The acclaimed pocket books from **Professor Smudge **Published by DEXTER Graphics

High res versions of some of our YouTube movies can be downloaded from this page.

Click on the movie's link, below, and then go to**Save Page As...** in the **File** menu.

Click on the movie's link, below, and then go to

The value of *x* changes - what happens to the shape of the object?

Pause the movie, to give you time to think.

(Note: the scale used for the drawing is not constant....)

What happens when *x* = 100, say? Would the object go off the screen?)

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It involves the familiar and age-old bar model, but not in the usual static way.

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The aim is to provide insight into the algebraic relations, rather than a quick-fix solution procedure.

Again, it matches our movie "A dynamic bar model for an equation (1)" about the equation 3a + 2 = 2a + 8.

Our sense is, the 1D model lends itself to an analytic approach, ie to transforming the given equation into something simpler.In the 2D model, it is perhaps easier to think about the relative values of the two expressions, as the value of 'a' varies ("Is the rectangle tall and thin or short and wide?").

As the variable 'e' changes, there is a value where the expressions 3e+2 and 2e+8 have the same value.

We then modify the scale factors used to determine the lengths of the rods and thus consider the modified equations 3e + 2 = 1.5e + 8 and 2.5e + 2 = 1.5e + 8.

The last equation turns out to have the same solution as the first. Why?

It might be argued that it is pushing the model rather far to use it for such an equation, thought it is a nice challenge to make sense of the model here.

[The equation is equivalent to e^2 + 3.5e – 36 = 0. There is of course another solution, not shown.]

The equation is equivalent to e^2 + 3.5e – 36 = 0 or (e+8)(e–4.5)=0.

In effect, we are transforming the equation into (e + 3)^2 = 135 + 9 = 144. Here 'e' only appears once, so we can solve the equation relatively easily by trial-and-improvement, by 'inspection' or the 'cover-up' method,

[Note: we only show the positive root here. The model can be used for the other root (e = -15) but the overlapping rectangles are difficult to construe especially if both roots are to be shown on the same screen.]