document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=95754>Question 95754</A>This question is from textbook <A HREF=http://www.amazon.com/exec/obidos/ASIN/0534944701/algebrahomeworkh></A><BR>\n" );
document.write( ": <I><SPAN STYLE=\"word-break: break-all;\"> It took Jerry 6 hr to ride his bicycle 30 mi against the wind, but took him only 2 hr to return to his starting point with the same wind. Find Jerry's average riding speed in still air, and find the average speed of the wind. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #69777 by </B><A HREF=/tutors/aboutme.mpl?userid=bucky><B>bucky(2189)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=bucky><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=69777\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> To solve this problem we will make use of two things. The first is that we will use the<BR>\n" );
document.write( "equation that says the distance traveled (D) is equal to the rate or speed of travel (R)<BR>\n" );
document.write( "times the time (T) that passes during the travel. In equation form this is:<BR>\n" );
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document.write( "D = R * T<BR>\n" );
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document.write( "The second thing we will use is that for an object that travels at a rate S in still air,<BR>\n" );
document.write( "if the wind is blowing in the direction of travel at a rate of W, it boosts the overall <BR>\n" );
document.write( "rate of the object to S + W. But if the wind is blowing in the direction opposite to the<BR>\n" );
document.write( "travel of the object, it lowers the speed of the object to S - W.<BR>\n" );
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document.write( "Now let's work the problem.<BR>\n" );
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document.write( "For the first trip, Jerry travels against the wind. He goes 30 miles in 6 hours. And since<BR>\n" );
document.write( "he is going against the wind, the wind is slowing him down. Therefore, if his rate in<BR>\n" );
document.write( "still air is S, his actual rate while going against the wind is S - W. If we substitute<BR>\n" );
document.write( "these values into the distance equation D = R*T the equation becomes:<BR>\n" );
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document.write( "30 = (S - W)*6 = 6S - 6W<BR>\n" );
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document.write( "Now for the trip back home.  Jerry travels with the wind this time. He goes the same 30 <BR>\n" );
document.write( "miles but does it in 2 hours. And since he is going with the wind, the wind is aiding him<BR>\n" );
document.write( "to go faster.  His rate is still air is S and the push of the wind raises his actual rate<BR>\n" );
document.write( "to S + W. Substituting these values into the equation results in the distance equation becoming:<BR>\n" );
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document.write( "30 = (S + W)* 2 = 2S + 2W<BR>\n" );
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document.write( "So now we have two equations:<BR>\n" );
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document.write( "30 = 6S - 6W and<BR>\n" );
document.write( "30 = 2S + 2W<BR>\n" );
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document.write( "We can solve these by variable elimination. Let's multiply both sides of the bottom equation <BR>\n" );
document.write( "(all terms) by 3 to get:<BR>\n" );
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document.write( "90 = 6S + 6W<BR>\n" );
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document.write( "This then makes our two equations:<BR>\n" );
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document.write( "30 = 6S - 6W and<BR>\n" );
document.write( "90 = 6S + 6W<BR>\n" );
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document.write( "Now notice that if we add these two equations together vertically the -6W and the +6W <BR>\n" );
document.write( "will cancel. The equation that results comes from 30 + 90 = 120 and 6S + 6S = 12S. So<BR>\n" );
document.write( "what we are left with is:<BR>\n" );
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document.write( "120 = 12S<BR>\n" );
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document.write( "We can solve this equation for S by dividing both sides by 12 to get:<BR>\n" );
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document.write( "10 = S<BR>\n" );
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document.write( "So we know that Jerry's speed in still air is 10 miles per hour.<BR>\n" );
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document.write( "We now can return to one of the early equations that contains both S and W and we can replace<BR>\n" );
document.write( "S with 10 to solve for W. For example, let's return to the equation:<BR>\n" );
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document.write( "30 = (S - W)*6<BR>\n" );
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document.write( "and when we replace S with 10, this equation becomes:<BR>\n" );
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document.write( "30 = (10 - W)*6<BR>\n" );
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document.write( "and this multiplies out on the right side to result in:<BR>\n" );
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document.write( "30 = 60 - 6W<BR>\n" );
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document.write( "Get rid of the 60 on the right side by subtracting 60 from both sides to get:<BR>\n" );
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document.write( "-30 = -6W<BR>\n" );
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document.write( "and solve for W by dividing both sides by -6 the multiplier of the W. That division<BR>\n" );
document.write( "results in:<BR>\n" );
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document.write( "W = -30/-6 = 5<BR>\n" );
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document.write( "So the wind goes at 5 miles per hour.  And Jerry's speed is 10 miles per hour.<BR>\n" );
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document.write( "Let's verify those answers. When Jerry is going against the wind his actual speed is<BR>\n" );
document.write( "10 miles per hour less the 5 miles per hour of the wind for a net rate of 5 miles per hour.<BR>\n" );
document.write( "In 6 hours at 5 miles per each hour he will go 30 miles. That checks. <BR>\n" );
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document.write( "Then going with the wind Jerry goes at 10 miles per hour and the wind aids him by adding<BR>\n" );
document.write( "5 miles per hour, so his rate is 15 miles per hour. In 2 hours at 15 miles per hour he<BR>\n" );
document.write( "will go 30 miles again. That also checks. So our answers of 10 mph for Jerry's speed in<BR>\n" );
document.write( "still air and 5 mph for the wind speed are good.<BR>\n" );
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document.write( "Hope this helps you to understand this problem and how you can work it to get an answer.<BR>\n" );
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document.write( " </SPAN>\n" );
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