document.write( "Question 731017: If $2000 is invested at 8% interest, how much must be invested at 11% interest so that the total return for both averages are 10%? \n" ); document.write( "
Algebra.Com's Answer #446952 by mananth(16946)\"\" \"About 
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2000------------------------8%\r
\n" ); document.write( "\n" ); document.write( "x---------------------------11%\r
\n" ); document.write( "\n" ); document.write( "(x+2000)-------------------10%\r
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\n" ); document.write( "\n" ); document.write( "2000*8%+11%x= (x+2000)*10%\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "multiply by 100
\n" ); document.write( "8*2000+11x+10(x+2000)\r
\n" ); document.write( "\n" ); document.write( "16000+11x=10x+20000\r
\n" ); document.write( "\n" ); document.write( "11x-10x= 20000-16000
\n" ); document.write( "x=4000
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