an investment of $50,000 was made by a busines club. the investment was split into three parts and lasted for one year.the first part of the investment earned 8% interest, the second 6%, and the third 9%. total interest from the investments was $3960. the interest from the first investment was two time the interest from the second. find the amounts of the three parts of the investment.
first part --- x
2nd part --- y
3rd part ---- 50000-x-y
interest on 1st = .08x
interest on 2nd = .06y
3rd interest = .09(50000-x-y)
but .08x = 2(.06)y
8x = 12y
y = 8x/12 = 4x/3
.08x + .06y + .09(50000-x-y) = 3960
sub in y = 4x/3 and solve
I am sure you can take it from here
To solve this problem, we can set up a system of equations based on the given information.
Let's represent the three parts of the investment as follows:
- First part: x dollars
- Second part: y dollars
- Third part: z dollars
According to the problem, the total investment amount is $50,000, so we can create our first equation:
x + y + z = 50000 ---(equation 1)
Next, we are given that the total interest earned from the three investments is $3960. We can calculate the interest earned from each part using their respective interest rates:
Interest from the first part (8%): 0.08x
Interest from the second part (6%): 0.06y
Interest from the third part (9%): 0.09z
The problem states that the interest from the first part was two times the interest from the second part, so we can write our second equation as:
0.08x = 2 * 0.06y ---(equation 2)
Lastly, we can create our third equation using the total interest earned:
0.08x + 0.06y + 0.09z = 3960 ---(equation 3)
Now we have a system of three equations (equations 1, 2, and 3) that we can solve simultaneously to find the values of x, y, and z.
To solve this system of equations, we can use substitution, elimination, or matrix methods. I will use the elimination method to solve the system.
First, we will multiply equation 2 by 0.08 to make the coefficients of x in equation 2 and equation 3 the same:
0.008x = 2 * (0.08 * 0.06)y
0.008x = 2 * 0.0048y
0.008x = 0.0096y ---(equation 4)
Now, let's subtract equation 4 from equation 3 to eliminate x:
0.08x + 0.06y + 0.09z = 3960
0.008x - 0.0096y = 0
Multiplying equation 4 by 10 to make the coefficients of x integers:
0.08x = 0.096y ---(equation 5)
- Subtracting equation 5 from equation 3:
0.08x + 0.06y + 0.09z = 3960
-0.08x + 0.096y = 0
Adding these two equations eliminates x:
0.156y + 0.09z = 3960 ---(equation 6)
Now we have two equations with two variables: equation 1 and equation 6.
Substituting equation 1 into equation 6:
(50000 - y - z) + 0.156y + 0.09z = 3960
50000 + 0.066y - 0.01z = 3960
0.066y - 0.01z = 3960 - 50000
0.066y - 0.01z = -46040 ---(equation 7)
Now we have two linear equations with two variables: equation 6 and equation 7.
Solving equations 6 and 7 simultaneously will give us the values of y and z.