babyagelesszj

Answered

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:

$4A\surd B-\surd B=3$

$\surd B(4A-1)=3$

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?

$\sqrt{B}=\frac{3}{(4A-1)}$

$B=(\frac{3}{(4A-1)}{)}^{2}$

$B=\frac{9}{(4A-1{)}^{2}}$

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.

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:

$4A\surd B-\surd B=3$

$\surd B(4A-1)=3$

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?

$\sqrt{B}=\frac{3}{(4A-1)}$

$B=(\frac{3}{(4A-1)}{)}^{2}$

$B=\frac{9}{(4A-1{)}^{2}}$

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.

Answer & Explanation

Ashley Parks

Expert

2022-07-09Added 11 answers

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 $\sqrt{B}(4A-1)$ you get indeed $4A\sqrt{B}-\sqrt{B}$. It's called "distributive law".

Second question: the B is still there: if you multiply $\sqrt{B}(4A-1)$ you get indeed $4A\sqrt{B}-\sqrt{B}$. It's called "distributive law".

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