document.write( "Question 1158346: Money is invested at two rates of interest. One rate is 6% and the other is
\n" ); document.write( "2%.If there is $1200 more invested at 6% than at 2%, find the amount invested at each rate if the total annual interest received is $890.Let x=amount invested at
\n" ); document.write( "6% and y= amount invested at 2%.Then the system that models the problem is
\n" ); document.write( "{x=y+1200
\n" ); document.write( "0.06x+0.02y=890.Solve the system by using the method of addition.
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Algebra.Com's Answer #781270 by josmiceli(19441)\"\" \"About 
You can put this solution on YOUR website!
(1) \"+x+-+y+=+1200+\"
\n" ); document.write( "(2) \"+.06x+%2B+.02y+=+890+\"
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\n" ); document.write( "(2) \"+6x+%2B+2y+=+89000+\"
\n" ); document.write( "(2) \"+3x+%2B+y+=+44500+\"
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\n" ); document.write( "Add (1) and (2)
\n" ); document.write( "(1) \"+x+-+y+=+1200+\"
\n" ); document.write( "(2) \"+3x+%2B+y+=+44500+\"
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\n" ); document.write( "\"+4x+=+45700+\"
\n" ); document.write( "\"+x+=+11425+\"
\n" ); document.write( "and
\n" ); document.write( "(1) \"+x+-+y+=+1200+\"
\n" ); document.write( "(1) \"+11425+-+y+=+1200+\"
\n" ); document.write( "(1) \"+y+=+10225+\"
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