From Law to STEM: A lawyer’s journey from ground zero (part 3)

Attorney Celeste
3 min readNov 19, 2020

This is a series about a lawyer navigating her way through math / maths / mathematics / quantitative analysis.

One of the great things about being a lawyer is that you learn to think critically about an argument. Algebra is similar, in that it forces you to think logically — balancing all of the variables or factors. A good attorney uses different facts, factors, and issues to solve legal problems.

Another way to use maths is in thinking about the attorney work week. For instance, is the 80 hour work week worth it — for $80,000 per year? What about $160,000 for 80 hours? 80 hours is common in the legal field (hence, the high rate of burnout).

Using maths, we see that 80 weeks for $80,000 is less than $20 per hour. For $160,000, that breaks down to less than $40.00.

Photo by Jonathan Brinkhorst on Unsplash

We can create a function to calculate the hourly breakdown.
h * w = s
Where…
* h = weekly hours
* w = weeks in year / period
* s = salary

Below, we will solve for “h” (hourly rate):
(a) For $80k salary: (80h) * 52 = 80000
<a1> (h)(80 * 52) = 80000
<a2> (h)(80 * 52) ÷ (80 * 52) = 80000 ÷ (80 * 52)
<a3> (h) = 80000 ÷ (80 * 52)
<a3> (h) = 80000 ÷ (4160)
<a4> (h) ≈ $19.23

(b) For $160k salary: (80h) * 52 = 160000
<b1> 2(h) = $38.46

Thus, as a lawyer, you can earn less per hour than a salesperson, a painter, or a preschool teacher. (https://www.indeed.com/q-Entry-Level-$30-jobs.html). So for good measure, we’ll look at someone working for $50k per year, at just 40 hours per week (lucky, lol). That breaks down to about $24.04 per hour. [$50000/(40*52) ≈ $24]

… Law school does not teach you numerical maths!

Today, we are going to discuss the following topics using the lawyer method of critical thinking:
* SOLVE AN EQUATION USING SUBTRACTION
* SOLVE AN EQUATION USING ADDITION
Think of these topics as mental tools to solve math problems.

SOLVE AN EQUATION USING SUBTRACTION
- Issue: How do we use subtraction to ensure the variable (literal factor / letter) is the same as the number on the other side of the equal sign?
- Rules: Get the variable alone. Subtract both sides in the equation by the same number. Subtract the side with the variable and the number on the other side of the equation.
- Application:
* D + 13 = 36
* D + 13 (-13) = 36 (-13)
* D = 23
- Conclusion:
* D = 23
* Validate / Check:
* (23) + 13 ≟ 36
* 36 = 36
* Validated because 23 plus 13 is 36.

SOLVE AN EQUATION USING ADDITION
- Issue: How do we use addition to ensure the variable (literal factor / letter) is the same as the number on the other side of the equal sign?
- Rules: Get the variable alone. Add both sides in the equation by the same number. Add the side with the variable and the number on the other side of the equation.
- Application:
* D — 12 = 36
* D — 12 (+12) = 36 (+12)
* D = 48
- Conclusion:
* D = 48
* Validate / Check:
* (48) — 12 ≟ 36
* 36 = 36
* Validated because 48 minus 12 is 36.

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