document.write( "Question 103338: WHILE ROAD TESTING A NEW MAKE OF CAR, THE EDITOR OF A CONSUMER MAGAZINE FINDS THAT HE CAN GO 12 MILES INTO A 2-MILE-PER-HOUR WIND IN THE SAME AMOUNT OF TIME HE CAN GO 13 MILES WITH A 2-MILE-PER-HOUR WIND BEHIND HIM. FIND THE SPEED OF THE CAR IN STILL AIR. \n" ); document.write( "
Algebra.Com's Answer #75493 by ankor@dixie-net.com(22740)\"\" \"About 
You can put this solution on YOUR website!
WHILE ROAD TESTING A NEW MAKE OF CAR, THE EDITOR OF A CONSUMER MAGAZINE FINDS THAT HE CAN GO 12 MILES INTO A 2-MILE-PER-HOUR WIND IN THE SAME AMOUNT OF TIME HE CAN GO 13 MILES WITH A 2-MILE-PER-HOUR WIND BEHIND HIM. FIND THE SPEED OF THE CAR IN STILL AIR.
\n" ); document.write( ":
\n" ); document.write( "Let s = speed of the car in still air
\n" ); document.write( "Then
\n" ); document.write( "(s+2) = speed with the wind
\n" ); document.write( "and
\n" ); document.write( "(s-2) = speed against the wind
\n" ); document.write( ":
\n" ); document.write( "Write a time equation: Time = Dist/speed\r
\n" ); document.write( "\n" ); document.write( "\"12%2F%28%28s-2%29%29\" = \"13%2F%28%28s%2B2%29%29\"
\n" ); document.write( ":
\n" ); document.write( "Cross multiply:
\n" ); document.write( "13(s-2) = 12(s+2)
\n" ); document.write( ":
\n" ); document.write( "13s - 26 = 12s + 24
\n" ); document.write( ":
\n" ); document.write( "13s - 12s = 24 + 26
\n" ); document.write( ":
\n" ); document.write( "s = 50 mph
\n" ); document.write( ":
\n" ); document.write( "Check solution by finding if, indeed, the times are equal
\n" ); document.write( "12/48 = .25 hr
\n" ); document.write( "13/52 = .25 hr \n" ); document.write( "
\n" );