Solve the following word problem.

Question

Money is invested at two rates of interest. One rate is 7%
and the other is 2%
. If there is $1400
more invested at 7%
than at 2%
, find the amount invested at each rate if the total annual interest received is $440
. Let x=
amount invested at 7%
and y=
amount invested at 2%
. Then the system that models the problem is {x=y+1400 0.07x+0.02y=440
. Solve the system by using the method of addition.

Answer 1

To solve the system of equations by the method of addition, we can add the two equations together to eliminate one variable.

x = y + 1400
0.07x + 0.02y = 440

First, we can multiply the second equation by 100 to get rid of the decimal points:

100(0.07x + 0.02y) = 100(440)
7x + 2y = 44000

Now we can add the two equations together:

(x) + (7x + 2y) = (y + 1400) + (44000)
8x + 2y = y + 45400

Next, we can simplify the equation by combining like terms:

8x + 2y - y = 45400
8x + y = 45400

Now we have eliminated one variable and can solve for x.

To do this, we can subtract y from both sides of the equation:

8x = 45400 - y

Now we can substitute the expression for x from the first equation into the second equation:

x = (45400 - y) / 8

Substituting this expression for x into the first equation gives:

(45400 - y) / 8 = y + 1400

Next, we can multiply both sides of the equation by 8 to clear the fraction:

45400 - y = 8(y + 1400)

Expanding the equation gives:

45400 - y = 8y + 11200

Now we can combine like terms:

45400 - 11200 = 8y + y

34200 = 9y

Next, we can divide both sides of the equation by 9 to solve for y:

y = 34200 / 9
y = 3800

Now, substituting this value of y back into the expression for x:

x = y + 1400
x = 3800 + 1400
x = 5200

Therefore, the amount invested at 7% is $5200, and the amount invested at 2% is $3800.