# Reasoning with linear inequalities

Alignments to Content Standards: A-REI.B.3 A-REI.A.1

The following is a student solution to the inequality $$\frac{5}{18} - \frac{x-2}{9} \leq \frac{x-4}{6}.$$ \begin{align} \frac{5}{18} - \frac{x-2}{9} & \leq \frac{x-4}{6} \newline \frac{5}{18} - \frac22 \frac{x-2}{9} & \leq \frac33 \frac{x-4}{6} \newline \frac{5}{18} - \frac{2x-2}{18} & \leq \frac{3x-4}{18} \newline 5 - (2x - 2) & \leq 3x - 4 \newline 5 - 2x + 2 & \leq 3x - 4 \newline 7 - 2x & \leq 3x - 4 \newline -5x & \leq -11 \newline x & \leq \frac{11}{5} \newline \end{align}

1. There are two mathematical errors in this work. Identify at what step each mathematical error occurred and explain why it is mathematically incorrect.

The first mathematical error occurred going from line ____ to line ____.

Why it is incorrect:

The second mathematical error occurred going from line ____ to line ____.

Why it is incorrect:

2. How would you help the student understand his mistakes?
3. Solve the inequality correctly.

## IM Commentary

This problem is intended to detect the ability of the student to identify errors in mathematical reasoning, and to help students see the process of solving a equation or inequality is a special kind of proof. Teachers should spend time emphasizing the latter of these, that the two mathematical errors made in the deduction are logical missteps—the line is not a logical consequence of the previous one—and not just a failure to follow rules. Conversation might also be productively had around the difference between (to borrow terms from the Algebra Initiative) mathematical errors and strategic errors, that is, mathematically correct steps (i.e. the first line implies the second) which are not necessarily the most efficient at the given stage of the derivation.

This problem is intended for assessment in the Chicago Algebra Initiative (a joint program of Depaul University, University of Chicago and University of Illinois in Chicago with the Chicago Public Schools (CPS)). It is slightly adapted from one of 13 items on a 2 hour test required for a teacher with a K-8 certication to obtain a CPS-credential to high school algebra in 8th grade in the Chicago Public Schools.

## Solution

The following is a student solution to the inequality \begin{align} \frac{5}{18} - \frac{x-2}{9} & \leq \frac{x-4}{6} \newline \frac{5}{18} - \frac22 \frac{x-2}{9} & \leq \frac33 \frac{x-4}{6} \newline \frac{5}{18} - \frac{2x-2}{18} & \leq \frac{3x-4}{18} \newline 5 - (2x - 2) & \leq 3x - 4 \newline 5 - 2x + 2 & \leq 3x - 4 \newline 7 - 2x & \leq 3x - 4 \newline -5x & \leq -11 \newline x & \leq \frac{11}{5} \newline \end{align}

a. There are two mathematical errors in this work. Identify at what step each mathematical error occurred and explain why they are mathematically incorrect.

The first mathematical error occurred going from line __ 2 __ to line __ 3 __.

Why it is incorrect: The error is an incorrect application of the distributive property in two places. The second term should be $\frac{2x-4}{18}$ and the term on the right hand side of the inequality should be $\frac{3x-12}{18}$.

The second mathematical error occurred going from line __ 7 __ to line __ 8 __.

Why it is incorrect: The student divided both sides of an inequality by -5; the direction of the inequality should be reversed when multiplying (dividing) by a negative number.

b. How would you help the student understand his mistakes?

First mistake:

Here are some possible responses.

I would remind the student of the statement of the distributive property $a(b + c) = ab + ac$ and recall that since $a - c$ is an abbreviation of $a(b + c)$, this also implies $a(b - c) = ab - ac$.

To illustrate the distributive property I would ask the student to evaluate 2(5 - 2) in two ways. Then I would point out that the rule still held when 5 was replaced by $x$.

To illustrate the distributive property I would ask the student to evaluate $2(x-2)$. Then I would point out that the same distribution takes place when the $x - 2$ is the numerator of a fraction.

The student may have a deep conceptual confusion or made have made a 'careless' mistake caused by misreading the problem. A blind application of PEMDAS might lead the student to think that multiplication before addition means the expression $2 x-2$ (ignoring the denominator to multiply the numerators) evaluates to $2x - 2$. We should remember that the fraction bar means that the numerator and the denominator are each a term. So, $$\frac{x-2}{9} = (x-2)(9^{-1}).$$

There are several ways in which the 'multiplication by 1' in line 2 might be written: \begin{align} & \frac22 \frac{x-2}{9}, \newline & \frac22 \left( \frac{x-2}{9} \right), \newline & \frac22 \cdot \frac{x-2}{9}. \end{align} Students may not recognize the equivalence of these various expressions.

Second mistake:

To illustrate the law that if $b \leq c$ and $a$ is negative then $ab \geq bc$, I would ask, what the result would be dividing from both sides of the inequality $6 \leq 12$ by -2. That is, which is bigger $\frac{6}{-2}$ or $\frac{12}{-2}$?

c. Solve the inequality correctly.

Method I. This method minimally corrects the students attempt. \begin{align} \frac{5}{18} - \frac{x-2}{9} & \leq \frac{x-4}{6} \newline \frac{5}{18} - \frac22 \left( \frac{x-2}{9} \right) & \leq \frac33 \left( \frac{x-4}{6} \right) \newline \frac{5}{18} - \frac{2x-4}{18} & \leq \frac{3x-12}{18} \newline 5 - (2x - 4) & \leq 3x - 12 \newline 5 - 2x + 4 & \leq 3x - 12 \newline 9 - 2x & \leq 3x - 12 \newline -5x & \leq -21 \newline x & \geq \frac{21}{5} \newline \end{align}

Method II. This is a slightly more efficient method. \begin{align} \frac{5}{18} - \frac{x-2}{9} & \leq \frac{x-4}{6} \newline 18 \left( \frac{5}{18} - \frac{x-2}{9} \right) & \leq 18 \left( \frac{x-4}{6} \right) \newline 5 - 2(x - 2) & \leq 3(x - 4) \newline 5 - 2x + 4 & \leq 3x - 12 \newline 9 - 2x & \leq 3x - 12 \newline -5x & \leq -21 \newline x & \geq \frac{21}{5} \newline \end{align}