document.write( "Question 844196: A new edition of a video game cost $60, but its previous version only cost $40. If 15 games are sold for $840, how many copies of the newest game sold?\r
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Algebra.Com's Answer #508601 by josgarithmetic(39618)\"\" \"About 
You can put this solution on YOUR website!
Maybe a mixture problem but also a linear system problem.\r
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\n" ); document.write( "\n" ); document.write( "n is how many new
\n" ); document.write( "p is how many previous versions\r
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\n" ); document.write( "\n" ); document.write( "Account for total versions sold: \"n%2Bp=15\"
\n" ); document.write( "Account for sales value: \"60n%2B40p=840\" which is simplifiable to \"3n%2B2p=42\"\r
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\n" ); document.write( "\n" ); document.write( "SOLVE THE SYSTEM:
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\n" ); document.write( "n+p=15
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\n" ); document.write( "3n+2p=42
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\n" ); document.write( "\n" ); document.write( "Choosing Elimination Method, you can use 2n+2p=30 and 3n+2p=42, and subtract first equation from second equation.
\n" ); document.write( "\"%283n%2B2p%29-%282n%2B2p%29=42-30\"
\n" ); document.write( "\"3n%2B2p-2n-2p=12\"
\n" ); document.write( "\"n%2B0=12\"
\n" ); document.write( "\"highlight%28n=12%29\" and this means \"highlight%28p=3%29\"
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