Based on Thales' second theorem, i.e. the similarity of triangles:

$$c:p=d:q \left(I\right)$$

$$12:8=\left(d+q\right):q \left(II\right)$$

$$12:x=\left(d+q\right):\left(q+\frac{d}{3}\right) \left(III\right)$$

$$x:8=\left(q+\frac{d}{3}\right):q \left(IV\right)$$

$$\left(II\right) \to 8·\left(d+q\right)=12·q$$

$$\left(III\right) \to x·\left(d+q\right)=12·\left(q+\frac{d}{3}\right)$$

Dividing the two equations above:

$$\frac{x}{8}=\frac{\bcancel{12}·\left(q+\frac{d}{3}\right)}{\bcancel{12}·q}$$

$$\frac{x}{8}=\frac{q}{q}+\frac{d}{3·q}$$

$$\frac{x}{8}=1+\frac{1}{3}·\frac{d}{q}$$

$$\left(II\right)\to 12·q=8·\left(d+q\right)$$

$$12=\frac{8·\left(d+q\right)}{q}$$

$$12=8·\left(1+\frac{d}{q}\right)$$

$$\frac{12}{8}=1+\frac{d}{q}$$

$$\frac{d}{q}=\frac{12}{8}-1$$

$$x=8·\left(1+\frac{1}{3}·\frac{12-8}{8}\right)$$

$$y=8+\frac{2·\left(12-8\right)}{3}$$

$$x=8+\frac{4}{3}$$

In the case of the y-section, the only difference is that the ratio is not one-third but two-thirds. Accordingly:

$$y=8·\left(1+\frac{2}{3}·\frac{12-8}{8}\right)$$

$$y=8\left(1+\frac{2}{3}·\frac{4}{8}\right)$$

$$y=8+\frac{2}{3}·4$$