document.write( "Question 34865: With the wind, an airplane travels 2200 miles in 4 hours. Against the wind, it takes 5 hours to travel 2350 miles. Find the rate of the plane in still air. \n" ); document.write( "
Algebra.Com's Answer #21091 by Paul(988)![]() ![]() ![]() You can put this solution on YOUR website! Let the rate of the wind be y \n" ); document.write( "Let the rate of the plane be x \n" ); document.write( "Equation 1: \n" ); document.write( "4x+4y=2200 \n" ); document.write( "y=550-x (subsittuion) \n" ); document.write( "Equation 2: \n" ); document.write( "5x-5y=2350 \n" ); document.write( "Apply subsitution: \n" ); document.write( "5x-5(550-x)=2350 \n" ); document.write( "10x=5100 \n" ); document.write( "x=510 \n" ); document.write( "y=550-510 \n" ); document.write( "y=40 \n" ); document.write( "Hence, the speed of the plane is 510mph and the speed of the wind is 40mph. \n" ); document.write( "Paul. \n" ); document.write( " \n" ); document.write( " |