document.write( "Question 31886This question is from textbook Beginning and Intermediate Algebra
\n" ); document.write( ": Boy do I need help! Here's the problem: An airplane can fly downwind a distance of 600miles in 2hrs. However, the return trip against the same wind takes three hours. Find the speed of the wind.\r
\n" ); document.write( "\n" ); document.write( "I tried to make a chart and ended up that the wind was going 250 miles an hour and I don't think that sounds right. Any help would be really appreciated. Thank you,Liz
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Algebra.Com's Answer #18471 by mbarugel(146)\"\" \"About 
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Hello!
\n" ); document.write( "When the airplane flies downwind, the speed of the wind is added to the one of the plane. When it's going against the wind, then it should be subtracted.\r
\n" ); document.write( "\n" ); document.write( "Let's call X to the speed of the plane (\"without wind\") and Y to the speed of the wind. So we get the following equations:\r
\n" ); document.write( "\n" ); document.write( "\"X+%2B+Y+=+600%2F2+=+300\"
\n" ); document.write( "When going downwind, the final speed of the plane (plane + wind) is 300 mph, because it travels 600 mph in 2 hours.\r
\n" ); document.write( "\n" ); document.write( "The other fact is that when going against the wind takes 3 hours, so its speed is 200 mph (600 miles in 3 hours). So we get the equation:\r
\n" ); document.write( "\n" ); document.write( "\"X+-+Y+=+200\"\r
\n" ); document.write( "\n" ); document.write( "Now let's find Y, which is what we're interested in. Isolate X from the 2nd equation:\r
\n" ); document.write( "\n" ); document.write( "\"X+=+200+%2B+Y\"
\n" ); document.write( "And then replace this X into the first equation:
\n" ); document.write( "\"X%2B+Y=300\"
\n" ); document.write( "\"200%2BY%2BY=300\"
\n" ); document.write( "\"2Y+=+300+-+200+=+100\"
\n" ); document.write( "\"Y+=+50\"\r
\n" ); document.write( "\n" ); document.write( "So the speed of the wind is 50 mph.\r
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\n" ); document.write( "\n" ); document.write( "I hope this helps!
\n" ); document.write( "Get more answers at Online Math Answers.com!
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