How can $12,600 be split between two investments, one paying 6% annually and one paying 8% annually so that the interest from the two accounts equals $864
If x is at 6%, then the rest (12600-x) is at 8%. So, add up the interest:
.06x + .08(12600-x) = 864.00
To solve this problem, we can set up a system of linear equations. Let's denote the amount of money invested at 6% by x and the amount invested at 8% by y.
The first equation represents the total amount invested:
x + y = $12,600
The second equation represents the total interest earned:
0.06x + 0.08y = $864
To solve this system of equations, we can use substitution or elimination method. Let's use the elimination method:
Multiply the first equation by 0.06:
0.06(x + y) = 0.06($12,600)
0.06x + 0.06y = $756
Now subtract this equation from the second equation:
(0.06x + 0.08y) - (0.06x + 0.06y) = $864 - $756
0.02y = $108
Divide both sides of the equation by 0.02:
y = $108 / 0.02
y = $5,400
Now substitute the value of y back into the first equation and solve for x:
x + $5,400 = $12,600
x = $12,600 - $5,400
x = $7,200
Therefore, $7,200 should be invested at 6% and $5,400 should be invested at 8% in order to earn a total of $864 in interest.