document.write( "Question 928376: An investor decides to invest some cash in the account paying 14% annual interest and to put the rest in a stock fund the ends up earning 10% over the course of a year. The investor puts $1500 more in the first account then in the stock fund, and at the end of the year finds the total interest from the 2 investments was $1020. how much money was invested at each of the two rates? \n" ); document.write( "
Algebra.Com's Answer #563587 by richwmiller(17219)\"\" \"About 
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We don't know the total invested.
\n" ); document.write( "We know the difference in the two accounts.
\n" ); document.write( "We know that the interest for the two accounts is $1020
\n" ); document.write( "0.14*x+0.1*y=1020
\n" ); document.write( "We know that the account at 14% has $1500 more.
\n" ); document.write( "x=1500+y
\n" ); document.write( "We substitute for x
\n" ); document.write( "0.14*(1500+y)+0.1*y=1020
\n" ); document.write( "We multiply out
\n" ); document.write( "210+0.14y+0.1*y=1020
\n" ); document.write( "We combine like terms.
\n" ); document.write( "0.24*y=810
\n" ); document.write( "Isolate y
\n" ); document.write( "y=$3375 at 10%
\n" ); document.write( "x=1500+y
\n" ); document.write( "Calculate x
\n" ); document.write( "x=$4875 at 14%
\n" ); document.write( "Now,we know the total invested is: 8250
\n" ); document.write( "Total invested $4875+$3375=$8250
\n" ); document.write( "We check
\n" ); document.write( "0.14*4875+0.1*3375=1020
\n" ); document.write( "682.5+337.5=1020
\n" ); document.write( "1020=1020
\n" ); document.write( "Since this statement is TRUE and neither x nor y is negative all is well.
\n" ); document.write( "codeintmt
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