How i get the $x$ of the equation I have this:
$\frac{32}{20 - \frac{16}{5-\frac{5}{x}}}$ = 1
My main problem is that I want to multiply by something on both sides.
I will move $(5-\frac{5}{x})$ to side of 1, then:
$\frac{32}{20-16} = 1(5-\frac{5}{x}$)
$8 = 5 -\frac{5}{x}$ , now multiply by $x$
$8x = 5x - 5 $
$3x = 5$
$x = 5/3$
But according to symbolab it's $x = 15/19$, could you help me what I'm wrong about? I have done a lot of exercises, but I have a hard time clearing the $ x $ when it is very low.

Unfortunately you can't move the $5-{5\over x}$ out of the left side because it is part of a "family" which is the denominator of the bigger fraction.
$$\frac{32}{20 - \frac{16}{5-\frac{5}{x}}}=1$$
If we were to put parenthesis around the terms, it would be clearer why we can't do what you did:
$$\frac{32}{\Big(20 - \frac{16}{\big(5-\frac{5}{x}\big)}\Big)}=1$$
As you can see, the expression $5-{5\over x}$ is "stuck" inside the bigger parenthesis on the bottom, and cannot be taken out.
Having that said, let's solve the equation correctly:
Multiply both sides by the bigger parenthesis denominator:
$$32=20 - \frac{16}{5-\frac{5}{x}}$$
Subtract both sides by 20:
$$12=- \frac{16}{5-\frac{5}{x}}$$
$$-12=\frac{16}{5-\frac{5}{x}}$$
Multiply by the "middle" denominator:
$$-12\bigg({5-\frac{5}{x}}\bigg)=16$$
$$-60+\frac{60}{x}=16$$
Add 60 to both sides:
$$\frac{60}{x}=76$$
Multiply by x:
$$76x=60$$
Divide by 76:
$$x=\frac{60}{76}$$
$$x=\frac{15}{19}$$
DONE! :D
