Solve the following word problem.
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.
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.