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A discussion with a frustrated 10th grade student sent me here. I had provided two linear function expressions, $f(x)=2x+2$ and $g(x)=-\frac{1}{2}x-2$, now find the intersection of the two lines!

The student seemed clear that there was the same $y$-value for both $f$ and $g$, thus the equation $f(x)=g(x)$ made sense, he suggested. But planning his work with this, he failed to understand, how he could use the equation, where he knew for sure that the $y=f(x)$ indeed was the same as the $y=g(x)$: Why couldn't he finish the task by solving $2x+2=-\frac{1}{2}x-2$, getting $x=-\frac{8}{5}$? So where did the $y$ value go? He just used it, but it is not there...

What I find interesting, is that he -- in a manner of speaking -- realises only one side of a dual problem. But this duality was hard to get to the students understanding. Too hard, actually...

Thus, my question may be phrased: What problems may build students' understanding of the duality of arguing for identity and establishing value?

In the example, the student was perfectly clear, that the two lines had same height in intersection. He failed, however to see that the height, or $y$ value, was not yet established.

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    $\begingroup$ I wonder, would it help to do the actions of solving $2x+2=-\frac{1}{2}x-2$, not to the pair but to the triple: $2x+2=y=-\frac{1}{2}x-2$? You would then get $x=\frac{2}{5}y+\frac{1}{5}x-\frac{4}{5} = -\frac{8}{5}$ so the $y$ didn't go anywhere, it just didn't end up being manipulated to something informative. $\endgroup$ – Adam Nov 6 '17 at 0:34
  • $\begingroup$ Nice twist! Certainly, I shall your perspective into consideration. $\endgroup$ – Engelsmann Nov 6 '17 at 11:01
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    $\begingroup$ I would just give a clear direction: "Express the answer as an $(x, y)$ point." $\endgroup$ – Daniel R. Collins Nov 6 '17 at 13:22
  • $\begingroup$ Could you clarify what steps the student could complete? Did he find a value for x? (I answered below, assuming that he did) $\endgroup$ – josinalvo Nov 6 '17 at 15:46
  • $\begingroup$ At this point, I shall only quote DUVAL “A Cognitive Analysis of Problems pf Comprehension in a Learning of Mathematics” 2006 for the following: “[???] representations can be individuals’ beliefs, conceptions or misconceptions to which one gets access through the individual’s verbal or schematic productions”. Page 104. Thus, my question might have been better off calling for representations rather than for exercises. $\endgroup$ – Engelsmann Nov 12 '17 at 15:34
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I had a similar interaction with a student this weekend. I tend to walk it back to talk about these objects are just ways we invented to talk about number and quantity. They translate into a language. So many students get focussed on the machine process of manipulation of quantity to get results. In that they lose the idea these expressions represent something.

In the case you have I would add some mechanical process to their work but explaining the what it means.

In cases of equal signs I use $y=f(x)$ means wherever we see $y$ we can replace it with $f(x)$.

So I would have the student set up what they know first:

$f(x)=2x+2$ $g(x)=-\frac{1}{2}x-2$

The student identified that:

$y=f(x)$ $y=g(x)$

Thus: $f(x)=g(x)$

Here I would tell the student if $f(x)=2x+2$ means wherever I see $f(x)$ I can replace it with $2x+2$. Until they can produce:

$2x+2=g(x)$

I would repeat the argument for $g(x)$ to produce:

$2x+2= -\frac{1}{2}x-2$

Then you would want to reflect that this statement says the same thing as $y=y$.

From here the discussion will be that this equality tells us the $x$ value where the $y$ values are equal, even though we don't know what the $y$ value is.

Once the $x$ value is found the question is well what was that $y$ value.

Maybe even presenting a problem of this nature and emphasizing what equality lets us know on each step, while tracking what still isn't known.

I do this in tutoring often. It helps students also avoid the habit of ending problems early if they can track what each step resolves for them.

As an extension in my Physics class I have some problems where we find a value for $v^2$ in on one step, and in the next step we can substitute $v^2$ directly in. Many student will solve for $v$ and square again. I try to emphasize that equality statements allow you to swap expressions no matter if it has been isolated to one variable or not.

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  • $\begingroup$ Valuabke guidance is your insisting on "cross-identity": I quote " Until they can produce 2x+2=g(x) I would ..." $\endgroup$ – Engelsmann Nov 6 '17 at 6:05
  • $\begingroup$ The quoted equality should indeed move the student"s understanding - and thus her learning - towards realizing the duality, I look for, by using identity without knowing the y value. Something to work with! $\endgroup$ – Engelsmann Nov 6 '17 at 6:25
  • $\begingroup$ You emphasize procedure in the sentence “Once the x value is found”. I agree as to the problem solving has both a procedural and a semantic side to it. $\endgroup$ – Engelsmann Nov 6 '17 at 6:43
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I'm not entirely sure I understand what the student can do and cannot do, but I believe something I sometimes used to do might help. You have two lines, one given by $y = 2x + 2$ and the other given by $y = -\frac{1}{2}x – 2,$ and you want to find the point that is on both lines. To keep from getting confused with "$x$ as a variable", and "$x$ as a value that is fixed by an equation", and other things, let $(a,b)$ be the coordinates of the point that is on both lines. Then what we want to find are the values of $a$ and $b.$ (See this 20 January 2011 ap-calculus post at Math Forum for a situation where using this method is even more helpful.)

Since $(a,b)$ is on the line $y = 2x + 2,$ we have $b = 2a + 2.$ Since $(a,b)$ is on the line $y = -\frac{1}{2}x – 2,$ we have $b = -\frac{1}{2}a – 2.$ Now we have two equations for the two unknowns. Solving this system of simultaneous equations gives us the values of $a$ and $b.$

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  • $\begingroup$ Admittedly, the cause of my student"s frustration is hard to conceptualize. But I tend to finding it in the fact that the $y$-s are equal but not known rather than your distinction between "variable" and "fixed by equation". It is somehow to distinguish between " fixed" in your sense, and *value isolated", alas, I am not a native English speaker... $\endgroup$ – Engelsmann Nov 5 '17 at 21:31
  • $\begingroup$ "equal but not known" Maybe an example like this? I am thinking of a number. If John multiplies the number by three, and Susan adds 6 to the number, then John and Susan get the same result. What is the number? $\endgroup$ – Dave L Renfro Nov 5 '17 at 21:38
  • $\begingroup$ Could you point me to exercises which students may realize (in reflection after their work) afe intended to exercise away their misconception? - Or even "forcing" them to better understand the mentioned duality? $\endgroup$ – Engelsmann Nov 5 '17 at 21:42
  • $\begingroup$ I don't think I've ever seen exercises that so specifically target the issues you are talking about. I think this is just something that students gradually internalize (i.e. make part of one's nature by learning, or by unconscious assimilation) when learning a new topic or skill. $\endgroup$ – Dave L Renfro Nov 5 '17 at 21:53
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(I assume the student found x but not y. Was that the case? Where did the student get stuck?)

I'd return to the concrete model (i.e.: the cartesian plane).

Take some squared paper, and show some points (this is point (3,4), this is point (8,-1))

Then, draw some lines that have an integer intersection, and ask for the point in question. (possibly marking the values of x and y points on the axis and drawing an dotted line between the intersection and each axis)

To end, draw the lines of the problem, show the student the x already discovered, and draw the dotted line to the y axis.

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  • $\begingroup$ Your procedure is fairly much what I ask for, only I would prefer a set of several exercises or the like where students could try out (and possibly discuss with fellow students) how the difference between identity and known value smells and feels. The good people contributing with comments and answers have send me down several paths to try out, and having tried them out several times, I hope to be see clearly the two sides of "value": the one giving identity and the one making up the known value. $\endgroup$ – Engelsmann Nov 8 '17 at 19:49

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