Answer
The solution set is $\{2\}$.
Work Step by Step
STEP 1: Restrictions. There must be no zeros in the denominator, so
$\left[\begin{array}{lll}
5z-11\neq 0 & 2z-3\neq 0 & 5-z\neq 0\\
z\neq 11/5 & z\neq 3/2 & z\neq 5
\end{array}\right]$
STEP 2: Fractions.
The LCM of all denominators is $(5z-11)(2z-3)(5-z)$
After multiplying both sides with the LCM,
$5(2z-3)(5-z) +4(5z-11)(5-z) =-3(5z-11)(2z-3)$
STEP 3: Remove all parentheses and simplify.
$5(-2z^{2}+13z-15)+4(-5z^{2}+36z-55)=-3(10z^{2}-37z+33)$
$-10z^{2}+65z-75-20z^{2}+144z-220=-30z^{2}+111z-99$
$-30z^{2}+209z-295=-30z^{2}+111z-99$
STEP 4: Variables on one side, all other terms on the other...
Add $30z^{2}-111z+295$ to both sides,
$98z=196$
STEP 5: Simplify and solve.
$98z=196\qquad/\div 98$
$z=2$
No restriction prevents $2$ being a solution, so the
solution set is $\{2\}$
