document.write( "Question 903434: a total of 4000 is invested:part at 11% the other at 13% . How much is invested at each if there is an annual rate of 450 \n" ); document.write( "

Algebra.Com's Answer #548010 by richwmiller(17219) $\"\"$ $\"About$

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We know the total amount of money invested. $4000
\n" ); document.write( "x+y=4000,
\n" ); document.write( "We know that the total interest for the year for the two accounts is $450
\n" ); document.write( "0.11*x+0.13*y=450
\n" ); document.write( "x=4000-y
\n" ); document.write( "We substitute for x
\n" ); document.write( "0.11*(4000-y)+0.13*y=450
\n" ); document.write( "We multiply out
\n" ); document.write( "440-0.11y+0.13*y=450
\n" ); document.write( "We combine like terms.
\n" ); document.write( "0.02*y=10
\n" ); document.write( "Isolate y
\n" ); document.write( "y=500 at 13%
\n" ); document.write( "x=4000-y
\n" ); document.write( "Calculate x
\n" ); document.write( "x=3500 at 11%
\n" ); document.write( "We check
\n" ); document.write( "0.11*3500+0.13*500=450
\n" ); document.write( "385+65=450
\n" ); document.write( "450=450
\n" ); document.write( "Since this statement is TRUE and neither amount is negative then it is ok
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