document.write( "Question 306555: Roberto invested some money at 7%, and then invested $2000 more than twice this amount at 10%.
\n" ); document.write( "His total annual income from the two investments was $3980. How much was invested at 10%?
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Algebra.Com's Answer #219365 by checkley77(12844)\"\" \"About 
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.07x+.10(2x+2,000)=3,980
\n" ); document.write( ".07x+.20x+200=3,980
\n" ); document.write( ".27x=3,980-200
\n" ); document.write( ".27x=3,780
\n" ); document.write( "x=3,780/.27
\n" ); document.write( "x=$14,000 amount invested @ 7%
\n" ); document.write( "2*14,000+2,000=28,000+2,000=$30,000 invested @ 10%
\n" ); document.write( "Proof:
\n" ); document.write( ".07*14,000+.10*30,000=3,980
\n" ); document.write( "980+3,000=3,980
\n" ); document.write( "3,980=3,980
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