SOLUTION: Steve traveled 200 miles at a certain speed. Had he gone 10 mph faster, the trip would have taken 1 hour less. Find the speed of his vehicle. How would you turn this into an alge

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: Steve traveled 200 miles at a certain speed. Had he gone 10 mph faster, the trip would have taken 1 hour less. Find the speed of his vehicle. How would you turn this into an alge      Log On


   

Question 144445: Steve traveled 200 miles at a certain speed. Had he gone 10 mph faster, the trip would have taken 1 hour less. Find the speed of his vehicle.
How would you turn this into an algebraic equation and solve it. I'm sorry if i seem to be a total idiot. But i have no CLUE. Please help me someone!!
Found 2 solutions by stanbon, Alan3354:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Steve traveled 200 miles at a certain speed. Had he gone 10 mph faster, the trip would have taken 1 hour less. Find the speed of his vehicle.
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Before speed change DATA:
distance = 200 miles ; speed = x mph ; time = d/r = 200/x hrs.
----------------------------
After speed change DATA:
distance = 200 miles ; speed = "x+10" mph ; time = d/r = 200/(x+10)
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EQUATIOn:
original time - new time = 1 hr
200/x - 200/(x+10) = 1
200x+ 2000 - 200x = x(x+10)
x^2+10x-2000 = 0
(x+50)(x-40) = 0
Positive solution:
x = 40 mph (original speed)
x + 10 = 50 mph (newspeed)
=============================
Cheers,
Stan H.

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Tricky problem.
Call the speed S and the time H (for hours).
The distance is 200, and is Speed x Hours, S times H.
Going faster takes less time. 200 = (S + 10) times (H-1)
We have 2 equations in 2 unknowns now.
SH = 200 eqn 1
(S+10)(H-1) = 200 eqn 2
SH + 10H -S -10 = 200, or
SH + 10H -S -10 = SH
10H -S -10 = 0
S = 200/H
10H - 200/H - 10 = 0
Multiply by H
10H%5E2+-10H+-200+=+0
Divide by 10
H%5E2+-H+-20+=+0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-1x%2B-20+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-1%29%5E2-4%2A1%2A-20=81.

Discriminant d=81 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--1%2B-sqrt%28+81+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-1%29%2Bsqrt%28+81+%29%29%2F2%5C1+=+5
x%5B2%5D+=+%28-%28-1%29-sqrt%28+81+%29%29%2F2%5C1+=+-4

Quadratic expression 1x%5E2%2B-1x%2B-20 can be factored:
1x%5E2%2B-1x%2B-20+=+%28x-5%29%2A%28x--4%29
Again, the answer is: 5, -4. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-1%2Ax%2B-20+%29

H = 5, or -4. -4 hours makes no sense, so it's discarded.
The time is 5 hours to go 200 miles. So the speed is 40 mph
Going 50 mph would take 4 hours, so the answer is correct.