document.write( "Question 867641: Cyril invested part of 185000 at 7% and the rest at 9%. If the interest from the 9% investment is 1450 more than the interest earned at the 7% investment, how much did he invest at the 9% rate? \n" ); document.write( "
Algebra.Com's Answer #523152 by checkley79(3341)![]() ![]() ![]() You can put this solution on YOUR website! .07x=.09(185,00-x)-1,450 \n" ); document.write( ".07x=16,650-.09x-1,450 \n" ); document.write( ".07x+.09x=16,650-1,450 \n" ); document.write( ".16x=15,200 \n" ); document.write( "x=15,200/.16 \n" ); document.write( "x=95,000 invested @ 7%. \n" ); document.write( "185,000-95,000=90,000 invested @ 9% \n" ); document.write( "Prof: \n" ); document.write( ".07*95,000=.09*90,000-1,450 \n" ); document.write( "6,650=8,100-1,450 \n" ); document.write( "6,650=6,650 \n" ); document.write( " |