document.write( "Question 190266: A person invested $16,000, part of 8% and the remainder at 6%. If the total yearly intrest from these investments was $1180, find the amount invested at each rate. \n" ); document.write( "

Algebra.Com's Answer #142791 by josmiceli(19441) $\"\"$ $\"About$

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Let $\"x\"$ = the amount invested @ 8%
\n" ); document.write( " $\"16000+-+x\"$ will be the amount invested @ 6%
\n" ); document.write( "given:
\n" ); document.write( " $\".08x+%2B+.06%2A%2816000+-+x%29+=+1180\"$
\n" ); document.write( "----------------------------------
\n" ); document.write( " $\".08x+%2B+960+-+.06x+=+1180\"$
\n" ); document.write( " $\"+.02x+=+220\"$
\n" ); document.write( " $\"x+=+11000\"$
\n" ); document.write( " $\"16000+-+x+=+16000+-+11000\"$
\n" ); document.write( " $\"16000+-+11000+=+5000\"$
\n" ); document.write( "$11000 was invested @ 8% and $5000 was invested @ 6%
\n" ); document.write( "check:
\n" ); document.write( " $\".08x+%2B+.06%2A%2816000+-+x%29+=+1180\"$
\n" ); document.write( " $\".08%2A11000+%2B+.06%2A5000+=+1180\"$
\n" ); document.write( " $\".08%2A11000+%2B+.06%2A5000+=+1180\"$
\n" ); document.write( " $\"880+%2B+300+=+1180\"$
\n" ); document.write( " $\"1180+=+1180\"$
\n" ); document.write( "OK \n" ); document.write( "

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