document.write( "Question 1176546: A tour bus normally leaves for its destination at
\n" ); document.write( "5:00 p.m. for a 300 mile trip. This week however, the bus leaves at 6:00 p.m. To arrive on time, the driver drives 10 miles per hour faster than usual. What is the bus` usual speed? \n" ); document.write( "
Algebra.Com's Answer #803692 by ewatrrr(24785)\"\" \"About 
You can put this solution on YOUR website!
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document.write( " d = rt \r\n" );
document.write( " 300 = rt  0r 300/r = t\r\n" );
document.write( " 300 = (r+10)(t-1)\r\n" );
document.write( "  300 = (r+10)(300/r -1 )\r\n" );
document.write( "  300 = (r+10)((300-r)/r \r\n" );
document.write( "  300r = (r+10)((300-r) \r\n" );
document.write( "  300r = 300r + 3000 - r^2 - 10r\r\n" );
document.write( "    r^2 + 10r + 3000 = 0\r\n" );
document.write( "    (r+60)(r-50) = 0\r\n" );
document.write( "            r = 50mph, the usual speed   |tossing out negative number for speed\r\n" );
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document.write( "50mph*6hr = 60mph*5hr = 300mi  checks\r\n" );
document.write( "Wish You the Best in your Studies.\r\n" );
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