an investment of $6600.00 is divided between two simple interest accounts. On one account the annual simple interest rate is 9% and on the second account the annual simple interest is 5.5% How much should be invested in each account so that the total interest from the two accounts is $468? (x at 9% and y at 5.5%)
first equation:
x+y = 6600 or y = 6600-x
.09x + .055y = 468
how about multiplying each term by 200
18x + 11y = 93600
use substitution:
18x + 11(6600-x) = 93600
carry on ....
To solve this problem, let's set up a system of equations based on the information given.
Let's say x represents the amount invested at 9%, and y represents the amount invested at 5.5%.
We know that the total investment is $6600. So, the sum of x and y is equal to $6600:
x + y = 6600 ----(equation 1)
We also know that the total interest earned from both accounts is $468. The interest earned from the first account at 9% is calculated by multiplying the amount invested (x) by the interest rate (9%). The interest earned from the second account at 5.5% is calculated by multiplying the amount invested (y) by the interest rate (5.5%). Adding these two amounts gives us the total interest of $468:
0.09x + 0.055y = 468 ----(equation 2)
Now, we can solve the system of equations by substitution or elimination method. Let's use the elimination method:
Multiply equation 1 by 0.055 to match the coefficients of y:
0.055x + 0.055y = 0.055 * 6600
0.055x + 0.055y = 363
Now, subtract this equation from equation 2:
0.09x + 0.055y - (0.055x + 0.055y) = 468 - 363
0.09x - 0.055x = 105
0.035x = 105
Next, solve for x:
x = 105 / 0.035
x = 3000
Now, substitute the value of x back into equation 1 to find y:
3000 + y = 6600
y = 6600 - 3000
y = 3600
Therefore, $3000 should be invested at 9% and $3600 should be invested at 5.5% in order to get a total interest of $468 from the two accounts.