document.write( "Question 965284: bob has $15000 to invest. he plans to invest part at 8%, with the remainder invested at 10%. find the amount invested at each rate if the total annual interest income is 1300 \n" ); document.write( "
Algebra.Com's Answer #590008 by Fombitz(32388)\"\" \"About 
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\"0.08X%2B0.10Y=1300\"
\n" ); document.write( "\"8X%2B10Y=130000\"
\n" ); document.write( ".
\n" ); document.write( ".\r
\n" ); document.write( "\n" ); document.write( "\"X%2BY=15000\"
\n" ); document.write( "Multiply the second equation by 8 and subtract from the first equation,
\n" ); document.write( "\"8X%2B10Y-8X-8Y=130000-120000\"
\n" ); document.write( "\"2Y=10000\"
\n" ); document.write( "\"Y=5000\"
\n" ); document.write( "\"X%2B5000=10000\"
\n" ); document.write( "$10000 at 8%, $5000 at 10%
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