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I was debating with a maths teacher about how to simplify algebraic fractions, and what their simplest form is. From the starting point of $\frac{100x+t(300-x)}{300}=75$, you can easily simplify down to $x=\frac{22500-300t}{100-t}$. This is where they advocated leaving it, as it's supposedly in simplest form. I think, however, that even though it can't be truly simplified, turning it into something such as $x=300-\frac{7500}{100-t}$ is much better, even if it does introduce an extra term.

It's important to note that this has nothing to do with marking. I learnt algebraic fractions five or six years ago. This is just a fun little debate about which form should be considered simplified.

Elliott Price
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    there will not generally be a ''simplest" form for intricate expressions. For your example, a Mobius Transformation is a useful shape since it is quickly inverted.... what do you get when solving for $t$ in terms of $x ; ? ; ; $ – Will Jagy Oct 25 '23 at 00:42
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    I would say whatever seems cleanest for the situation; there are no hard and fast rules. When teachers absolutely insist on a certain final form and are too firm on it, it can lead to needless errors and wasted time. Like when we were kids and $7/3$ just had to be $2\frac13$ or even forcing to rationalize things like $3/\sqrt{5}$ else lose points. If you use or know Delta Math for HS HW it takes answers with great flexibility which is actually a nice feature. Hope this helps. – AlgTop1854 Oct 25 '23 at 00:46
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    "Simplest" is in the eye of the beholder. (Mathematica and I often have very different ideas about what the Simplify[] command should yield, especially with trig functions.) I happen to like your form better ... better still: $$x=300+\frac{7500}{t-100}$$ This eliminates an unnecessary subtraction. Note also that the implicit form $$(x-300)(t-100)=7500$$ provides a bit of insight about the relation's symmetry; in particular, one may recognize this as representing a "rectangular hyperbola" with asymptotes $x=300$ and $t=100$. ... Use whatever form best suits your (and your readers') needs. – Blue Oct 25 '23 at 01:16
  • As a practical example ... In this recent answer of mine, I worked a bit to find the most useful form of a rather convoluted expression for the value $(a')^2$, and then realized that subtracting-off $a^2$ left something that reduced very nicely. See Equation (3) ... and observe that, to aid in the discussion going forward, I chose to express the relation as "$(a')^2-a^2=\text{nice fraction}$" rather than to keep $(a')^2$ by itself. ... Mathematics, as with any form of communication, allows for differences in presentational style. – Blue Oct 25 '23 at 01:32

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