document.write( "Question 325153: Roberto invested some money at 7%, and then invested $2000 more than twice this amount at 9%. His total annual income from the two investments was $2930. How much was invested at 9%?
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Algebra.Com's Answer #232928 by checkley77(12844)\"\" \"About 
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.09(2x+2,000)+.07x=2,930
\n" ); document.write( ".18x+180+.07x=2,930
\n" ); document.write( ".25x=2,930-180
\n" ); document.write( ".25x=2,750
\n" ); document.write( "x=2,750/.25
\n" ); document.write( "x=11,000 ans. for the amount invested @ 7%.
\n" ); document.write( "2*11,00+2,000
\n" ); document.write( "22,000+2,000=24,000 ans. for the amount invested @ 9%.
\n" ); document.write( "Proof:
\n" ); document.write( ".09*24,000+.07*11,000=2,930
\n" ); document.write( "2,160+770=2,930
\n" ); document.write( "2,930=2,930
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