Martha's $18,750 is invested in two accounts, one earning 12% interest and the other earning 10%. After 1 year, her combined interest income is $2,117. How much has she invested at each rate?
let $ be the amount at 12 percent
$*0.12+ (18750-$)*0.10=2117
solve for $
$*.02=2117-1875
$=12,100 check that, so then the other is 18750-12,100=6,650
If x is invested at 12%, then the rest (18750-x) is at 10%
So, just add up the interest:
0.12x + 0.10(18750-x) = 2117
To find out how much Martha has invested at each rate, let's use a system of equations.
Let's assume Martha has invested an amount x at 12% interest and an amount y at 10% interest.
According to the problem, the total interest earned after 1 year is $2,117. This can be expressed as the sum of interest earned from each account:
0.12x + 0.10y = 2,117 ...(equation 1)
We also know that the total amount invested is $18,750. So, the sum of the amounts invested in each account should be equal to this total:
x + y = 18,750 ...(equation 2)
We now have a system of two equations with two variables: x and y.
To solve this system of equations, we can use the substitution method or the elimination method. In this case, the elimination method is more convenient.
Let's multiply both sides of equation 2 by 0.10 to make the coefficients of y in both equations the same:
0.10(x + y) = 0.10(18,750)
0.10x + 0.10y = 1,875 ...(equation 3)
Now, subtract equation 3 from equation 1 to eliminate y:
(0.12x + 0.10y) - (0.10x + 0.10y) = 2,117 - 1,875
0.12x - 0.10x = 242
0.02x = 242
x = 242 / 0.02
x = 12,100
Now, substitute the value of x back into equation 2 to find y:
12,100 + y = 18,750
y = 18,750 - 12,100
y = 6,650
Therefore, Martha has invested $12,100 at a 12% interest rate and $6,650 at a 10% interest rate.