document.write( "Question 990855: Annette drive to Shawnee in 4 hours and drove back in three hours. What were her speeds of her speed coming back was 11 miles per hour greater that her speed going? \n" ); document.write( "
Algebra.Com's Answer #610861 by macston(5194)\"\" \"About 
You can put this solution on YOUR website!
.
\n" ); document.write( "D=distance one way; R=Annette's speed going to Shawnee
\n" ); document.write( "R+11mph=speed returning
\n" ); document.write( ".
\n" ); document.write( "4(R)=3(R+11mph)
\n" ); document.write( "4R=3R+33mph
\n" ); document.write( "R=33mph
\n" ); document.write( "ANSWER 1: Her speed going was 33 mph.
\n" ); document.write( ".
\n" ); document.write( "Return speed=R+11mph=33mph+11mph=44mph
\n" ); document.write( "ANSWER 2: Her return speed was 44 mph.
\n" ); document.write( ".
\n" ); document.write( "CHECK:
\n" ); document.write( "D=R(4hrs)
\n" ); document.write( "D=33mph(4hrs)=132 miles
\n" ); document.write( ".
\n" ); document.write( "D=R+11(3hrs)
\n" ); document.write( "132mi=(33mph+11mph)(3hrs)
\n" ); document.write( "132mi=(44mph)(3hrs)
\n" ); document.write( "132mi=132mi
\n" ); document.write( "
\n" );