document.write( "Question 512387: Hi! This is a problem from me test. I really need help. Thank you
\n" ); document.write( "On a trip Jennifer noticed that her car averaged 21mi/gal of gas except for the days she used the air conditioning, and the it averaged only 16mi/gal. If she used 91 gal of gas to drive 1751 mi, how many of those miles did she use the air conditioning?
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Algebra.Com's Answer #342619 by Sarah900(12)\"\" \"About 
You can put this solution on YOUR website!
(mileage with AC (mi/gal))*(gas used with AC(gal))+ (mileage without AC (mi/gal))*(gas used without AC(gal))=total miles traveled (miles)
\n" ); document.write( "Note how the units cancel out to leave you with miles
\n" ); document.write( "Assign x for one of the missing elements. We'll say the gas Jennifer used when she was using AC= x. Since she used 91 gallons total the expression that would tell you what she used without AC= 91-x.

\n" ); document.write( "Plug in what you know and solve for x
\n" ); document.write( "(21)*(x)+(16)*(91-x)=1751
\n" ); document.write( "21x+1456-16x=1751
\n" ); document.write( "5x+1456=1751
\n" ); document.write( "5x=295
\n" ); document.write( "x=59

\n" ); document.write( "So she used 59 gallons when she had AC. To find how much she used without
\n" ); document.write( "91-59=32

\n" ); document.write( "You can find how many miles she drove during each by multiplying the mileage with the corresponding number of gallons. So...
\n" ); document.write( "21*59=1239
\n" ); document.write( "16*32=512
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