Harnessing misuse of equals sign

Question

Students often misuse the equals sign to indicate "I've done this operation" rather than the proper use indicating numerical equivalence.

Eg. Tax is paid using the rule: \$3 572 plus 32.5c per \$1 over $37 000. How much tax is paid on \$78 540?

Answer 1: 78 540-37 000 = 41 540 x 32.5% = 13 500.50 + 3572 = $17 072.50

Answer 2: 3 572+(78 540-37 000)x 32.5% = 3 572+41 540 x 32.5% = 3 572+13 500.50 = $17 072.50

Both methods produce the correct answer. But Answer 1 misuses "=" to mean "now we do this", perhaps it also shows a lack of understanding of order of operations and algebra, but fundamentally it is a good method.

How could this method be written to not misuse the equals sign?

How could this method be used to introduce the correct use of the equals sign?

Observation: What they are doing is writing the expression (a+(b-c) x d) using a form of postfix notation: bc-dxa+ which is quite a rational thing to do.

Answer 1

I would write this as:

Earnings over $\$37,000$: $\$78,540-\$37,000 = \$41,540$

Base tax: $\$3,572$

Additional tax: $\$41,540\times 0.325 = \$13,500.50$

Total tax: $\$3,572+\$13,500.50 = \$17 072.50$

Answer 2

Answer 1,

$$78 540-37 000 = 41 540 \times 32.5\% = 13 500.50 + 3572 = $17 072.50$$

could be rewritten as: \begin{align*} 78540-37000 &=41540\\ 41540\times 32.5\% & =13500.50\\ 13500.50+3572 & =17072.50 \end{align*}

The use of $\leadsto$ has been suggested in the comments below to indicate each next step in the reasoning. At the level where a student makes this kind of mistake, they might not be ready for the introduction of further notation (if they get confused with $=$, they might get even more confused with some new symbol). As "guest" has suggested below, simply writing each step on a different line should be sufficient.

Answer 3

Update. It appears that the intent of this answer was not clear to some readers, so I will elaborate. The most common way to fix the student's computations is simply to break out the subexpressions onto separate lines (as in the other answers). There are however other noteworthy approaches that offer some advantages. Below we briefly present one such approach.

Observe that the student uses the equal sign as it functions on an (infix) calculator, i.e. to evaluate the preceding expression. Using variables (and disambiguating parentheses) the student's expression is essentially the first one displayed below, where the equal sign has been (ab)used to annotate (label) subexpressions with their values $\,\color{#0a0}n,\color{#c00}m.\,$

$$\begin{align} ((a\!-\!b =\color{#0a0}n) \times c = \color{#c00}m) + d &\\[.2em] \underbrace{\overbrace{(a\ \ -\ \ b)}^{\Large\color{#0a0} n}\ \ \times\ \ c}_{\Large\color{#c00} m}\ \ +\ \ d\ \ & \end{align}$$

We can greatly improve this by using better ways to annotate the values $\,\color{#0a0}n,\color{#c00}m.\,$ e.g. as above using under/overbraces. Notice how this helps clarify the algebraic structure of the expression, e.g. now we can see with a single glance that it depends linearly on $c$ (interest rate) so we can make inferences about how its value changes with changes in the interest rate. But this innate algebraic structure is greatly obfusctated if instead we dissect the expression into many subexpressions and break them out onto separate lines (into a "straight line computation").

So my point is that we should teach students various ways to present such computations - not only the common straight-line methods. When one encounters more complex expressions the structure-preserving approaches can make a huge difference in reducing complexity. For a less trivial example, below is an excerpt from one of my MSE posts on telescopic induction.. Notice how the over/underline annotations below allow one to comprehend in a single glance the effect of the telescopic cancellations - which would be greatly obfuscated if the intermediate annotated expressions were broken out onto separate lines.

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Hint $\: $ First trivially inductively prove the Fundamental Theorem of Difference Calculus

$$\rm\ F(n)\ =\ \sum_{i\: =\: 1}^n\:\ f(i)\ \ \iff\ \ \ \color{#c00}{F(1)=f(1)},\,\ \ \color{#0a0}{F(n) - F(n\!-\!1)\ =\ f(n)}\ \ {\rm for}\ \ n> 1$$

whose proof is simply a rigorous inductive proof of the following telescopic cancellation

$$\rm \underbrace{\overbrace{\color{#c00}{F(1)}}^{\large \color{#c00}{f(1)}}\phantom{-\color{#c00}{F(1)}}}_{\large =\ 0}\!\!\!\!\!\!\!\!\!\!\!\!\!\overbrace{\color{#0a0}{-\,F(1)\!+}\!\phantom{F(2)}}^{\large\color{#0a0}{ f(2)}}\!\!\!\!\!\!\!\!\! \underbrace{\color{#0a0}{F(2)} -F(2)}_{\large =\ 0}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\overbrace{\phantom{-F(2)}+ F(3)}^{\large f(3)}\!\!\!\!\!\!\!\!\!\!\underbrace{\phantom{F(3)}-F(3)}_{\large =\ 0}+\,\underbrace{\cdot\ \cdot\ }_{\large =\,0\,}\overbrace{\cdot\ +F(n)}^{\large f(n)}\ =\ F(n) $$