \quad\text{so}\quad m \lt 10$
Test one value less than the boundary and one value greater than it in the original inequality. At 5 minutes the tank has 80 liters left, which is more than 40. At 12 minutes it has 24 liters left, which is not more than 40.
$120 - 8(5) = 120 - 40 = 80 \qquad 120 - 8(12) = 120 - 96 = 24$
Interpret the result. The symbol excludes the boundary: at exactly 10 minutes the tank has exactly 40 liters left, which is not more than 40. Time since the valve opened cannot be negative, so m must be zero or greater. Time can include parts of a minute, so m does not have to be a whole number.
$m \ge 0 \quad\text{and}\quad m \lt 10$

Answer

The tank has more than 40 liters left at any time from $0$ minutes up to, but not including, $10$ minutes. A time such as $9.5$ minutes counts, because time does not have to be a whole number.

Here is the idea to keep: the answer to a word problem with a limit is a set of values, and the situation decides which values belong.

To write the inequality, name your variable and its unit, build the total, and let the words of the problem choose the symbol. To solve it, undo the addition or subtraction and then the multiplication. Reverse the symbol whenever you multiply or divide both sides by a negative number.

To interpret the result, read it back into the situation. The symbols $\le$ and $\ge$ include the boundary, and $\lt$ and $\gt$ exclude it. The situation may also limit the solutions to whole numbers or to nonnegative numbers.

Next, you will study solved inequalities that contain mistakes, and you will use these same checks to find and fix the errors.