document.write( "Question 904176: Two investments totaling $38,500 produce an annual income of $2805. One investment yields 3% per year, while the other yields 9% per year. How much is invested at each rate? \n" ); document.write( "

Algebra.Com's Answer #548555 by richwmiller(17219) $\"\"$ $\"About$

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We know the total amount of money invested. $38500
\n" ); document.write( "x+y=38500,
\n" ); document.write( "We know that the total interest for the year for the two accounts is $2805
\n" ); document.write( "0.03*x+0.09*y=2805
\n" ); document.write( "x=38500-y
\n" ); document.write( "We substitute for x
\n" ); document.write( "0.03*(38500-y)+0.09*y=2805
\n" ); document.write( "We multiply out
\n" ); document.write( "1155-0.03y+0.09*y=2805
\n" ); document.write( "We combine like terms.
\n" ); document.write( "0.06*y=1650
\n" ); document.write( "Isolate y
\n" ); document.write( "y=27500 at 9%
\n" ); document.write( "x=38500-y
\n" ); document.write( "Calculate x
\n" ); document.write( "x=11000 at 3%
\n" ); document.write( "We check
\n" ); document.write( "0.03*11000+0.09*27500=2805
\n" ); document.write( "330+2475=2805
\n" ); document.write( "2805=2805
\n" ); document.write( "Since this statement is TRUE and neither amount is negative then it is ok
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