Let line L1 be the graph of 5x + 8y= -13. Line L2 is perpendicular to line L1 and passes through the point (10, -10). If line L2 is the graph of the equation y = mx + b, then find m + b.

Rearrange the L1 equation as \(y=-\frac{5}{8}x-\frac{13}{8}\)

The slope of L2 must be \(m=\frac{8}{5}\) so its equation will be \(y=\frac{8}{5}x+b\)

We know it goes through point (10, -10), so \(-10=\frac{8}{5}\times10+b\)

Can you take it from here?