Question

word problem attached

Answer 1

The hourly rate each of them work is 1/time
Let E be time for Enrique
Let H be time for Heather

Working together means you add the hourly rates to get the group rate
\[\frac{1}{e} + \frac{1}{h} = \frac{1}{10}\]
we know that E can finish 15 hrs faster than H.
\[e = h -15\]
Substitute
\[\frac{1}{h-15} + \frac{1}{h} = \frac{1}{10}\]
solve for h
\[\frac{2h-15}{h(h-15)} =\frac{1}{10}\]
\[20h -150 = h^2 -15h\]
\[h^2 -35h +150 = 0\]
\[(h-30)(h-5) = 0\]
h = 30 because its impossible for heather to do it in 5 hours and be 15 hours slower than Enrique

Therefore
E = 30-15 = 15