document.write( "Question 364336: Roberto invested some money at 6%, and then invested $4000 more than twice this amount at 12%.
\n" ); document.write( "His total annual income from the two investments was $3480. How much was invested at 12%?
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Algebra.Com's Answer #259834 by mananth(16946)\"\" \"About 
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$x @ 6 %
\n" ); document.write( "2x+4000 @ 12%
\n" ); document.write( "interest = 3480
\n" ); document.write( "...
\n" ); document.write( "0.06x+0.12(2x+4000) = 3480
\n" ); document.write( "multiply by 100
\n" ); document.write( "6x+12(2x+4000)=348000
\n" ); document.write( "6x+24x+48000=348000
\n" ); document.write( "30x = 300,000
\n" ); document.write( "x= 30,000 investment @ 6%
\n" ); document.write( "2x+4000= 60,000+4000= $64,000 @12%
\n" ); document.write( "...
\n" ); document.write( "m.ananth@hotmail.ca
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