Solve the following

$4-\frac{2(z-4}{3}=\frac{1}{2}(2 z+5)$

$\Rightarrow \frac{12-2(z-4)}{3}=\frac{1(2 z+5)}{2} \quad$ (L. C.M. of 1 and 3 is 3$)$

$\Rightarrow \frac{12-2 z+8}{3}=\frac{2 z+5}{2}$

$\Rightarrow \frac{20-2 z}{3}=\frac{2 z+5}{2}$

$\Rightarrow 2(20-2 z)=3(2 z+5) \quad$ (by cross multiplication)

$\Rightarrow 40-4 z=6 z+15$

$\Rightarrow 40-15=6 z+4 z$

$\Rightarrow 25=10 z$

$\Rightarrow 10 z=25 \quad$ (by transposition)

$\Rightarrow z=\frac{25}{10}=\frac{5}{2}$

$\therefore z=\frac{5}{2}$