document.write( "Question 331202: barbara drove 330 miles from home to tuscan on her round trip she was able to increase her speed by 11miles per hour.If the return trip took 1 hour less than find her original speed to tuscan \n" ); document.write( "

Algebra.Com's Answer #237411 by mananth(16946) $\"\"$ $\"About$

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let the original speed be x mph
\n" ); document.write( "on her return it was x+11 mph
\n" ); document.write( "..
\n" ); document.write( "distance is constant = 330 miles.
\n" ); document.write( "..
\n" ); document.write( "return was 1 hour less.
\n" ); document.write( "time taken while going - time returning = 1 hour
\n" ); document.write( "..
\n" ); document.write( "time while going = 330/x
\n" ); document.write( "time while returning = 330/x+11
\n" ); document.write( "..
\n" ); document.write( "330/x-330/(x+11)=1
\n" ); document.write( "LCD = x(x+11)
\n" ); document.write( "330(x+11)-330x/ x (x+11)=1
\n" ); document.write( "330x+3630-330x=x(x+11)
\n" ); document.write( "3630= x^2 +11x
\n" ); document.write( "x^2+11x-3630=0
\n" ); document.write( "x^2+66x-55x-3630=0
\n" ); document.write( "x(x+66)-55(x+66)=0
\n" ); document.write( "(x-55)(x+66)=0
\n" ); document.write( "x=55 ignore negative value \r
\n" ); document.write( "\n" ); document.write( "original speed = 55 mph \n" ); document.write( "

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