document.write( "Question 137110: A man invests part of $6,000 at 4% interest and the rest at 3% interest. His total annual income is $225. How much has he invested at each rate? \n" ); document.write( "
Algebra.Com's Answer #100392 by checkley77(12844)![]() ![]() ![]() You can put this solution on YOUR website! .04X+.03(6,000-X)=225 \n" ); document.write( ".04X+180-.03X=225 \n" ); document.write( ".01X=225-180 \n" ); document.write( ".01X=45 \n" ); document.write( "X=4500 AMOUNT INVWESTED @ 4% \n" ); document.write( "6000-4500=1500 INVESTED @ 3% \n" ); document.write( "PROOF: \n" ); document.write( ".04*4500+.03*1500=225 \n" ); document.write( "180+135=225 \n" ); document.write( "225=225 \n" ); document.write( " |