document.write( "Question 140729: a pickup can travel 300 miles in the same time that a car going 10 miles per hour faster can travel 350 miles. How fast is the car traveling? \n" ); document.write( "
Algebra.Com's Answer #102422 by solver91311(24713)\"\" \"About 
You can put this solution on YOUR website!
distance equals rate times time, or \"d=rt\", so \"t=d%2Fr\"\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "rate of the car: r (I chose the car's rate to be r rather than the pickup's rate because the question asks for the car's rate)
\n" ); document.write( "distance for the car: 350\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "rate of the pickup: r - 10 (the pickup is going 10 mph slower)
\n" ); document.write( "distance for the pickup: 300\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "time for the pickup: \"t=300%2F%28r-10%29\"
\n" ); document.write( "time for the car: \"t=350%2Fr\"\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "But the time for the pickup and car are the same, so:\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"350%2Fr=300%2F%28r-10%29\"\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Cross multiply, collect like terms, and solve for r
\n" ); document.write( "
\n" );