So I have this equation $$\frac1x - \frac{x}6 = \frac23$$
from what I learned and what I need to do, apparently I need to find the LCM of the $3$ denominators which I've thought up of $6x$ then if I multiply it to every term:
$$6x\left(\frac1x\right) - 6x\left(\frac{x}6\right) = 6x\left(\frac23\right)$$
then I cancel the factors(?)
$$6\bcancel{x}\left(\frac1{\bcancel{x}}\right) - \bcancel{6}x\left(\frac{x}{\bcancel6}\right) = (\bcancel3)(2)x\left(\frac2{\bcancel3}\right)$$
that leaves:
$$6(1) - x(x) = 2x(2)$$
simplify:
$$-x^2 -4x + 6 = 0$$
then I multiply everything to $-1$ to remove the negative on A
$$x^2 + 4x - 6 = 0$$
I do not believe this is the correct way to do it so can you guys spot anything that might be wrong in my end?
multiplying your equation by $6x$ we get $6-x^2=4x$ or $$0=x^2+4x-6$$ and it must be $x\ne 0$