SOLUTION: How do I solve using a system of two equations in two variables?
When a plane flies into the wind, it can travel 3000km in 6 hours. When it flies with the wind, it can travel th
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When a plane flies into the wind, it can travel 3000km in 6 hours. When it flies with the wind, it can travel th
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Question 141075This question is from textbook Prentice Hall Algebra 1
: How do I solve using a system of two equations in two variables?
When a plane flies into the wind, it can travel 3000km in 6 hours. When it flies with the wind, it can travel the same distance in 5 hours. Find the rate of the plane in still air and the rate of the wind. This question is from textbook Prentice Hall Algebra 1
You can put this solution on YOUR website! When a plane flies into the wind, it can travel 3000km in 6 hours. When it flies with the wind, it can travel the same distance in 5 hours. Find the rate of the plane in still air and the rate of the wind.
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Let plane in still air rate be "s" and wind rate be "w".
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EQUATIONS:
Against the wind : s - w = 3000/6 = 500
With the wind : s + w = 3000/5 = 600
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Add the two equations to solve for "s":
2s = 1100
s = 550 mph (plane speed is still wind)
550 + w = 600
w = 50 mph (wind speed)
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Cheers,
Stan H.
You can put this solution on YOUR website! See the solution I gave for this problem --> http://www.algebra.com/algebra/homework/quadratic/Quadratic_Equations.faq.question.137499.html
Then look at your problem and find out what two equations you can make to describe what you are given.
We know the distance is the same going both ways.
Let s be the speed of the plane and w be the wind.
Then D = velocity * time.
3000 = (s-w) * 6 or 500 = s-w
3000 = (s+w) * 5 or 600 = s+w
Now use the info from the link above and you can do the rest
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