document.write( "Question 197730: A car rental company has two rental rates. Rate 1 is $35 per day plus $0.16 per mile. Rate 2 is $50 per day and $0.09 per mile. If you plan to rent for one week, how many miles would you need to drive to pay less by taking rate 2? \n" ); document.write( "

Algebra.Com's Answer #148444 by josmiceli(19441) $\"\"$ $\"About$

You can put this solution on YOUR website!
Let $\"d\"$ = number of days rented
\n" ); document.write( "Let $\"m\"$ = miles driven
\n" ); document.write( "given:
\n" ); document.write( "(1) $\"35d+%2B+.16m\"$ = rate 1
\n" ); document.write( "(2) $\"50d+%2B+.09m\"$ = rate 2
\n" ); document.write( " $\"d+=+7\"$
\n" ); document.write( "First find the break-even point by making rate 1 = rate 2
\n" ); document.write( "(1) $\"35d+%2B+.16m+=+50d+%2B+.09m\"$
\n" ); document.write( " $\"35%2A7+%2B+.16m+=+50%2A7+%2B+.09m\"$
\n" ); document.write( " $\"245+%2B+.16m+=+350+%2B+.09m\"$
\n" ); document.write( " $\".07m+=+105\"$
\n" ); document.write( " $\"m+=+1500\"$ mi
\n" ); document.write( " $\"245+%2B+.16%2A1500+=+245+%2B+240\"$
\n" ); document.write( "rate 1 = $\"485\"$
\n" ); document.write( " $\"350+%2B+.09%2A1500+=+350+%2B+135\"$
\n" ); document.write( "rate 2 = $\"485\"$
\n" ); document.write( "-----------------
\n" ); document.write( "If I drive 1 extra mile, which rate is lower?
\n" ); document.write( "(1) $\"35d+%2B+.16m\"$ = rate 1
\n" ); document.write( " $\"245+%2B+.16%2A1501\"$
\n" ); document.write( " $\"245+%2B+240.16+=+485.16\"$
\n" ); document.write( "and
\n" ); document.write( "(2) $\"50d+%2B+.09%2A1501\"$ = rate 2
\n" ); document.write( " $\"350+%2B+135.09+=+485.09\"$
\n" ); document.write( "Any milage over 1500 will make rate 2 less than rate 1
\n" ); document.write( "I'll plot the lines:
\n" ); document.write( " $\"+graph%28+800%2C+500%2C+-300%2C+2000%2C+-30%2C+600%2C+245+%2B+.16x%2C+350+%2B+.09x%29+\"$ \n" ); document.write( "

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