If $x=\frac{8ab}{a+b}$, then find the value of:
$$\frac{x+4a}{x-4a}+\frac{x+4b}{x-4b}$$

Given $x=\frac{8ab}{a+b}$. We can split this expression to apply Componendo and Dividendo.
First, divide by $4a$:

$$\frac{x}{4a}=\frac{2b}{a+b}$$

Applying Componendo and Dividendo:

$$\frac{x+4a}{x-4a}=\frac{2b+(a+b)}{2b-(a+b)}=\frac{3b+a}{b-a}\text{— (Equation 1)}$$

Next, divide by $4b$:

$$\frac{x}{4b}=\frac{2a}{a+b}$$

Applying Componendo and Dividendo:

$$\frac{x+4b}{x-4b}=\frac{2a+(a+b)}{2a-(a+b)}=\frac{3a+b}{a-b}=-\frac{3a+b}{b-a}\text{— (Equation 2)}$$

Adding Equation 1 and Equation 2:

$$\frac{x+4a}{x-4a}+\frac{x+4b}{x-4b}=\frac{3b+a}{b-a}-\frac{3a+b}{b-a}$$

$$=\frac{(3b+a)-(3a+b)}{b-a}=\frac{2b-2a}{b-a}=\frac{2(b-a)}{b-a}=2$$