The distance between the lines $3x+4y=9$ and $6x+8y=15$ is:

Let $d_1$ and $d_2$ be the distances of two lines $3x+4y-9=0$ and $6x+8y-15=0$ respectively from origin.

$\therefore {\mathrm{d}}_1=\frac{\left|3\right(0)+4(0)-9|}{\sqrt{3^2+4^2}}\Rightarrow {\mathrm{d}}_1=\frac{9}{5}$

and ${\mathrm{d}}_2=\frac{\left|6\right(0)+8(0)-15|}{\sqrt{36+64}}=\frac{15}{10}=\frac{3}{2}$

$\therefore$ distance between these lines is $\mathrm{d}={\mathrm{d}}_1-{\mathrm{d}}_2$

$\Rightarrow \mathrm{d}=\frac{9}{5}-\frac{3}{2}=\frac{18-15}{10}=\frac{3}{10}$