Perform the operations in the correct order: 12-6/15+6*5-9.

Set up the form for the partial fraction decomposition. Do not solve for A, B, C, and so on.
${\displaystyle \frac{x-14}{x^2-3x}}$

How do you solve this fraction?
$\frac{2}{3}-\frac{1}{16}$
My answer: ${\displaystyle \frac{1}{48}}$. I believe that it's incorrect.
Three does not divide into 16, so I cross multiplied. What am I doing wrong?

The intial fraction is:
${\displaystyle \frac{3x^3-x+4}{x^3+2x^2+6x}}$
How should to make the decomposition for the denominator?
Thanks!

Is it possible to have a fraction wherein the numerator and denominator are also fractions?
For example:
${\displaystyle \frac{\frac{3}{5}}{\frac{7}{8}}}$
I was wondering if such a situation had a name, wherein both the numerator and the denominator of a fraction consist of fractions themselves. I was also wondering if this was something common, or at least in some way useful.

Solve Equations with Fraction or Decimal Coefficients. In the following exercises, solve each equation by clearing the fractions.
${\displaystyle \frac{1}{2}(k+3)=\frac{1}{3}(k+16)}$

Solve: ${\displaystyle \frac{x+3}{4}=\frac{x-2}{3}+\frac{1}{4}}$.

Subtract fractions. Simplify the answer.
${\displaystyle \frac{x-9}{6x}-\frac{x+9}{6x}=?}$