document.write( "Question 159477: Sally invested $4500.00. Part of it was invested at 7.5% and part of it was 6%. After one year her interest was $303.00. How much was invested at each rate? \n" ); document.write( "
Algebra.Com's Answer #117625 by checkley77(12844) ![]() You can put this solution on YOUR website! .075X+.06(4500-X)=303 \n" ); document.write( ".075X+270-.06X=303 \n" ); document.write( ".015X=303-270 \n" ); document.write( ".015X=33 \n" ); document.write( "X=33/.015 \n" ); document.write( "X=2,200 INVESTED @ 7.5% \n" ); document.write( "4500-2200=2300 INVESTED @ 6% \n" ); document.write( "PROOF: \n" ); document.write( ".075*2200+.06*2300=303 \n" ); document.write( "165+138=303 \n" ); document.write( "303=303 \n" ); document.write( " \n" ); document.write( " |