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)\"\" \"About 
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.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( "
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