Sign Errors in 5 − 2x = 13
Mr. Bhatt's Grade 9 applied (MFM1P) class of 22 has just written an exit card: solve 5 − 2x = 13. Three responses stand out. S1: 2x = 13 − 5 = 8, so x = 4. S2: −2x = 13 + 5 = 18, so x = −9. S3: builds a table of values for the expression, finds that x = −4 gives 13, writes 'x = −4, I checked,' and shows no method. A common department quiz covering inverse operations and the distributive property is in two days. Six students are on IEPs, and the department head has asked that every class write the same quiz on the same day.
1. Diagnose each response and choose the single error most worth making public tomorrow. Justify the choice against what the quiz will assess.
Read the worked answer
- S1 deleted the negative coefficient. The student is operating on '2x' as a label for the term rather than on the whole signed quantity −2x, so the sign never enters the calculation.
- S2 isolated the term correctly but chose the wrong inverse for the constant: added 5 instead of subtracting 5. The student never checked, or would have found 5 + 18 = 23 ≠ 13.
- S3 shows the concept of a solution (a value making the equation true) and good number sense, but no efficient procedure.
Make S2 public. It is the failure the quiz will punish most widely, because the same 'wrong inverse operation' move reappears in the distributive-property items, and it is the one error a thirty-second check repairs. S1's error is not a lesson topic but a tracking issue that needs the balance model and individual follow-up; putting it on the board invites the class to memorise 'don't lose the minus,' which is not a repair. S2's error, by contrast, can be resolved by a generalisable habit: substitute and verify.
2. S3's table is correct. Explain how you would use it in tomorrow's lesson and what it would and would not accomplish before the quiz.
Read the worked answer
How I would use it: open with it, because it establishes the definition of a solution before any procedure arrives. Then convert it into a recording method: from x = −4, work forwards in words - 'double it, negate it, add five' - and then undo those steps in reverse. The table becomes the evidence that the undoing is right, not a rival method.
What it accomplishes: it gives the whole class, including students who are not fluent with signed numbers, a way to check any solution the quiz asks for. That closes S2's error and S1's, since substitution catches both.
What it does not accomplish: it does not scale. The quiz will include items such as 3(x − 4) = 15 and equations with non-integer solutions, where a table is laborious or impossible to bracket. It also gives no language for why the operation must be applied to both sides, which is what the next course assumes. So S3's table is the entry and the check - never the destination.
3. A colleague proposes teaching 'move it across and change the sign' as the dependable method for this class. How would you respond, and what specifically would you teach instead?
Read the worked answer
I would agree that the shortcut is real mathematics - it is a compression of subtracting the same quantity from both sides - and then show where it breaks for this class. A rule about symbols moving across an equals sign is detached from equality, so it gives no reason to keep −2x intact (S1) and no reason to prefer subtracting 5 to adding 5 (S2). It also fails the moment students meet variables on both sides or multiply by a negative in an inequality, where something genuinely new happens.
What I would teach instead: a two-column record where every line names the operation applied to both sides.
5 − 2x = 13 · subtract 5 from both sides → −2x = 8 · divide both sides by −2 → x = −4 · check: 5 − 2(−4) = 13 ✓
Then, in the last five minutes, I would show the shortcut as the compression of that record - offered after the reasoning, so it is a convenience students can justify rather than a substitute for justification.