document.write( "Question 128975: A man invested a sum of money at 6%. A second sum $400 more than the first sum is invested at 3%. A third sum 4 times as much as the first sum is invested at 4%. The total annual incomr is $237, How much has he invested at each rate? \n" ); document.write( "
Algebra.Com's Answer #94320 by checkley71(8403)\"\" \"About 
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.06X+.03(X+400)+.04(4X)=237
\n" ); document.write( ".06X+.03X+12+.16X=237
\n" ); document.write( ".25X+12=237
\n" ); document.write( ".25X=237-12
\n" ); document.write( ".25X=225
\n" ); document.write( "X=225/.25
\n" ); document.write( "X=900 INVESTED @ 6%.
\n" ); document.write( "900+400=1300 INVESTED @ 3%
\n" ); document.write( "900*4=3600 INVESTED @ 4%.
\n" ); document.write( "PROOF
\n" ); document.write( ".06*900+.03*1300+.04*3600=237
\n" ); document.write( "54+39+144=237
\n" ); document.write( "237=237\r
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