Question 224164
The wind is blowing at a steady rate of 10 mph. It takes a truck a total of 5 hours to travel 120 miles west against the wind and then 120 miles back with the wind. What is the trucks speed in stil
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Applying the distance formula: d = rt
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Let x = speed of truck in still wind
and y = time traveling against the wind
then
"distance traveling against wind"
(x-10)y = 120
"distance traveling with wind"
(x+10)(10-y) =120
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So, now we have two equations and two unknowns:
(x-10)y = 120  
(x+10)(10-y) =120
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Our system of equations:
(x-10)y = 120  (equation 1)
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(x+10)(10-y) =120
10x-xy-10y+100 = 120 (equation 2)
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Solve equation 1 for y:
y = 120/(x-10)
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Plug the definition of y (above) into equation 2 and solve for x:
10x-xy-10y+100 = 120
10x+100-xy-10y = 120
10x+100-(xy+10y) = 120
10x+100-y(x+10) = 120
10x+100-(120/(x-10))(x+10) = 120
Multiply BOTH sides by (x-10) to get rid of denominators:
10x(x-10)+100(x-10)-120(x+10) = 120(x-10)
10x^2-100x+100x-1000-120x-1200 = 120x-1200
10x^2-120x-2200 = 120x-1200
10x^2-240x-1000 = 0
x^2-24x-100 = 0
Since we can't factor, we must resort to the quadratic equation.  Doing so will yield:
x = {27.62, -3.62}
A negative speed does NOT make sense, so toss it out leaving:
x = 27.62 mph
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Details of quadratic equation to follow:
*[invoke quadratic "x", 1, -24, -100 ]