document.write( "Question 911700: You have already invested $300 in a stock with an annual return of 11%. How much of an additional $1,300 should be invested at 12% and how much at 6% so that the total return on the entire $1,600 is 9%? (Round each answer to the nearest cent.) \n" ); document.write( "
Algebra.Com's Answer #553287 by JoelSchwartz(130)\"\" \"About 
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x=amount invested at twelve percent
\n" ); document.write( "y=is equal to the amount at six percent.
\n" ); document.write( "I know that the amount at twelve percent plus the amount at six percent is equal to 1300.
\n" ); document.write( "x+y=1300
\n" ); document.write( "We also know that three hundred dollars is invested plus thirteen hundred dollars which equals 1600 dollars. The interest rate times sixteen hundred is the total interest of the investment.
\n" ); document.write( ".09*1600=144
\n" ); document.write( "I know that three hundred times eleven percent is equal to 33
\n" ); document.write( ".11*300=33
\n" ); document.write( "33+.12x+.06y=144
\n" ); document.write( ".12x=144-33-.06y
\n" ); document.write( ".12x=111-.06y
\n" ); document.write( "x=925-1/2y
\n" ); document.write( "x+y=1300
\n" ); document.write( "925-1/2y+y=1300
\n" ); document.write( "925+1/2y=1300
\n" ); document.write( "1/2y=375
\n" ); document.write( "y=750
\n" ); document.write( "x=925-1/2*750
\n" ); document.write( "x=925-375
\n" ); document.write( "x=550
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