Stating identity is not the same as knowing value

Question

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.

Answer 1

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.

Answer 2

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 $b.$

Answer 3

(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.