STUDENT: I think it’s x times 3 minus 3, so x 3 minus 3, yeah, because, because, then.
STUDENT: Yeah. It works! Yeah!
STUDENT: No, because
STUDENT: Times 3 minus 3..
STUDENT: minus 3 would be negative 3…
STUDENT: Oh, yeah, so it is impossible!
STUDENT: Because, because, 0 times 3
STUDENT: Yeah, because anything times 0
STUDENT: and then minus 3, you’re going negative, and we’re only…
STUDENT: Yeah, yeah. I wonder why he said 0? He said that 0 could be a possibility.
STUDENT: I kind of understand where Griffin was coming from, though.
STUDENT: Yeah, like if he thought, that…
STUDENT: Yeah, it’s kind of fun.
OBSERVING TEACHER: Are you agreeing?
STUDENT: Then it’s minus 3.
FRAN DICKINSON: So tell me. What were you hearing at your table? Kristin , what did you hear at your table?
STUDENT: Um, well, we thought that it could work, because your parameter was greater than or equal to 0
FRAN DICKINSON: All right. Christina, what did you hear at your table?
STUDENT: Well, we thought it wouldn’t work, because, well, we think it would be negative 3. Instead of 0. Like, we didn’t think that 0 would work.
FRAN DICKINSON: What would be negative 3? Can you be more specific?
STUDENT: The, the…
STUDENT: y… the...
STUDENT: the x value…
STUDENT: it would be the y. The, um, output. Because, like, 0 wouldn’t be positive.
FRAN DICKINSON: 0, so, let me get some clarification now. So Griffin’s guess was, his guess was a y value. He guessed 0. So maybe I should write that on the board here. So. And my question to you was, is this possible? So I asked a very vague question there. What do you think, Robby?
STUDENT: It’s kind of, because, I kind of noticed this, it’s because the y, in every one, is bigger than the x, and your parameters say that x has to be greater than or equal to 0, so if the x is smaller than the y, it can’t be smaller than 0, so it wouldn’t work.
FRAN DICKINSON: Can someone put in different words what they hear Robby saying? Kylie.
STUDENT: Um, with all the numbers we’ve done so far, the y is bigger than x. so if the y is 0, then x would have to be lower than 0. But the parameters say that x has to be equal to or greater than 0. So it wouldn’t work.
FRAN DICKINSON: I see that I have some comments around the room. Eric, would you like to make a comment?
STUDENT: I think it works. I think, can I say what the x would be?
FRAN DICKINSON: Sure!
STUDENT: I think it would be 1, because, um. The other ones work that x times 3 minus 3, equals, um, y? and 1, so 1 times 3 is equal to 3, minus 3 is zero. So I thought it would work.
FRAN DICKINSON: I heard some “Ahhhhs.” Did I hear that somewhere? I see some agreement around the room as well. Teo, what are you agreeing with?
STUDENT: I’m agreeing, because we talked about the same thing at our table.Because I was thinking negative at first but then I saw it as a y value, not the x value. So…
STUDENT: And I thought, I don’t know. And then Griffin told me what he was thinking. About what he said.
FRAN DICKINSON: Very good! Can I see agreement in the room? Or disagreement. This point here, is 1, 0.