#2020 06 06#1?

$\frac{2x-3}{6}-\frac{4-x}{36}=\frac{x+2}{9}$ Multiply both sides by $36$ $\red{36}\times\frac{(2x-3)}{\red{6}}-\cancel{36}\times\frac{(4-x)}{\cancel{36}}=\red{36}\times\frac{(x+2)}{\red{9}}$ $12x-18-4+x=4x+8$ $13x-4x=8+18+4=8+22=30$ $9x=30$ $x=\frac{30}{9}=\boxed{\red{\frac{10}{3}}}$

$\begin{aligned} \dfrac{2x-3}{6} - \dfrac{4-x}{36} &= \dfrac{x+2}{9} \\ \\6(2x-3) - (4-x) &= 4(x+2) &\quad \color{#3D99F6} ()\times 36 \\ \\ 9x &= 30 &\quad \color{#3D99F6}\text{Simplify} \\ \\ x&= \dfrac{10}{3} = \dfrac{a}{b} \end{aligned}$

$2(10) - 3(3) = 11 \implies \boxed{2a - 3b =11}$

$\dfrac{12x-18-4+x}{36}=\dfrac{x+2}{9}$

$\implies \dfrac{13x-22}{4}=x+2$

$\implies 13x-22=4x+8$

$\implies x=\dfrac{10}{3}$

$\implies a=10, b=3$

$\implies 2a-3b=11$

$\implies a+b\neq 37$