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?

$$y = 20 / 1 - 5 x$$

The second problem:

$$4A√B - √B = 3$$

$$√B(4A-1) = 3$$

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.


1 Answer 1


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".

  • 1
    $\begingroup$ That did it! Awesome. I do know the distributive law. The letters must have shaken me up or something.. Thanks! $\endgroup$
    – Cro-Magnon
    Jul 29, 2015 at 8:56

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