Just wondering, did you completely make up the "If 3x−y=12, what is the value of 8 x /2 y ?" question? Because I'm fairly certain that for one, it doesn't have enough information to solve for either x or y. If 8x/2y equaled something without introducing a third variable, then you could solve for both x and y, but that isn't the question. Also, I don't know why he would ever answer with 2^12 instead of just 4096. Or I've just forgotten how to do that kind of problem maybe. Your second math problem is completely valid though, and could be written out as 800/x=y and 800/(x-2)=y+20, where x is the number of people at the start and y is the cost at the start, solve for x (if you don't trial and error it). Since there are 2 variables and two equations, it can be solved for. Math: 800/x=y and 800/(x-2)- 20= y; 800/x = 800/(x-2) -20 => 800(x-2)/x= 800 - 20(x-2) => 800(x-2)=800x-20x^2+40x => 800x - 1600 = -20x^2+840x => 20x^2 -40x - 1600 =0 => x^2 - 2x - 80 = 0 => (x+8) (x-10) = 0 so the answer to x is -8 or 10. But since we're counting people, the answer isn't going to be -8 people so the answer is 10.
Sorry I'm such a math nerd. I do like your story, it's really cool! Everything else was just so good I didn't have anything to add.
"haha I sure hope no math nerds read this fic and tear my math apart" *Checks comments* "WHAT"
To be honest, I just googled both problems and hoped the Internet was right. I haven't done math in four years and I don't plan to start now 😎 Hmm, I might just find a different math problem and switch it in, then. Thanks for reviewing!
You’re welcome! Sorry to dismantle something more minor in the story. A hard algebra problem would be one that requires the pythagorean theorem, though of course that would be harder to write because the answer would have a square root. If you want, I could alter the problem you have slightly so it works (and isn’t hard to write), though I’m sure you can find one yourself too.
And don’t feel bad about not knowing it, I’ve tutored multiple people in algebra (as well as significantly harder math subjects) so y’know. I’m just unrealistically interested in math. Plato (philosophy) briefly mentions squares in one of his books (Meno) as an analogy for something and it was my favorite part because it was math and square roots and stuff. Most others reading that book for the class liked that part least.
hey, if you wouldn't mind, I'd be happy to take that altered problem! I don't trust strangers on the internet as much as I trust you (a stranger on the internet) lol.
You can do “If 3x−y=12 and 8x /2y=4, then what does x and y equal? Where the answer is x=6 and y=6
Math: 3x-12=y and 8x/4=2x=2y=>x=y; 3x-12=x => 2x=12 => x=6 and since x=y y=6
I don’t know if this is a particularly hard problem but it reuses what you already had.
The nice thing about all of these problems is that it’s easier to check the answer than it is to solve for the answer. It’s pretty easy to realize that 800/8=100 and 800/10=80 and 80 is 20 less than 100, it’s harder to actually solve for that 10. Same with above in that you can just do the math (3*6=18, 18-6=12 and (8*6)/(2*6)=8/2=4), and you can for below too. So for the math teacher she can check Leo’s work more easily (computation-wise) than Leo can solve for it.
If you wanted a harder problem, maybe
2x-y=8 and x^2-7x-18=y
Where the answers are x=10,y=12 and x=-1,y=-10
Math: 2x-8=y and x^2-7x-18; x^2-7x-18=2x-8 => x^2-9x-10=0 => (x+1)(x-10)=0 so x=-1 or 10, plug that back in to get y and 2(-1)-8=-2-8=-10 so when x=-1, y=-10 and 2(10)-8=20-8=12 so when x=10, y=12.
You’re welcome! I mean, if I’m wrong, worst case scenario is some other math nerd comes and tells you I’m wrong so you aren’t risking much. But I don’t think I am wrong, and I’m glad to help!
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