document.write( "Question 26284: An investment advisor invested $14,000 in two accounts. One investment earned 8% annual simple interest and the other investment earned 6.5% annual simple interest. The amount of interest earned for 1 year was $1024. How much was invested in each account? \n" ); document.write( "
Algebra.Com's Answer #14128 by stanbon(75887)\"\" \"About 
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Let amount invested at 6.5% be \"x\"
\n" ); document.write( "The amount invested at 8% is \"14000-x\"
\n" ); document.write( "Interest on 6.5% money is 6.5%x= 0.065x
\n" ); document.write( "Interest on 8% money is 8%(14000-x)=0.08(14000)-0.08x=1120-0.08x
\n" ); document.write( "EQUATION:
\n" ); document.write( "Int. + Int. = $1024
\n" ); document.write( "0.065x+1120-0.08x=1024
\n" ); document.write( "-.015x= -96
\n" ); document.write( " x= 6400 dollars (amount invested at 6.5%
\n" ); document.write( "14000-6400=7600 (amount invested at 8%\r
\n" ); document.write( "\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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