document.write( "Question 638642: An investor has part of her money in an account that pays 9% annual interest, and the rest in an account that pays 11% annual interest. If she has $8000 less in the higher paying account than in the lower paying one and her total annual interest income is $2010, how much does she have in each account? \n" ); document.write( "
Algebra.Com's Answer #402324 by mananth(16946)\"\" \"About 
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9%--- x\r
\n" ); document.write( "\n" ); document.write( "11%---- x-8000\r
\n" ); document.write( "\n" ); document.write( "Interest = 2010\r
\n" ); document.write( "\n" ); document.write( "9%*x+11%(x-8000)= 2010\r
\n" ); document.write( "\n" ); document.write( "multiply by 100\r
\n" ); document.write( "\n" ); document.write( "9x+11(x-8000)= 201000\r
\n" ); document.write( "\n" ); document.write( "9x+11x-88000=201000\r
\n" ); document.write( "\n" ); document.write( "20x= 201000+88000\r
\n" ); document.write( "\n" ); document.write( "x= (201000+88000)/20\r
\n" ); document.write( "\n" ); document.write( "x=14450\r
\n" ); document.write( "\n" ); document.write( "Investment at 9%= $14,450
\n" ); document.write( "Investment at 11%= $6,450\r
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