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)![]() ![]() 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( " |