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)\"\" \"About 
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.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( "
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