document.write( "Question 144557: I need help with the following problem...I have NO idea what to do!! Thanks in advance for your help!\r
\n" ); document.write( "\n" ); document.write( "Roberto invested some money at 6%, and then invested $3000 more than twice this amount at 10%. His total annual income from the two investments was $4200. How much was invested at 10%?
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Algebra.Com's Answer #105305 by checkley77(12844)\"\" \"About 
You can put this solution on YOUR website!
.06x+.10(3,000+2x)=4,200
\n" ); document.write( ".06x+300+.20x=4,200
\n" ); document.write( ".26x=4,200-300
\n" ); document.write( ".26x=3,900
\n" ); document.write( "x=3,900/.26
\n" ); document.write( "x=$15,000 invested @ 6%.
\n" ); document.write( "3,000+2*15,000
\n" ); document.write( "3,000+$30,000=$33,000 invested @ 10%.
\n" ); document.write( "proof:
\n" ); document.write( ".06*15,000+.10*33,000=4,200
\n" ); document.write( "900+3,300=4,200
\n" ); document.write( "4,200=4,200\r
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