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)![]() ![]() ![]() You can put this solution on YOUR website! .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 \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |