document.write( "Question 77448: A six passenger plane cruises at 180 mph in calm air. If the plane flies 7 miles with the wind in the same amount of time as it flies 5 miles against the wind, then what is the wind speed? \n" ); document.write( "
Algebra.Com's Answer #55567 by ptaylor(2198)\"\" \"About 
You can put this solution on YOUR website!
Distance (d)= rate(r) times time (t) or d=rt; t=d/r \r
\n" ); document.write( "\n" ); document.write( "Let x=rate of speed of wind\r
\n" ); document.write( "\n" ); document.write( "Speed of plane with wind =180+x
\n" ); document.write( "Distance travelled with wind is 7 mi
\n" ); document.write( "Time to travel 7 mi with wind is 7/(180+x)-----Time #1
\n" ); document.write( "Speed of plane against wind=180-x
\n" ); document.write( "Distance travelled against wind is 5 mi
\n" ); document.write( "Time to travel 5 mi against wind is 5/(180-x)----Time #2\r
\n" ); document.write( "\n" ); document.write( "Now we are told that Time #1 = Time #2 so our equation to solve is:\r
\n" ); document.write( "\n" ); document.write( "7/(180+x)=5/(180-x) Multiply both sides by (180+x)(180-x) to get rid of the fractions:\r
\n" ); document.write( "\n" ); document.write( "7(180+x)(180-x)/(180+x)=5(180+x)(180-x)/(180-x) Simplify by cancelling\r
\n" ); document.write( "\n" ); document.write( "7(180-x)=5(180+x) get rid of parens
\n" ); document.write( "1260-7x=900+5x subtract 1260 and also 5x from both sides\r
\n" ); document.write( "\n" ); document.write( "1260-1260-7x-5x=900-1260+5x-5x collect like terms\r
\n" ); document.write( "\n" ); document.write( "-12x=-360 divide both sides by -12\r
\n" ); document.write( "\n" ); document.write( "x=30 mph--------------------speed of the wind\r
\n" ); document.write( "\n" ); document.write( "CK
\n" ); document.write( " t=d/r
\n" ); document.write( "7/(180+30)=5/(180-30)
\n" ); document.write( "7/210=5/150
\n" ); document.write( "1/30=1/30\r
\n" ); document.write( "\n" ); document.write( "Hope this helps----ptaylor\r
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