SOLUTION: Doug and Austin both run on their junior high school cross country team. Doug was not feeling well and ran 2 miles per hour slower than Austin. Doug finished the four mile race 10

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Question 614456: Doug and Austin both run on their junior high school cross country team. Doug was not feeling well and ran 2 miles per hour slower than Austin. Doug finished the four mile race 10 minutes after Austin. How fast was Austin running? How long did it take Doug to finish the race?
this is what I have but it doesn't seem to be working out
D = R x T
Doug ... 4 ... x-2 ... [4/(x-2)]+10
Austin ... 4 ... x ... 4/x
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +s+ = Austin's speed
Let +t+ = Austin's time
+s+-+2+ = Doug's speed
+t+%2B+1%2F6+ = Doug's time ( 1/6 hr = 10 min )
+d+=+4+ It's a 4 mile race
----------------------
Austin's equation:
(1) +4+=+s%2At+
Doug's equation:
(2) +4+=+%28+s+-+2+%29%2A%28+t+%2B+1%2F6+%29+
-------------------------
(2) +4+=+s%2At+-+2t+%2B+s%2F6+-+1%2F3+
(2) +s%2F6+-+2t+=+4+%2B+1%2F3+-+s%2At+
Substitute (1) into (2)
(2) +s%2F6+-+2t+=+4+%2B+1%2F3+-+4+
(2) +s%2F6+-+2t+=+1%2F3+
(2) +s+-+12t+=+2+
And also from (1):
(1) +t+=+4%2Fs+
Substituting:
(2) +s+-+12%2A%284%2Fs%29+=+2+
(2) +s%5E2+-+48+=+2s+
(2) +s%5E2+-+2s+-+48+=+0+
Use quadratic formula
s+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
+a+=+1+
+b+=+-2+
+c+=+-48+
s+=+%28-%28-2%29+%2B-+sqrt%28+%28-2%29%5E2-4%2A1%2A%28-48%29+%29%29%2F%282%2A1%29+
s+=+%28+2+%2B-+sqrt%28+4+%2B+192+%29%29+%2F+2+
s+=+%28+2+%2B-+sqrt%28+196+%29%29+%2F+2+
s+=+%28+2+%2B+14%29+%2F+2+
+s+=+16%2F2+
+s+=+8+
Austin's speed was 8 mi/hr
From (1):
(1) +t+=+4%2Fs+
(1) +t+=+4%2F8+
(1) +t+=+1%2F2+
+t+%2B+1%2F6+ = Doug's time
+t+%2B+1%2F6+=+1%2F2+%2B+1%2F6+
+t+%2B+1%2F6+=+2%2F3+
and
+%282%2F3%29%2A60+=+40+
Doug's time is 2/3 hr, or 40 min
check:
(2) +4+=+%28+s+-+2+%29%2A%28+t+%2B+1%2F6+%29+
(2) +4+=+%28+8+-+2+%29%2A%28+1%2F2+%2B+1%2F6+%29+
(2) +4+=+6%2A%28+4%2F6+%29+
(2) +4+=+4+
OK