Error Analysis: Writing and Graphing Inequalities | Grade 7 Math | LevelKite

An inequality's graph makes a claim about its solutions, the values of the variable that make the inequality true. The circle sits on the boundary, the number where the two sides of the inequality are equal. A filled circle claims the boundary is a solution, and a hollow circle claims it is not. The ray, the thick arrow from the circle, claims that every number on its side is a solution and that every number on the other side is not.

In error analysis, you get to be the checker: you find a mistake in someone's work, name it and fix it. A graph is a claim, so you can test it with numbers.

Key concept

Test the boundary and one value on each side

Test numbers in the original inequality or words, not in the answer you are checking, since that answer may hold the mistake.

The boundary decides the circle. If it makes the inequality true, the circle should be filled, as with $\le$ and $\ge$ , which include the boundary. If not, it should be hollow, as with $<$ and $>$ , which exclude it.
One value on each side decides the ray. The ray belongs on the side whose value makes the inequality true. In a one-step inequality, a test result can change only at the boundary, so one test per side is enough.

When solving, multiplying or dividing both sides by a negative number reverses the symbol; a positive number never does. A context may limit solutions to whole or nonnegative (zero or positive) numbers.

Start with a fault in a graph. Dana solved $-2x > 6$ , where $x$ is an unknown number, and wrote $x < -3$ . The number line below shows values of $x$ . Dana's graph was meant to show $x < -3$ , but it is flawed: it has a filled circle at $-3$ and a ray to the left.

Dana's flawed graph: a filled circle at

-3

and a ray to the left.

Worked example

Test Dana's flawed graph above using the original inequality, $-2x > 6$ . Name her fault and give the corrected graph.

Test the boundary, $-3$ . Multiply first: $-2(-3) = 6$ . Then compare: $6 > 6$ is false, because $6$ is equal to $6$ , not greater. The symbol $>$ excludes the boundary, so $-3$ is not a solution and the circle must be hollow.
$-2(-3) = 6, \quad 6 > 6 \text{ is false}$
Test $-4$ , a value left of the boundary. Since $-2(-4) = 8$ and $8 > 6$ is true, $-4$ is a solution, so the ray belongs on the left side.
$-2(-4) = 8, \quad 8 > 6 \text{ is true}$
Test $-2$ , a value right of the boundary. Since $-2(-2) = 4$ and $4 > 6$ is false, the right side holds no solutions, and Dana's ray to the left is correct. This also confirms her reversed symbol, which dividing by the negative number $-2$ requires.
$-2(-2) = 4, \quad 4 > 6 \text{ is false}$
Name the fault: Dana drew a filled circle at an excluded boundary, since $>$ and $<$ both exclude it. The fix is a hollow circle at $-3$ with the ray still running left, as in the corrected graph below.

Answer

Dana's only mistake is the filled circle: the graph of $x < -3$ needs a hollow circle at $-3$ and a ray to the left.

Corrected: a hollow circle at

-3

and a ray to the left show

x < -3

.

Some faults start earlier, when words become symbols. A club sells tickets for 5 dollars each and must raise at least 180 dollars. Let $t$ be the number of tickets sold, so the club raises $5t$ dollars. The graph should show every value of $t$ for which $5t$ is at least $180$ . Leo wrote $5t < 180$ and divided both sides by $5$ to get $t < 36$ . His flawed graph below, on a number line of $t$ -values, has a hollow circle at $36$ and a ray to the left. His inequality may itself be wrong, so test values against the words, not his symbols.

Leo's flawed graph: a hollow circle at

36

and a ray to the left.

Worked example

Test Leo's flawed graph above against the words 'at least 180 dollars'. Name his fault and give the corrected inequality and graph.

Test the boundary, $36$ . Selling $36$ tickets raises $5(36) = 180$ dollars, and 180 dollars is at least 180 dollars, since 'at least' counts 180 itself. So $36$ is a solution, and the circle must be filled, not hollow.
$5(36) = 180, \quad 180 \ge 180 \text{ is true}$
Test $35$ on the left and $37$ on the right. Selling $35$ tickets raises $175$ dollars, which is not enough; selling $37$ raises $185$ dollars, which is enough. So the ray must run to the right, not the left.
$5(35) = 175 < 180, \quad 5(37) = 185 \ge 180$
Name the fault: Leo wrote $<$ for 'at least'. 'At least 180' means 180 or more, which is $\ge$ . That one wrong symbol caused both graph faults: $<$ excludes the boundary, and $t < 36$ takes the values below $36$ instead of the values above it.
$\text{Leo: } 5t < 180 \qquad \text{correct: } 5t \ge 180$
Solve the corrected inequality by dividing both sides by $5$ . Dividing both sides by the same positive number keeps their order, so the symbol stays $\ge$ .
$\frac{5t}{5} \ge \frac{180}{5}, \quad t \ge 36$
Apply the context: $t$ counts tickets, so it must be a whole number. The corrected graph below shows $t \ge 36$ , with a filled circle at $36$ because $\ge$ includes the boundary, and a ray to the right. Only the whole numbers on the ray, $36$ , $37$ , $38$ and so on, make sense here.

Answer

Leo should have written $5t \ge 180$ , which gives $t \ge 36$ : the club must sell $36$ or more tickets, graphed with a filled circle at $36$ and a ray to the right.

Corrected: a filled circle at

36

and a ray to the right show

t \ge 36

.

To check inequality work, test it with numbers instead of trusting how it looks, and use the original inequality or words. The boundary decides the circle: filled for $\le$ or $\ge$ , which include it, and hollow for $<$ or $>$ , which exclude it. In a one-step inequality, one value on each side decides where the ray runs. Together, these tests catch a wrong circle, a wrong side, a misread phrase such as 'at least' and a symbol that was not reversed after multiplying or dividing both sides by a negative number. In a context, keep only the values that make sense, such as whole or nonnegative numbers.