document.write( "Question 339965: an investment club invested part of $10,000 at 9% annual interest and the rest at 8%. If the annual income from these investments was $860, how much was invested at 8% \n" ); document.write( "

Algebra.Com's Answer #243642 by josmiceli(19441) $\"\"$ $\"About$

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Let $\"a\"$ = dollars invested at 9%
\n" ); document.write( "Let $\"b\"$ = dollars invested at 8%
\n" ); document.write( "given:
\n" ); document.write( "(1) $\"a+%2B+b+=+10000\"$
\n" ); document.write( " $\".09a+%2B+.08b+=+860\"$
\n" ); document.write( "------------------
\n" ); document.write( "(2) $\"9a+%2B+8b+=+86000\"$
\n" ); document.write( "Multiply both sides of (1) by $\"8\"$ and
\n" ); document.write( "subtract (1) from (2)
\n" ); document.write( "(2) $\"9a+%2B+8b+=+86000\"$
\n" ); document.write( "(1) $\"-8a+-+8b+=+80000\"$
\n" ); document.write( " $\"a+=+6000\"$
\n" ); document.write( "And, from (1)
\n" ); document.write( "(1) $\"6000+%2B+b+=+10000\"$
\n" ); document.write( " $\"b+=+4000\"$
\n" ); document.write( "$6000 was invested at 9% and
\n" ); document.write( "$4000 was invested at 8%
\n" ); document.write( "check:
\n" ); document.write( " $\".09a+%2B+.08b+=+860\"$
\n" ); document.write( " $\".09%2A6000+%2B+.08%2A4000+=+860\"$
\n" ); document.write( " $\"540+%2B+320+=+860\"$
\n" ); document.write( " $\"860+=+860\"$
\n" ); document.write( "OK \n" ); document.write( "

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