Answer
$x+y=38$
Work Step by Step
$16=2^{4}, \quad 27=3^{3}$
$16^{x-3}=(2^{4})^{x-3}=2^{4x-12}$
$27^{x}=(3^{3})^{x}=3^{3x}$
Apply the exponential equality principle on both equations
$\left[\begin{array}{lll}
2^{y}=2^{4x-12} & ... & 3^{y+2}=3^{3x}\\
y=4x-12 & & y+2=3x
\end{array}\right]$
Substitute $y=4x-12$ into the other equation.
$4x-12+2=3x$
$4x-10=3x$
$x-10=0$
$x=10$
Back-substitute
$y=4(10)-12=40-12=28$
$x=10$
$y=28$
$x+y=38$
