You invested $25,000 into two accounts: one that has simple 3% interest, the other with 2% interest. If your total interest payment after one year was $600, how much was in each account after the year passed?
If x is at 3%, then
.03x + .02(25000-x) = 600
Having found x, then the two accounts hold
1.03x and 1.02(25000-x)
To solve this problem, we need to set up a system of equations based on the information provided.
Let's say the amount of money invested in the account with 3% interest is x (in dollars), and the amount invested in the account with 2% interest is y (also in dollars).
According to the problem, the total amount invested is $25,000. Therefore, we have the equation:
x + y = $25,000 ---(Equation 1)
Now, let's calculate the interest earned from each account:
The account with 3% interest will earn interest equal to 3% of the amount invested, which is 0.03x.
The account with 2% interest will earn interest equal to 2% of the amount invested, which is 0.02y.
The total interest earned is given as $600. So, we have the equation:
0.03x + 0.02y = $600 ---(Equation 2)
To solve this system of equations, we can use the method of substitution or elimination.
Let's solve it using the substitution method:
From Equation 1, we can express x in terms of y as:
x = $25,000 - y
Now substitute this value of x in Equation 2:
0.03($25,000 - y) + 0.02y = $600
750 - 0.03y + 0.02y = $600
-0.01y = $600 - $750
-0.01y = -$150
y = -$150 / -0.01
y = $15,000
Now, substitute the value of y in Equation 1 to find x:
x + $15,000 = $25,000
x = $25,000 - $15,000
x = $10,000
So, after one year, $10,000 was in the account with 3% interest, and $15,000 was in the account with 2% interest.