A total of $12,000 is invested in two funds paying 9% and 11% simple interest. If the yearly interest is $1,180, how much of the $12,000 is invested at each rate?
There seems to be a decimal in this problem if so how do I get rid of it verifying x,y?
If x is invested at 9%, 12000-x is invested at 11%. So, adding up the interest,
.09x + .11(12000-x) = 1180
x = 7000
To solve this problem, we can use a system of equations. Let's assume that x dollars is invested at 9% interest rate, and y dollars is invested at 11% interest rate.
The first equation can be made by adding the two investments together: x + y = 12,000 (since the total amount invested is $12,000).
The second equation can be formed by calculating the interest earned from each investment and adding them together: 0.09x + 0.11y = 1,180 (since the 9% investment yields 9% of x dollars, and the 11% investment yields 11% of y dollars, which give a total interest of $1,180).
To solve this system of equations, we can use either substitution or elimination. In this case, we'll use substitution to solve for one variable in terms of the other.
From the first equation, we can isolate x: x = 12,000 - y.
Substituting this expression for x into the second equation, we have: 0.09(12,000 - y) + 0.11y = 1,180.
Expanding the multiplication and simplifying the equation, we get: 1,080 - 0.09y + 0.11y = 1,180.
Combining like terms, we have: 0.02y = 100.
Dividing both sides of the equation by 0.02, we find: y = 5,000.
Now that we know the value of y, we can substitute it back into the first equation to find x: x + 5,000 = 12,000.
Subtracting 5,000 from both sides of the equation, we get: x = 7,000.
Therefore, $7,000 is invested at 9% interest rate, and $5,000 is invested at 11% interest rate.