---

Solution

Rewrite second line: $6x - 3y - 5 = 0$ Divide by 3: $2x - y - \dfrac{5}{3} = 0$ First line: $y = 2x + 4 \Rightarrow 2x - y + 4 = 0$ Distance between parallel lines: $d = \dfrac{|c_2 - c_1|}{\sqrt{a^2 + b^2}}$ Here: $c_1 = 4$, $c_2 = -\dfrac{5}{3}$, $a=2,\ b=-1$ Compute: $d = \dfrac{\left|4 - \left(-\dfrac{5}{3}\right)\right|}{\sqrt{4+1}} = \dfrac{\left|\dfrac{12}{3}+\dfrac{5}{3}\right|}{\sqrt{5}} = \dfrac{17/3}{\sqrt{5}} = \dfrac{17\sqrt{5}}{15}$