document.write( "Question 696497: A total of $7000 is invested: part at 7% and the remainder at 11%. How much is invested at each rate if the annual interest is $760?
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Algebra.Com's Answer #429085 by checkley79(3341)\"\" \"About 
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.07x+.11(7000-x)=760
\n" ); document.write( ".07x+770-.11x=760
\n" ); document.write( "-.04x=760-770
\n" ); document.write( "-.04x=-10
\n" ); document.write( "x=-10/-.04
\n" ); document.write( "x=$250 invested @ 7%.
\n" ); document.write( "7000-250=$6750 invested @ 11%.
\n" ); document.write( "Proof:
\n" ); document.write( ".07*250+.11*6750=760
\n" ); document.write( "17.50+742.50=760
\n" ); document.write( "760=760
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