document.write( "Question 634366: A man makes three investments totaling $24000 with the interest rates at 6%, 7% and 8%. The total annual income is $1720, and the income from the 7% investment is $4o less than the combined income from the other investments. Find the amount of each investment. \n" ); document.write( "
Algebra.Com's Answer #399531 by sachi(548)\"\" \"About 
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A man makes three investments totaling $24000 with the interest rates at 6%, 7% and 8%. The total annual income is $1720, and the income from the 7% investment is $4o less than the combined income from the other investments. Find the amount of each investment.
\n" ); document.write( "ans:-
\n" ); document.write( "let his investment & income out of that is as follows
\n" ); document.write( " investment$ rate income$
\n" ); document.write( " say x 6%=.06 .06x
\n" ); document.write( " say y 7%=.07 .07y
\n" ); document.write( " 24000-(x+y)=z say 8%=.08 .08z
\n" ); document.write( "----------------------------------------
\n" ); document.write( "total=24000 1720
\n" ); document.write( "so .06x+.07y+.08z=1720------eqn 1
\n" ); document.write( "& .07y+40=.06x+.08z ------eqn 2
\n" ); document.write( "from above 2 eqns .06x+.08z =.07y+40=1720-.07y
\n" ); document.write( "or 2 * .07y=1720-40=1680
\n" ); document.write( "or y=1680/.14=12000
\n" ); document.write( "now from eqn 1
\n" ); document.write( ".06x+.07*12000+.08(24000-x-12000)=1720
\n" ); document.write( "or(.06-.08)x=1720-.07*12000-.08*12000=1720-1800=-80
\n" ); document.write( "or -.02x=-80
\n" ); document.write( "or x=80/.02=4000
\n" ); document.write( "now z=24000-12000-4000=8000
\n" ); document.write( "so his invesments are $4000,$12000,$8000 (ans) \n" ); document.write( "
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