Solving One-Step Inequality Word Problems | Grade 7 Math | LevelKite

Some questions have many correct answers. If tickets cost 8 dollars each and you have 50 dollars, you could buy 1 ticket, or 3, or 6. You can describe all the possible answers at once with an inequality: a statement that compares two quantities with $<$ (less than), $>$ (greater than), $\le$ (less than or equal to) or $\ge$ (greater than or equal to).

A variable is a letter that stands for an unknown number. A solution of an inequality is any value of the variable that makes the inequality true.

You can already write an inequality for a situation. In this lesson you will solve one-step inequalities, which need one operation (adding, subtracting, multiplying or dividing) to get the variable alone. Then you will decide which solutions fit the situation.

Solving an inequality means rewriting it as a simpler inequality with exactly the same solutions. You do that by doing the same operation to both sides (never multiplying or dividing by zero). What matters is what the operation does to the order of the two sides, meaning which side is less.

Adding or subtracting the same number changes both sides by the same amount, so the side that was less is still less.
Multiplying or dividing by the same positive number (a number greater than zero) can stretch or shrink the gap between the sides, but it never changes which side is less.
Multiplying or dividing by the same negative number (a number less than zero) switches the order. Multiplying by $-2$ is the same as multiplying by $2$ and then taking the opposite of each side, such as $-8$ for $8$ . Taking opposites swaps which number is less: $8 < 20$ , but $-8 > -20$ , just as $-8$ degrees is warmer than $-20$ degrees.

The table tests each case on the true statement $4 < 10$ . Watch whether the symbol stays the same or must reverse (turn around, so that $<$ becomes $>$ ) to keep the statement true.

Do each operation to both sides of the true statement: 4 is less than 10

Operation on both sides	Left side	Right side	True statement	What the symbol does
Add $3$	$4 + 3 = 7$	$10 + 3 = 13$	$7 < 13$	Stays $<$
Subtract $3$	$4 - 3 = 1$	$10 - 3 = 7$	$1 < 7$	Stays $<$
Add $-3$	$4 + (-3) = 1$	$10 + (-3) = 7$	$1 < 7$	Stays $<$
Multiply by $2$	$4 \times 2 = 8$	$10 \times 2 = 20$	$8 < 20$	Stays $<$
Divide by $2$	$4 \div 2 = 2$	$10 \div 2 = 5$	$2 < 5$	Stays $<$
Multiply by $-2$	$4 \times (-2) = -8$	$10 \times (-2) = -20$	$-8 > -20$	Reverses to $>$
Divide by $-2$	$4 \div (-2) = -2$	$10 \div (-2) = -5$	$-2 > -5$	Reverses to $>$

Key concept

One operation on both sides, then watch the symbol

To solve a one-step inequality, do the same operation to both sides so that the variable is alone.

Add or subtract the same number: the symbol stays the same.
Multiply or divide by the same positive number: the symbol stays the same.
Multiply or divide by the same negative number: the symbol reverses ( $<$ and $>$ trade places, and so do $\le$ and $\ge$ ).

The boundary is the value of the variable that makes the two sides equal. With $<$ or $>$ , the boundary is not a solution. With $\le$ or $\ge$ , it is a solution.

In a word problem, the situation can limit the solutions further. A count of objects must be a whole number ( $0, 1, 2, 3, \dots$ ), and many quantities cannot be negative.

Before you trust an answer, run two checks. To substitute means to replace the variable with a number.

Check the boundary. Substitute the boundary into the original inequality. The two sides should come out equal. Then look at the symbol: $\le$ and $\ge$ include the boundary as a solution, and $<$ and $>$ leave it out.
Check a test value. A test value is a number from your solutions that is not the boundary. Substitute it into the original inequality. It should make a true statement. In a one-step inequality, if the boundary checks out but the test value makes a false statement, your symbol points the wrong way.

Worked example

Example 1: Packing under a weight limit

An airline lets a packed suitcase weigh at most $50$ pounds. Your empty suitcase weighs $9$ pounds. How many pounds of belongings can you pack?

Name the unknown and its unit of measure: let p be the weight of your belongings in pounds. The packed suitcase weighs p plus 9 pounds. At most 50 pounds means this total is less than or equal to 50.
$p + 9 \le 50$
Subtract 9 from both sides to undo the addition and get p alone. Subtracting 9 lowers both sides by the same amount, so the comparison does not change and the symbol stays the same. Only multiplying or dividing by a negative number reverses the symbol.
$p + 9 - 9 \le 50 - 9$
Simplify each side. On the left, 9 minus 9 is 0, which leaves p. On the right, 50 minus 9 is 41. The boundary is 41.
$p \le 41$
Check the boundary: substitute 41 for p in the original inequality. Both sides equal 50. The symbol is less than or equal to, so the boundary 41 is a solution.
$41 + 9 = 50, \quad 50 \le 50 \ \text{is true}$
Check a test value: choose a solution that is not the boundary, such as 30, and substitute it in the original inequality. The statement is true, so the symbol points the right way.
$30 + 9 = 39, \quad 39 \le 50 \ \text{is true}$
Fit the answer to the situation. A weight cannot be negative, so p must be nonnegative: 0 or greater. A weight does not have to be a whole number, so a value such as 20.5 pounds works.
$p \ge 0 \quad \text{and} \quad p \le 41$

Answer

You can pack any weight of belongings from $0$ pounds up to and including $41$ pounds.

Worked example

Example 2: Buying tickets on a budget

Ride tickets cost $8$ dollars each. You can spend at most $50$ dollars. How many tickets can you buy?

Let t be the number of tickets you buy. Each ticket costs 8 dollars, so t tickets cost 8 times t dollars. At most 50 dollars means this cost is less than or equal to 50.
$8t \le 50$
Divide both sides by 8 to undo the multiplication and get t alone. Because 8 is positive, dividing does not change which side is less, so the symbol stays the same. Dividing by a negative number would reverse it.
$\frac{8t}{8} \le \frac{50}{8}$
Simplify each side. On the left, 8 times t divided by 8 is t. On the right, 50 divided by 8 is 6.25, because 8 times 6 is 48 and the 2 left over is one fourth of 8. The boundary is 6.25.
$t \le 6.25$
Check in the original inequality. At the boundary 6.25, both sides equal 50, and less than or equal to includes the boundary. The test value 6 gives 48, and 48 is less than 50, so the symbol points the right way.
$8 \times 6.25 = 50, \quad 50 \le 50 \ \text{is true}; \qquad 8 \times 6 = 48, \quad 48 \le 50 \ \text{is true}$
Fit the answer to the situation. You cannot buy part of a ticket or a negative number of tickets, so t must be a whole number. The boundary 6.25 makes the inequality true, but it is not a possible number of tickets. The whole numbers that are 6.25 or less are 0 through 6.
$t = 0,\ 1,\ 2,\ 3,\ 4,\ 5,\ \text{or}\ 6$

Answer

You can buy any whole number of tickets from $0$ through $6$ . The most you can buy is $6$ tickets, which cost $48$ dollars.

Worked example

Example 3: A submarine dive

Elevation tells how far something is above or below the ocean surface, and it is negative below the surface. A submarine starts at an elevation of $0$ feet. While it dives, its elevation changes by $-25$ feet each minute. For how many minutes must it dive to be lower than $-300$ feet?

Let m be the number of minutes the submarine dives. Starting from 0 feet, each minute changes the elevation by negative 25 feet, so after m minutes the elevation is negative 25 times m feet. Lower than negative 300 feet means less than negative 300. Exactly negative 300 does not count.
$-25m < -300$
Divide both sides by negative 25 to undo the multiplication and get m alone. Dividing by a negative number switches which side is less, so reverse the symbol from less than to greater than.
$\frac{-25m}{-25} > \frac{-300}{-25}$
Simplify each side. On the left, negative 25 times m divided by negative 25 is m. On the right, negative 300 divided by negative 25 is 12, because negative 25 times 12 is negative 300. The boundary is 12.
$m > 12$
Check the boundary in the original inequality. At 12 minutes the elevation is exactly negative 300 feet, so the two sides are equal. Less than and greater than leave the boundary out, so 12 is not a solution.
$-25 \times 12 = -300, \quad -300 < -300 \ \text{is false}$
Check a test value greater than 12, such as 13, in the original inequality. Negative 325 is less than negative 300, so the statement is true and the symbol points the right way.
$-25 \times 13 = -325, \quad -325 < -300 \ \text{is true}$
Fit the answer to the situation. Time must be nonnegative, and every number greater than 12 already is. Time does not have to be a whole number, so 12.5 minutes is also a solution.
$-25 \times 12.5 = -312.5, \quad -312.5 < -300 \ \text{is true}$

Answer

The submarine must dive for more than $12$ minutes. Exactly $12$ minutes is not enough, because that puts it at $-300$ feet, which is not lower than $-300$ feet.

Common mistake: dividing by a negative number but keeping the symbol. Suppose you divide both sides of $-25m < -300$ by $-25$ and leave the symbol alone. You get $m < 12$ . Test $m = 10$ in the original inequality: $-25 \times 10 = -250$ , and $-250 < -300$ is false, because $-250$ feet is higher than $-300$ feet.

Here is why it fails. More minutes give a lower elevation, so a greater $m$ goes with a lesser $-25m$ . Dividing by the negative number switches the order, and the symbol must reverse, which gives $m > 12$ .

Do not reverse just because you see a negative number. Adding or subtracting any number keeps the symbol, and so does multiplying or dividing by a positive number, even when a side is negative.

Takeaway. To solve a one-step inequality word problem, define the variable with its unit of measure, write the inequality, and do one operation to both sides. Adding or subtracting keeps the symbol, and so does multiplying or dividing by a positive number. Multiplying or dividing by a negative number reverses it. Check the boundary and one test value in the original inequality: $<$ and $>$ leave the boundary out, and $\le$ and $\ge$ include it. Then keep only the solutions that fit the situation, which may have to be whole numbers or nonnegative.

Next, you will use these checks for error analysis: finding and fixing mistakes in work with inequalities.