You can
put this solution on YOUR website!time = distance / rate
(250 / A) + 3 = 440 / (A + 5)
multiplying by A(A + 5) ___ 250A + 1250 + 3A^2 + 15A = 440A
3A^2 - 175A + 1250 = 0
factoring ___ (3A - 25)(A - 50) = 0
3A - 25 = 0 ___ A = 25/3 ___ B = 40/3
A - 50 = 0 ___ A = 50 ___ B = 55
You can
put this solution on YOUR website!You picked a tough problem but we can do it.
a) what formula are you going to use? Ans.: st=d, speed * time=distance
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b) speed
Let s-5=car A's speed
Let s=car B's speed
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c)time
Let t=car B's time
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We can make a system of equations now. Look at the formula for B 1st because it is the simplest.
A) (s-5)(t-3)=250
B) st=440
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B) s=440/t
A)
(440/t -5)(t-3)=250 Substitute 440/t for s
(440-5t)(t-3)=250t multiply t times each side to eliminate the fraction.
-5t^2+455t-1320=250t
t^2-91t+264=-50t Divide each side by -5
t^2-41t+264=0
(t-8)(t-33)=0 Factor
t=8 hrs, t=33 hrs
If we use t=8 then for B: s=440/8=55mph
Then for A: s=50mph
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Check:
A travels 250 mi in 5h
B travels 440 mi in 8h which is 3hrs longer than A traveling 250mi.
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Ed
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