document.write( "Question 819906: A truck can travel 315 miles in the same time a car travels 585. If the trucks milage is 30 mph slower than the cars, what is the average rate for each? \n" ); document.write( "
Algebra.Com's Answer #493265 by josgarithmetic(39620)\"\" \"About 
You can put this solution on YOUR website!
You want to find the rate for each.
\n" ); document.write( "r= rate for the CAR.\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Vehicle__________speed____________time_______________distance
\n" ); document.write( "TRUCK____________r-30_____________t__________________315
\n" ); document.write( "CAR______________r________________t__________________518\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Using rate*time=distance, get the time expressions for each:\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Vehicle__________speed____________time_______________distance
\n" ); document.write( "TRUCK____________r-30_____________\"315%2F%28r-30%29\"__________________315
\n" ); document.write( "CAR______________r________________\"518%2Fr\"__________________518\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Remember from description, those times are equal.
\n" ); document.write( "\"highlight_green%28315%2F%28r-30%29=518%2Fr%29\"
\n" ); document.write( "\"518%28r-30%29=315r\"
\n" ); document.write( "\"518r-518%2A30=315r\"
\n" ); document.write( "\"518r-315r=518%2A30\"
\n" ); document.write( "\"203r=15540\"
\n" ); document.write( "\"r=15540%2F203\"
\n" ); document.write( "\"highlight%28r=76.55%29\" or close enough to 77 miles per hour for the car.
\n" ); document.write( "The truck goes at 47 mph.
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