Solve absolute value equation or indicate the equation has no

Find a recurrence relation for the number of bit sequences of length n with an even number of 0s.

Solve, please:

1) $\frac{8}{x+1}=\frac{4}{3}$
2) $\frac{x+3}{3x-6}>0$
3) $2x+3=\frac{x}{4}$
4) $\frac{1}{x^2-4}\le 0$
5) $\frac{4}{x}+\frac{1}{3x}=9$
6) $\frac{x+32}{x+6}\le 6$
7) $\frac{4}{x}+\frac{1}{x^2}=\frac{1}{5x^2}$
8) $1+\frac{2}{x+1}<\frac{2}{x}$
9) $\frac{x-3}{2x+10}+2x-12=\frac{x^2+3x-18}{2x+10}$
10) $\frac{12x^3+16x^2-3x-4}{8x^3+12x^2+10x+15}<0$

Prove that ${\displaystyle f\left(f\left(x\right)\right)\ge 0}$ for all real x

To insert: Appropriate signs (<, >, =) in each statement to make them true: |-3| |-2|

Find the vertex and intercepts to sketch the graph of each quadratic formula. Give the equation of the parabolas

Which quadrant contains no solutions to the system
If the system of inequalities
$y\ge 2x+1$ and $y>0.5x-1$

Solution of a system of differential equations
$$\begin{gathered} x^{\prime }=-y+x{y}^2 \\ {y}^{\prime }=4x-4x^{2}y \end{gathered}$$
$$\begin{gathered} x^{\prime }=-y+x{y}^2 \\ {y}^{\prime }=4x-4x^{2}y \end{gathered}$$proposed solution to this which uses the fact that
$\frac{dy}{dx}=\frac{y^{\prime }}{x^{\prime }}=\frac{4x(1-xy)}{y(xy-1)}=-\frac{4x}{y}$
and then solves this as a separable equation, which is straight forward enough. What I am struggling to understand is why one chooses to calculate $\frac{dy}{dx}$. It seems to me quite arbitrary to calculate this derivative. What is the connection between this derivative and the original system?