SOLUTION: 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 tota
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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. Answer by lwsshak3(11628) (Show Source):
You can put this solution on YOUR website! 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.
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Let x=amount invested at 8%
Let y=amount invested at 2%
..
.08x+.02y=490
x-y=800
..
8x+2y=49000 (mult. by 10)
-8x+8y=-6400 (mult. by -8)
add
10y=42600
y=4260
x=800+y=5060
amount invested at 8%=$5060
amount invested at 2%=$4260
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