document.write( "Question 214655: If you increase your rate by 25 mph and you are going 25 miles you get there 10 min earlier, what was your initial rate? \n" ); document.write( "
Algebra.Com's Answer #162092 by Alan3354(69443) You can put this solution on YOUR website! If you increase your rate by 25 mph and you are going 25 miles you get there 10 min earlier, what was your initial rate? \n" ); document.write( "--------------------- \n" ); document.write( "d = rt \n" ); document.write( "25 = rt \n" ); document.write( "25 = (r+25)*(t- 1/6) [10 mins = 1/6 hour] \n" ); document.write( "rt = (r+25)*(t- 1/6) = rt + 25t - r/6 - 25/6 \n" ); document.write( "6rt = 6rt + 150t - r - 25 \n" ); document.write( "150t = r+25 \n" ); document.write( "r = 25/t \n" ); document.write( "150t = (25/t) + 25 \n" ); document.write( "150t^2 = 25 + 25t \n" ); document.write( "6t^2 - t - 1 = 0 \n" ); document.write( "(3t + 1)*(2t - 1) = 0 \n" ); document.write( "t = 1/2 hour (Ignore negative solution) \n" ); document.write( "r = 50 mph \n" ); document.write( " \n" ); document.write( " |