Question

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.

Answer 1

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.