Find the distance between the lines 3x+4y = 9 and 6x+8y = 15

The lines are parallel with common slope of $\frac34$ .

For line $3x+4y=9$ the distances from the origin is: $\dfrac9{\sqrt{9+16}}=\dfrac95$

For line $6x+8y=15$ the distances from the origin is: $\dfrac{15}{\sqrt{36+64}}=\dfrac32$

The difference between these distances is $\dfrac95-\dfrac32=\dfrac3{10}$

The answer is $10\times\dfrac3{10}=\boxed3$

Since the two lines are parallel, the (perpendicular) distance between them are always the same. Hence, we only need get one point on one line, say from $3x+4y=9$ , and find its distance to the other line, say $6x+8y=15$ .

One point on the line $3x+4y=9$ is $(3,0)$ . Therefore, the distance between the two lines is the distance from $(3,0)$ to $6x+8y=15$ : $a=\frac{6(3)+8(0)-15}{\sqrt{6^2 + 8^2}} = \frac{3}{10}$ Therefore $10a=\boxed{3}$ .

The distance between two lines $Ax+By+C_1=0$ and $Ax+By+C_2=0$ is $a=\left|\dfrac{C_1-C_2}{\sqrt{A^2+B^2}}\right|$ .

Divide the second equation on both sides by $2$ to get $3x+4y-\dfrac{15}{2}=0$

So we have,

$a=\left|\dfrac{-9+\dfrac{15}{2}}{\sqrt{3^2+4^2}}\right|=\dfrac{1.5}{5}=0.3$

Finally,

$10a=10(0.3)=$ $\boxed{3}$