Error Analysis: Checking Solved Inequalities | Grade 7 Math | LevelKite

You already know how to solve a two-step inequality and how to read its answer in a real situation. In this lesson you take a new job: the checker.

Error analysis means finding and fixing mistakes in a finished solution. You audit the work, which means you check it one line at a time. You look for the first line that does not follow from the line above it, because every later line is built on that mistake. Then you explain why the line fails and write the correct line.

This skill matters because a wrong answer to an inequality can look almost perfect. Every number can be right while the inequality symbol faces the wrong way, or while the or equal to part has quietly disappeared. So you need a method that tests the work instead of trusting how it looks.

An inequality compares two quantities. Its solution set is every value of the variable that makes the comparison true. In this lesson, $x$ stands for an unknown number unless a problem names a different variable.

Solving means rewriting the inequality in simpler lines until the variable stands alone. Each line comes from one move, which is one operation done to both sides. Each new line must have exactly the same solution set as the line above it. A line that lets in a value that does not belong, or leaves out a value that does, is an error, even when its numbers are right.

So to judge a line, you need to know what each move does to the order of two numbers.

Each move, tested on the true statement that 2 is less than 6

Move on both sides	Result of the test	Symbol	Why
Add or subtract the same number	Add $4$ : $6 < 10$ . Subtract $5$ : $-3 < 1$ .	Stays the same	Both sides change by the same amount, so the smaller side stays smaller.
Multiply or divide by the same positive number	Multiply by $3$ : $6 < 18$ . Divide by $2$ : $1 < 3$ .	Stays the same	A positive factor stretches or shrinks both sides the same way, so the smaller side stays smaller.
Multiply or divide by the same negative number	Multiply by $-3$ : $-6 > -18$ . Divide by $-2$ : $-1 > -3$ .	Reverses	Multiplying by $-3$ is multiplying by $3$ , then taking opposites. That reverses order: $6 < 18$ , but $-6 > -18$ .

Auditing lines shows you where an error is. Substitution shows you whether the final answer holds up. To substitute, replace the variable with a number and work out both sides. Always use the original inequality, because a later line may already contain the error.

Make two tests:

Boundary test. Substitute the boundary. In the two-step inequalities of this topic, the correct boundary makes the two sides equal, so unequal sides mean that a move or its arithmetic is wrong. Equal sides make $\le$ and $\ge$ true, and they make $<$ and $>$ false. So the answer must include the boundary exactly when the original symbol is $\le$ or $\ge$ .
Direction test. Substitute one other value from the answer, such as $5$ for the answer $x > 4$ . If the original inequality comes out false, the answer is wrong. When the boundary test has passed, look for a line that multiplied or divided by a negative number without reversing the symbol, or that reversed it after a different move.

One value that works cannot prove an answer by itself, but one value that fails proves it wrong.

Key concept

Audit each line, then test the answer

In a correct solution, every line has the same solution set as the line above it. For each line, ask:

Move. Does it do the same operation to both sides, with correct arithmetic?
Direction. Does the symbol reverse when the line multiplies or divides both sides by a negative number, and stay the same when it adds or subtracts a number, or multiplies or divides by a positive number?
Boundary. Does the symbol keep its kind? $<$ and $>$ leave the boundary out, and $\le$ and $\ge$ keep it in.

Then test the answer by substituting values into the original inequality. In a word problem, also ask whether the situation allows only whole numbers ( $0$ , $1$ , $2$ and so on) or only nonnegative numbers (zero or more).

Worked example

Audit 1: find the faulty line

A student solved $-3x + 5 > 20$ in two lines. Line 1: $-3x > 15$ . Line 2: $x > -5$ . Find the first line that does not follow from the line above it, explain the error, and give the correct solution set.

Audit Line 1. The student subtracted 5 from both sides to undo adding 5. Subtracting the same number from both sides keeps the order, so the symbol stays the same. The arithmetic is right. Line 1 is correct.
$-3x + 5 - 5 > 20 - 5 \text{, so } -3x > 15$
Audit Line 2. To undo multiplying by -3, the student divided both sides by -3, and the number is right. But dividing both sides by a negative number reverses the order, so the greater-than symbol must become a less-than symbol. The student kept greater than. Line 2 is the error.
$15 \div (-3) = -5$
Write the correct line. Divide both sides by -3 and reverse the symbol.
$\frac{-3x}{-3} < \frac{15}{-3} \text{, so } x < -5$
Boundary test. Substitute -5 for x in the original inequality. Both sides equal 20, so the boundary number is right. The original symbol is greater than, and 20 is not greater than 20, so -5 is not a solution. The corrected answer leaves -5 out, so it agrees.
$-3(-5) + 5 = 15 + 5 = 20$
Direction test. Substitute a value from the corrected answer, such as -6. The original inequality comes out true, so the corrected symbol faces the right way.
$-3(-6) + 5 = 18 + 5 = 23, \quad 23 > 20 \text{ is true}$
Now test the student's answer. The number 0 is greater than -5, so it belongs to the student's solution set. In the original inequality it gives a false statement, so that set contains a number that is not a solution.
$-3(0) + 5 = 0 + 5 = 5, \quad 5 > 20 \text{ is false}$

Answer

Line 2 is the first incorrect line: the student divided both sides by $-3$ without reversing the symbol. The correct solution set is $x < -5$ .

Watch out: do not reverse the symbol just because you see a negative number. Suppose a solution reads $4x - 7 \le -19$ , then $4x \le -12$ (adding $7$ ), then $x \le -3$ (dividing by $4$ ). A checker sees negative numbers and says the answer should be $x \ge -3$ . That is the mistake. Adding a number keeps the order, and so does dividing by the positive number $4$ . The reversal rule is about the number you multiply or divide by, not about negative numbers that appear in the inequality.

Substitution settles it. The value $x = 0$ belongs to $x \ge -3$ , but $4(0) - 7 = 0 - 7 = -7$ , and $-7 \le -19$ is false. The value $x = -4$ belongs to $x \le -3$ : $4(-4) - 7 = -16 - 7 = -23$ , and $-23 \le -19$ is true. The original work was correct.

Worked example

Audit 2: an inequality with division

A student solved $\frac{x}{-4} - 2 \ge 1$ in two lines. Line 1: $\frac{x}{-4} \ge 3$ . Line 2: $x < -12$ . Find the first line that does not follow from the line above it, explain the error, and give the correct solution set.

Audit Line 1. The student added 2 to both sides to undo subtracting 2. Adding the same number to both sides keeps the order, so the symbol stays the same. The arithmetic is right. Line 1 is correct.
$\frac{x}{-4} - 2 + 2 \ge 1 + 2 \text{, so } \frac{x}{-4} \ge 3$
Audit Line 2. To undo dividing by -4, the student multiplied both sides by -4, and the number is right. Multiplying both sides by a negative number reverses the direction of the symbol, and reversing keeps the or-equal-to part. The student reversed the direction but dropped that part. Line 2 is the error.
$3 \cdot (-4) = -12$
Write the correct line. Multiply both sides by -4 and reverse the symbol, keeping its or-equal-to part.
$\frac{x}{-4} \cdot (-4) \le 3 \cdot (-4) \text{, so } x \le -12$
Boundary test. Substitute -12 for x in the original inequality. Both sides equal 1, and the original symbol counts equal sides as true, so -12 is a solution. The student's answer leaves -12 out. The corrected answer keeps it in.
$\frac{-12}{-4} - 2 = 3 - 2 = 1, \quad 1 \ge 1 \text{ is true}$
Direction test. Substitute a value from the corrected answer, such as -16. The original inequality comes out true, so the corrected symbol faces the right way.
$\frac{-16}{-4} - 2 = 4 - 2 = 2, \quad 2 \ge 1 \text{ is true}$

Answer

Line 2 is the first incorrect line: multiplying both sides by $-4$ reverses $\ge$ to $\le$ , not to $<$ . The correct solution set is $x \le -12$ , which includes $-12$ .

A word problem adds one more thing to audit: the final sentence. The variable now stands for a real quantity, and the situation may limit its values. A count of games, people or tickets must be a whole number. Many quantities, such as a count or a length, cannot be negative. So the answer to the inequality can include values that make no sense in the situation, such as half a game or a negative number of tickets.

Keep one more habit in mind. A good checker also says when a line is correct. Sometimes every algebra line is right and the error is in the conclusion.

Worked example

Audit 3: a word problem

Devon's arcade card holds $50$ credits. Each game uses $8$ credits, and he wants at least $14$ credits left. Let $g$ be the number of games he plays. A student wrote $50 - 8g \ge 14$ , then Line 1: $-8g \ge -36$ , then Line 2: $g \le 4.5$ . The student rounded up and concluded that Devon can play up to $5$ games. Find the error and give the correct answer.

Audit the setup. Each game uses 8 credits, so g games use 8g credits and leave 50 - 8g credits on the card. At least 14 means 14 or more, so the symbol must be greater than or equal to. The setup is correct.
$50 - 8g \ge 14$
Audit Line 1. The student subtracted 50 from both sides. Subtracting the same number from both sides keeps the order, so the symbol stays the same. Line 1 is correct.
$50 - 8g - 50 \ge 14 - 50 \text{, so } -8g \ge -36$
Audit Line 2. The student divided both sides by -8. Dividing both sides by a negative number reverses the direction of the symbol and keeps its or-equal-to part, and the student did both. Line 2 is correct.
$\frac{-8g}{-8} \le \frac{-36}{-8} \text{, so } g \le 4.5$
Audit the conclusion. The variable g counts games, so it must be a whole number such as 0, 1 or 2. Negative values and values such as 4.5 make no sense here. The whole numbers that are at most 4.5 are 0, 1, 2, 3 and 4. The number 5 is greater than 4.5, so rounding up leaves the solution set. The conclusion is the error.
$g = 0, 1, 2, 3 \text{ or } 4$
Confirm with substitution. Put 5 in place of g in the original inequality. Only 10 credits are left, which is fewer than 14, so 5 games is not a solution.
$50 - 8(5) = 50 - 40 = 10, \quad 10 \ge 14 \text{ is false}$
Test the greatest allowed value, 4. Then 18 credits are left, which is at least 14, so 4 games is a solution.
$50 - 8(4) = 50 - 32 = 18, \quad 18 \ge 14 \text{ is true}$

Answer

The algebra lines are correct, and the error is in the conclusion. Because $g$ must be a whole number that is $0$ or more, Devon can play $0$ , $1$ , $2$ , $3$ or $4$ games, so at most $4$ games.

Takeaway. To check a solved inequality, do not trust how the work looks. Ask whether each line keeps the same solution set as the line above it. Adding or subtracting a number keeps the symbol, and so does multiplying or dividing by a positive number. Multiplying or dividing by a negative number reverses it. At every line, $<$ and $>$ leave the boundary out, and $\le$ and $\ge$ keep it in. Then substitute the boundary and one other value into the original inequality. In a word problem, keep only the values that fit the situation, which may mean whole numbers or numbers that are zero or more.