document.write( "Question 134468: Sam invested $7,500 for one year, part at 12% annual interest and the rest at 10% annual interest. His total interest for the year was $890. How much money did he invest at 12%. \n" ); document.write( "

Algebra.Com's Answer #98373 by josmiceli(19441) $\"\"$ $\"About$

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$\"x\"$ = amount invested at 12%
\n" ); document.write( " $\"y\"$ = amount invested at 10%
\n" ); document.write( " $\".12x+%2B+.1y+=+890\"$
\n" ); document.write( " $\"x+%2B+y+=+7500\"$
\n" ); document.write( " $\"x+=+7500+-+y\"$
\n" ); document.write( " $\".12%2A%287500+-+y%29+%2B+.1y+=+890\"$
\n" ); document.write( " $\"900+-+.12y+%2B+.1y+=+890\"$
\n" ); document.write( " $\"-.02y+=+-10\"$
\n" ); document.write( " $\"y+=+10%2F.02\"$
\n" ); document.write( " $\"y+=+1000%2F2\"$
\n" ); document.write( " $\"y+=+500\"$
\n" ); document.write( " $\"x+=+7500+-+y\"$
\n" ); document.write( " $\"x+=+7000\"$
\n" ); document.write( "He invested $7000 at 12%
\n" ); document.write( "check:
\n" ); document.write( " $\".12%2A%287500+-+y%29+%2B+.1y+=+890\"$
\n" ); document.write( " $\".12%2A%287500+-+500%29+%2B+.1%2A500+=+890\"$
\n" ); document.write( " $\".12%2A7000+%2B+50+=+890\"$
\n" ); document.write( " $\"840+%2B+50+=+890\"$
\n" ); document.write( " $\"890+=+890\"$
\n" ); document.write( "OK \n" ); document.write( "

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