document.write( "Question 112427This question is from textbook Introductory Algebra
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document.write( ": Roberto invested some money at 7%, and then invested $2000 more than twice this amount at 11%. His total annual income from the two investments was $3990. How much was invested at 11%? \n" );
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Algebra.Com's Answer #82029 by checkley71(8403)![]() ![]() ![]() You can put this solution on YOUR website! .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 amount invested @ 7% \n" ); document.write( "2*13,000+2000=26,000+2,000=$28,000 amount invested @ 11%. \n" ); document.write( "proof \n" ); document.write( ".07*13,000+.11*28,000=3990 \n" ); document.write( "910+3080=3990 \n" ); document.write( "3990=3990 \n" ); document.write( " |