babyagelesszj

2022-07-08

Weird square root disappearing and flipping fraction upside down?
So here I was, making 2 math problems, I was able to solve them, but 2 operations seem a bit intractable to me. Maybe you can help me understand why this is true:
The first problem:
$x=\frac{1}{5}-\frac{4}{y}$
$\frac{4}{y}=\frac{1}{5}-x$
$\frac{4}{y}=\frac{1-5x}{5}$
$\frac{y}{4}=\frac{5}{1-5x}$
Why is it possible to turn $\frac{y}{4}$ upside down?
$y=20/1-5x$
The second problem:
$4A\surd B-\surd B=3$
$\surd B\left(4A-1\right)=3$
Where does the -√B go? I understand that the -1 comes from the - sign in front of the square root. But where does the other √B go?
$\sqrt{B}=\frac{3}{\left(4A-1\right)}$
$B=\left(\frac{3}{\left(4A-1\right)}{\right)}^{2}$
$B=\frac{9}{\left(4A-1{\right)}^{2}}$
Everything, except above the bold text I understand. Maybe I do not understand the full extent of a certain rule which I am familiar with in simpler situations. That's why I think an example would be very useful. I really want to have a deep understand of why these things are true.
Greetings, Bowser.

Ashley Parks

Expert

First question: if two fractions are equal, then their reciprocals are equal too (the reciprocal of a fraction is the same fraction "turned upside down", using your terminology).
Second question: the B is still there: if you multiply $\sqrt{B}\left(4A-1\right)$ you get indeed $4A\sqrt{B}-\sqrt{B}$. It's called "distributive law".

Do you have a similar question?