A total of $7000 is invested in three accounts. Some was invested at 6%, some at 8%, and the rest at 9%. find the amount invested at each rate if the total interest for one year was $555 and the amount invested at 8% was three times the amount invested at 9%. (assume simple interest). show what each variable stands for, show each equation, show how you solved it and indicate the solution.
555 = .09 x + .08 (3x) + .06 (7000 - 4x)
Let's denote the amount invested at 6%, 8%, and 9% as follows:
Let x = amount invested at 6%
Let y = amount invested at 8%
Let z = amount invested at 9%
We are given the following information:
1) The total amount invested is $7000:
x + y + z = 7000
2) The total interest earned is $555:
0.06x + 0.08y + 0.09z = 555
3) The amount invested at 8% is three times the amount invested at 9%:
y = 3z
To solve the system of equations, we will use the substitution method.
Step 1: Solve equation 3) for y in terms of z:
y = 3z
Step 2: Substitute y = 3z into equation 1):
x + 3z + z = 7000
x + 4z = 7000 ----(Equation 4)
Step 3: Substitute y = 3z into equation 2):
0.06x + 0.08(3z) + 0.09z = 555
0.06x + 0.24z + 0.09z = 555
0.06x + 0.33z = 555 ----(Equation 5)
Step 4: Solve equations 4) and 5) simultaneously.
Multiply equation 4) by -0.06 and equation 5) by 0.06 to eliminate x:
-0.06(x + 4z) = -0.06(7000)
-0.06x - 0.24z = -420 ----(Equation 6)
0.06(0.06x + 0.33z) = 0.06(555)
0.06x + 0.0198z = 33.3 ----(Equation 7)
Adding equations 6) and 7):
-0.24z + 0.0198z = -420 + 33.3
-0.2202z = -386.7
z = -386.7 / -0.2202
z ≈ 1752.08
Step 5: Substitute z = 1752.08 into equation 3) to find y:
y = 3z
y ≈ 3 * 1752.08
y ≈ 5256.24
Step 6: Substitute z = 1752.08 and y ≈ 5256.24 into equation 1) to find x:
x + y + z = 7000
x + 5256.24 + 1752.08 = 7000
x + 7008.32 = 7000
x ≈ -8.32
The solution is approximately:
x ≈ -8.32
y ≈ 5256.24
z ≈ 1752.08
Since the amounts invested cannot be negative, we can conclude that the amount invested at each rate is approximately:
x ≈ $0
y ≈ $5256.24
z ≈ $1752.08
Therefore, approximately $0 is invested at 6%, approximately $5256.24 is invested at 8%, and approximately $1752.08 is invested at 9%.
To solve this problem, we need to set up a system of equations based on the given information. Let's define the variables:
Let x be the amount invested at 6%.
Let y be the amount invested at 8%.
Let z be the amount invested at 9%.
According to the problem, the total investment is $7000, so our first equation is:
x + y + z = 7000 (Equation 1)
The total interest earned from these investments is $555, and the interest on each investment can be calculated as follows:
Interest on x at 6%: 0.06x
Interest on y at 8%: 0.08y
Interest on z at 9%: 0.09z
The second equation is:
0.06x + 0.08y + 0.09z = 555 (Equation 2)
The amount invested at 8% is three times the amount invested at 9%. This can be written as:
y = 3z (Equation 3)
Now, we have a system of three equations (Equations 1, 2, and 3) in three variables (x, y, and z). Let's solve it.
First, let's substitute Equation 3 into Equation 1:
x + (3z) + z = 7000
x + 4z = 7000 (Equation 4)
Now, let's substitute Equation 3 into Equation 2:
0.06x + 0.08(3z) + 0.09z = 555
0.06x + 0.24z + 0.09z = 555
0.06x + 0.33z = 555 (Equation 5)
We now have a system of two equations (Equations 4 and 5) in two variables (x and z). Let's solve it using elimination.
Multiply Equation 4 by 0.06 to make the coefficients of x in both equations equal:
0.06(x + 4z) = 0.06(7000)
0.06x + 0.24z = 420 (Equation 6)
Now, subtract Equation 6 from Equation 5:
0.06x + 0.33z - (0.06x + 0.24z) = 555 - 420
0.33z - 0.24z = 135
0.09z = 135
z = 135 / 0.09
z = 1500
Now, substitute the value of z back into Equation 4 to solve for x:
x + 4z = 7000
x + 4(1500) = 7000
x = 7000 - 6000
x = 1000
Finally, substitute the values of x and z into Equation 3 to solve for y:
y = 3z
y = 3(1500)
y = 4500
So, the amount invested at each rate is:
$1000 at 6%, $4500 at 8%, and $1500 at 9%.