document.write( "Question 897910: Money is invested at two rates of interest. One rate is 8% and the other is 2%. If there is $800 more invested at 8% than at 2%, find the amount invested at each rate if the total annual interest received is $490. Let x = amount invested at 8% and y = amount invested at 2%. Then the system that models the problem is {x 0.08x + 0.02y == y + 800 490. Solve the system by using the method of addition. \n" ); document.write( "
Algebra.Com's Answer #544454 by lwsshak3(11628)\"\" \"About 
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Money is invested at two rates of interest. One rate is 8% and the other is 2%. If there is $800 more invested at 8% than at 2%, find the amount invested at each rate if the total annual interest received is $490. Let x = amount invested at 8% and y = amount invested at 2%. Then the system that models the problem is {x 0.08x + 0.02y == y + 800 490. Solve the system by using the method of addition.
\n" ); document.write( "***
\n" ); document.write( "Let x=amount invested at 8%
\n" ); document.write( "Let y=amount invested at 2%
\n" ); document.write( "..
\n" ); document.write( ".08x+.02y=490
\n" ); document.write( "x-y=800
\n" ); document.write( "..
\n" ); document.write( "8x+2y=49000 (mult. by 10)
\n" ); document.write( "-8x+8y=-6400 (mult. by -8)
\n" ); document.write( "add
\n" ); document.write( "10y=42600
\n" ); document.write( "y=4260
\n" ); document.write( "x=800+y=5060
\n" ); document.write( "amount invested at 8%=$5060
\n" ); document.write( "amount invested at 2%=$4260\r
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "
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