Steve has $24,000 invested in two accounts. One earns 6% annual interest and the other earns 7%. After 1 year, his combined interest is $1635. How much was invested in each account?
.06 x + .07 (24,000 -x) = 1635
-.01 x + 1680 =1635
-.01 x = - 45
x = 4500
24000 - x = 19,500
To solve this problem, we can set up a system of equations based on the information given.
Let's say Steve invested x dollars in the account earning 6% interest and y dollars in the account earning 7% interest.
The interest earned in the account earning 6% interest can be calculated by multiplying the principal amount (x) by the interest rate (6%) and converting it to a decimal (0.06). So the interest earned in that account would be 0.06x.
Similarly, the interest earned in the account earning 7% interest would be 0.07y.
According to the problem, after one year, the combined interest earned in both accounts is $1635. Therefore, we can write the equation:
0.06x + 0.07y = 1635 ----(Equation 1)
Steve has $24,000 invested in total, so the sum of the amounts invested in both accounts should be equal to $24,000. This gives us the second equation:
x + y = 24000 ----(Equation 2)
Now we have a system of two equations with two variables. We can solve this system using substitution or elimination method.
Let's solve this system of equations using the substitution method. Solve Equation 2 for x in terms of y:
x = 24000 - y
Substitute this value of x into Equation 1:
0.06(24000 - y) + 0.07y = 1635
Now, simplify and solve for y:
1440 - 0.06y + 0.07y = 1635
0.01y = 195
y = 195 / 0.01
y = 19500
Now substitute this value of y back into Equation 2 to find x:
x + 19500 = 24000
x = 24000 - 19500
x = 4500
So Steve invested $4,500 in the account earning 6% interest and $19,500 in the account earning 7% interest.