a total of 10,000 is invested in two mutual funds. the first account yields 5% and the second account yields 6%. how much is invested in each account if the interest in a year in $575?
$X invested in 1st acc.
(10,000-x) invested in 2nd acc.
I = I1 + I2 = X*r1*t + (10,000-x)*r2*t = 575.
X*.05*1 + (10,000-x)*.06*1 = 575.
X = ?.
To solve this problem, we can set up a system of equations. Let's use variables to represent the amounts invested in each account. Let x represent the amount invested in the first account (yielding 5%) and y represent the amount invested in the second account (yielding 6%).
According to the problem, the total amount invested is $10,000. Therefore, our first equation is:
x + y = 10,000
The total interest earned in a year is $575. Since the interest is calculated by multiplying the amount invested by the interest rate, we can write the second equation as follows:
0.05x + 0.06y = 575
Now we have a system of two equations:
x + y = 10,000 ----(1)
0.05x + 0.06y = 575 ----(2)
To solve this system, we can use any method of solving linear equations. In this case, let's solve the system using the substitution method:
From equation (1), we have x + y = 10,000.
Solving for x, we get x = 10,000 - y.
Now, substitute x into equation (2):
0.05(10,000 - y) + 0.06y = 575
Expand and simplify:
500 - 0.05y + 0.06y = 575
0.01y = 75
y = 75 / 0.01
y = 7,500
Now, plug the value of y back into equation (1) to find x:
x + 7,500 = 10,000
x = 10,000 - 7,500
x = 2,500
Therefore, $2,500 is invested in the first account (yielding 5%), and $7,500 is invested in the second account (yielding 6%).