Cole's two student loans totaled $31000. One of his loans was at 2.8% simple interest and the other at 4.5%. After one year, Cole owed $1024.40 in interest. What was the amount of each loan?
First loan : x
second loan : 31000-x
.028x + .045(31000-x) = 1024.40
solve for x
I would first multiply by 1000 to get rid of the decimals,
take it from there
15,000
To solve this problem, let's assume the amount of the loan with a 2.8% interest rate is x, and the amount of the loan with a 4.5% interest rate is y.
We are given two pieces of information. First, the total amount of the loans is $31000, so we can write the equation: x + y = 31000.
Second, we are told that after one year, Cole owed $1024.40 in interest. The interest on the loan with a 2.8% interest rate is calculated as 2.8% of x, which is 0.028x. Similarly, the interest on the loan with a 4.5% interest rate is calculated as 4.5% of y, which is 0.045y. The sum of these two interests is $1024.40, so we can write another equation: 0.028x + 0.045y = 1024.40.
Now we have a system of two equations with two variables. We can solve this system to find the values of x and y.
To solve the system, we can use the substitution method or the elimination method. Since the first equation, x + y = 31000, is already solved for x, we can substitute this value into the second equation:
0.028x + 0.045y = 1024.40
0.028(31000 - y) + 0.045y = 1024.40
Multiplying out the parentheses, we get:
868 - 0.028y + 0.045y = 1024.40
Combining like terms, we have:
0.045y - 0.028y = 1024.40 - 868
0.017y = 156.40
To solve for y, we divide both sides of the equation by 0.017:
y = 156.40 / 0.017
y ≈ 9200
Now we can substitute this value of y back into the first equation to solve for x:
x + 9200 = 31000
x = 31000 - 9200
x ≈ 21800
Therefore, the amount of the loan with a 2.8% interest rate is approximately $21,800, and the amount of the loan with a 4.5% interest rate is approximately $9,200.