Solve the following word problem.
Money is invested at two rates of interest. One rate is 8%
and the other is 2%
. If there is $600
more invested at 8%
than at 2%
, find the amount invested at each rate if the total annual interest received is $470
. Let x=
amount invested at 8%
and y=
amount invested at 2%
. Then the system that models the problem is {x=y+600 0.08x+0.02y=470
. Solve the system by using the method of addition.
To solve the system of equations, we can use the method of addition.
First, we can rewrite the system of equations in standard form:
x - y = 600
0.08x + 0.02y = 470
To eliminate the variable y, we can multiply the first equation by 0.02 and the second equation by -1:
0.02(x - y) = 0.02(600)
-0.08x - 0.02y = -470
Simplifying these equations, we have:
0.02x - 0.02y = 12
-0.08x - 0.02y = -470
Next, we can add the two equations together:
0.02x - 0.08x - 0.02y - 0.02y = 12 - 470
-0.06x = -458
Dividing both sides by -0.06, we get:
x = 458 / 0.06
x = $7633.33
Substituting this value of x back into the first equation, we can solve for y:
7633.33 - y = 600
y = 7633.33 - 600
y = $7033.33
Therefore, the amount invested at 8% is $7633.33 and the amount invested at 2% is $7033.33.