document.write( "Question 524424: I have been stuck on this one for a while now. Trying to define variables, write and equation and solve. Please help me!!\r
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document.write( "The distance going up and down a hill is 15 miles. It takes a total of 2 hours to complete the trip. Going down the hill is 20 mph faster than going up. How fast is the trip up the hill in mph? \n" );
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Algebra.Com's Answer #347734 by nerdybill(7384)![]() ![]() You can put this solution on YOUR website! The distance going up and down a hill is 15 miles. It takes a total of 2 hours to complete the trip. Going down the hill is 20 mph faster than going up. How fast is the trip up the hill in mph? \n" ); document.write( "Let y = time (hours) going uphill \n" ); document.write( "then \n" ); document.write( "2-y = time (hours) going downhill \n" ); document.write( ". \n" ); document.write( "Let x = speed (mph) going uphill \n" ); document.write( "then \n" ); document.write( "x+20 = speed (mph) going downhill \n" ); document.write( ". \n" ); document.write( "applying distance formula of d=rt \n" ); document.write( "xy = 15 (equation 1) \n" ); document.write( "(x+20)(2-y) = 15 (equation 2) \n" ); document.write( ". \n" ); document.write( "solving equation 1 for y: \n" ); document.write( "y = 15/x \n" ); document.write( ". \n" ); document.write( "plug into equation 2 and solve for x: \n" ); document.write( "(x+20)(2-y) = 15 \n" ); document.write( "(x+20)(2-15/x) = 15 \n" ); document.write( "multiplying both sides by x: \n" ); document.write( "(x+20)(2x-15) = 15x \n" ); document.write( "2x^2-15x+40x-300 = 15x \n" ); document.write( "2x^2+25x-300 = 15x \n" ); document.write( "2x^2+10x-300 = 0 \n" ); document.write( "x^2+5x-150 = 0 \n" ); document.write( "(x+15)(x-10) = 0 \n" ); document.write( "x = {-15, 10} \n" ); document.write( "throw out negative solution leaving \n" ); document.write( "x = 10 mph (speed going up the hill) \n" ); document.write( " |