Then, 15 - x is the distance ran in the second part.
The time, t1, for the first part is t1 = xmi / 8mi/h
The time, t2, for the second part is t2 = (15 - x)mi / 20mi/h
The total time is t1 + t2 = 1.125 h
Then x/8 + (15 - x) / 20 = 1.125
To solve for x, multiply both sides by 40 (this is the least common multiple)
5x + 30 - 2x = 45
3x = 15
to find x divide: 15/3=5
And 15 - x = 10 mi.
First part:
Speed: 8mi/h
Distance: 5 mi
Time: 5mi/8mi/h = 5/8 h = 37.5 minutes
Second part
Speed: 20 mi/h
Distance: 10 mi
Average speed: 15mi/1.125h =13.33 mi/h
Distance: 15mi
Time: 1.125 h = 67.5 minutes
Check the picture below.
![\bf A=\cfrac{h(a+b)}{2}\quad \begin{cases} A=20\\ a=6z-1\\ b=2z+3\\ h=z \end{cases}\implies 20=\cfrac{z[(6z-1)~+~(2z+3)]}{2} \\\\\\ 20=\cfrac{z(8z+2)}{2}\implies 20=\cfrac{2z(4z+1)}{2}\implies 20=z(4z+1) \\\\\\ 20=4z^2+z\implies 0=4z^2+z-20](https://tex.z-dn.net/?f=%5Cbf%20A%3D%5Ccfrac%7Bh%28a%2Bb%29%7D%7B2%7D%5Cquad%20%0A%5Cbegin%7Bcases%7D%0AA%3D20%5C%5C%0Aa%3D6z-1%5C%5C%0Ab%3D2z%2B3%5C%5C%0Ah%3Dz%0A%5Cend%7Bcases%7D%5Cimplies%2020%3D%5Ccfrac%7Bz%5B%286z-1%29~%2B~%282z%2B3%29%5D%7D%7B2%7D%0A%5C%5C%5C%5C%5C%5C%0A20%3D%5Ccfrac%7Bz%288z%2B2%29%7D%7B2%7D%5Cimplies%2020%3D%5Ccfrac%7B2z%284z%2B1%29%7D%7B2%7D%5Cimplies%2020%3Dz%284z%2B1%29%0A%5C%5C%5C%5C%5C%5C%0A20%3D4z%5E2%2Bz%5Cimplies%200%3D4z%5E2%2Bz-20)

since the height is just a length unit, it can't be -2.3646.