document.write( "Question 906683: How can $60,000 be invested, part at 11% annual simple interest and the remainder at 9% annual simple interest, so that the interest earned by the two accounts will be equal?
\n" ); document.write( "The amount to invest at 11% is? $ \n" ); document.write( "

Algebra.Com's Answer #549999 by richwmiller(17219) $\"\"$ $\"About$

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We know the total amount of money invested. $60000
\n" ); document.write( "x+y=60000,
\n" ); document.write( "We know that the difference in interest earned by the two accounts is $0
\n" ); document.write( "0.11*x-0.09*y=0
\n" ); document.write( "x=60000-y
\n" ); document.write( "We substitute for x
\n" ); document.write( "0.11*(60000-y)-0.09*y=0
\n" ); document.write( "We multiply out
\n" ); document.write( "6600-0.11y-0.09*y=0
\n" ); document.write( "We combine like terms.
\n" ); document.write( "6600=0.2*y
\n" ); document.write( "Isolate y
\n" ); document.write( "y=6600/0.2
\n" ); document.write( "y=33000 at 9%
\n" ); document.write( "Calculate x
\n" ); document.write( "x=60000-33000
\n" ); document.write( "x=27000 at 11%
\n" ); document.write( "Interest earned at 11% is 2970
\n" ); document.write( "Interest earned at 9% is 2970
\n" ); document.write( "We check
\n" ); document.write( "0.11*27000-0.09*33000=0
\n" ); document.write( "2970-2970=0
\n" ); document.write( "0=0
\n" ); document.write( "Since this statement is TRUE and neither amount is negative then all is well
\n" ); document.write( " \n" ); document.write( "

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