Line $11x+3y-48 = 0$ tangent a graph $f(x) = \frac{4x + 3}{3x - 6}$ at $(a,b)$ Line $11x+3y-48 = 0$ tangent a graph $f(x) = \frac{4x + 3}{3x - 6}$ at (a,b) when $a<b$
$a-b = ...$
Find gradient of the line, df(x) / dx
$\frac{(4x+3)3 - (3x-6)(4)} {(3x-6)^2}$
Which the same as $ -11/3$
$(12x+9 - 12x + 24 ) 3= -11(3x-6)^2$
If i solve x, literally i solve $a $ right?
Is there less complicated way to solve it?

Rewrite 
$$f(x)=\frac43+\frac{11}{3x-6}.$$
Now 
$$f'(x)=-\frac{11}{(3x-6)^2}\cdot3$$
and $f'(x)=-11/3$ gives $(3x-6)=\pm1$, that is $x=7/3$ and $x=5/3$.
