Timeline for How to respond to “solve this equation” in a basic algebra class
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 24, 2023 at 23:37 | comment | added | Martin Kealey | In my opinion “𝑥∈{2,-3}” doesn't even fully capture the sense of THE solution, since without “∀𝑥” it can reasonably be read as “∃𝑥”: some solution in 𝑥 exists, and that it's either 2 or -3, but a full solution must assert that both values would make the original quadratic true. | |
| Apr 10, 2017 at 13:57 | comment | added | Henry Towsner | @slebetman: I included all three for a reason, and I don't agree that the third alone is enough. There are times where we should demand formality of students that they'll later be allowed, and we do this precisely when the stricter notation helps them understand concepts. We teach notations that are customary or useful; the failure here is that it's neither. | |
| Apr 10, 2017 at 7:00 | comment | added | slebetman | The last two points are most important: guide students to understand what they will encounter later in life - don't disagree with the rest of the world. When my son was in first grade he failed a test that asked something like what's the total number of oranges in five boxes of oranges if each box contains ten oranges. He answered 5x10=50, the teacher wanted him to answer 10x5=50 so gave him zero marks for all questions of that form. | |
| Apr 8, 2017 at 21:17 | comment | added | Veedrac | Perhaps this is my inexperience talking, but I would consider $x = 3 - 2$ to be a solution, for the same reason I'd consider $x = \sqrt{2}$, $x = 4 \times 10^3$, or $x = 2\pi$ to be solutions to other equations. This holds even more with the "solution set" formulation, where $\{1\}$ and $\{3 - 2\}$ express exactly the same thing. One normally assumes that the answer is wanted in its maximally-simplified form, of course, and is not so pedantic as to cause a fuss. | |
| Apr 7, 2017 at 22:20 | comment | added | Andreas Blass | +1 I'd add that the objection to "an equivalent equation" doesn't make sense to me. If an equivalent equation explicitly exhibits the value of the unknown, then that equation constitutes a solution. | |
| Apr 7, 2017 at 16:56 | history | answered | Henry Towsner | CC BY-SA 3.0 |