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 by addition, we want to eliminate one variable by adding the equations together. Let's start by multiplying the first equation by -0.02 to make the coefficients of y in both equations the same.
-0.02(x) + (-0.02)(y) = (-0.02)(y + 600)
-0.02x - 0.02y = -0.02y - 12
Now we can add the two equations together:
0.08x + 0.02y + (-0.02x) + (-0.02y) = 470 + (-12)
0.06x = 458
Divide both sides of the equation by 0.06:
x = 458 / 0.06
x = 7633.33
Now we can substitute this value of x back into the first equation to solve for y:
y = x - 600
y = 7633.33 - 600
y = 7033.33
Therefore, the amount invested at the rate of 8% is $7633.33, and the amount invested at the rate of 2% is $7033.33.