Find the slope of the tangent line to the curve
\[
5 \sin (x)+4 \cos (y)-3 \sin (x) \cos (y)+x=7 \pi
\]
at the point $(7 \pi, 7 \pi / 2)$.

Step 1 ：Given the curve equation: \(x - 3\sin(x)\cos(y) + 5\sin(x) + 4\cos(y) - 7\pi = 0\)

Step 2 ：Differentiate with respect to x: \(\frac{\partial}{\partial x} = -3\cos(x)\cos(y) + 5\cos(x) + 1\)

Step 3 ：Differentiate with respect to y: \(\frac{\partial}{\partial y} = 3\sin(x)\sin(y) - 4\sin(y)\)

Step 4 ：Find the derivative of y with respect to x: \(\frac{dy}{dx} = \frac{3\sin(x)\sin(y) - 4\sin(y)}{-3\cos(x)\cos(y) + 5\cos(x) + 1}\)

Step 5 ：Substitute the given point \((7\pi, \frac{7\pi}{2})\) into the equation: \(\frac{dy}{dx} = \frac{3\sin(7\pi)\sin(\frac{7\pi}{2}) - 4\sin(\frac{7\pi}{2})}{-3\cos(7\pi)\cos(\frac{7\pi}{2}) + 5\cos(7\pi) + 1}\)

Step 6 ：Simplify the expression: \(\frac{dy}{dx} = -1\)

Step 7 ：\(\boxed{-1}\) is the slope of the tangent line to the curve at the point \((7\pi, \frac{7\pi}{2})\)