document.write( "Question 130342: Roberto invested some money at 7%, and then invested $2,000 more than twice his amount at 11%. His total annual income from the two investemnts was $3,990. How much was invested at 11%. \n" ); document.write( "
Algebra.Com's Answer #95209 by checkley71(8403)\"\" \"About 
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.07X+.11(2X+2000)=3990
\n" ); document.write( ".07X+.22X+220=3990
\n" ); document.write( ".29X=3990-220
\n" ); document.write( ".29X=3770
\n" ); document.write( "X=3770/.29
\n" ); document.write( "X=13,000 INVESTED @ 7%.
\n" ); document.write( "2*13000+2000=26000+2000=28,000 IS THE AMOUNT INVESTED @ 11%.
\n" ); document.write( "PROOF:
\n" ); document.write( ".07*13000+.11*28000=3990
\n" ); document.write( "910+3080=3990
\n" ); document.write( "3990=3990
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