document.write( "Question 936926: If 15 oxen or 20 cows can eat the grass of a field in 80 days, then in how many days will 6 oxen and 2 cows eat the same grass?
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Algebra.Com's Answer #570407 by josmiceli(19441) $\"\"$ $\"About$

You can put this solution on YOUR website!
$\"+15+\"$ oxen eat ( 1 field ) / ( 80 days )
\n" ); document.write( " $\"+1+\"$ ox eats ( 1 field ) / ( 15*80 days )
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\n" ); document.write( " $\"+20+\"$ cows eat ( 1 field ) / ( 80 days )
\n" ); document.write( " $\"+1+\"$ cow eats ( 1 field ) / ( 20*80 days )
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\n" ); document.write( "Add their rates of eating to get their rate eating together
\n" ); document.write( "Let their rate eating together = ( 1 field ) / ( d days )
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\n" ); document.write( " $\"+6%2A%28+1%2F1200+%29+%2B+2%2A%28+1%2F1600+%29+=+1%2Fd+\"$
\n" ); document.write( "Multiply both sides by $\"+4800d+\"$
\n" ); document.write( " $\"+24d+%2B+6d+=+4800+\"$
\n" ); document.write( " $\"+30d+=+4800+\"$
\n" ); document.write( " $\"+d+=+160+\"$
\n" ); document.write( "6 oxen and 2 cows eat 1 field of grass in 160 days
\n" ); document.write( "check:
\n" ); document.write( " $\"+6%2A%28+1%2F1200+%29+%2B+2%2A%28+1%2F1600+%29+=+1%2Fd+\"$
\n" ); document.write( " $\"+6%2A%28+1%2F1200+%29+%2B+2%2A%28+1%2F1600+%29+=+1%2F160+\"$
\n" ); document.write( "Multiply both sides by $\"+40+\"$
\n" ); document.write( " $\"+6%2F30+%2B+2%2F40+=+1%2F4+\"$
\n" ); document.write( " $\"+1%2F5+%2B+1%2F20+=+5%2F20+\"$
\n" ); document.write( " $\"+4%2F20+%2B+1%2F20+=+5%2F20+\"$
\n" ); document.write( "OK\r
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