variable. b$.$ Keeping the restrictions in mind, solve the equation.

$\frac{4}{x+3}+\frac{3}{x-3}=\frac{18}{(x+3)(x-3)}$

a$.$ What is$/$are the value or values of the variable that make$($s$)$ the denominators zero?

$x=$

$($Simplify your answer. Use a comma to separate answers as needed.$)$