solving applied problems two equations (Coles two student loans totaled $31,000. 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?)
loan at 2.8% --- $ x
loan at 4.5 % ---- $ 31000-x
solve for x ....
.028x + .045(31000-x = 1024.40
times 1000
28x + 45(31000-x) = 1024400
etc
To solve this problem, we can set up a system of equations based on the given information. Let's assign variables to the unknowns:
Let's assume the amount of the loan with 2.8% interest is "x" (in dollars).
Let's assume the amount of the loan with 4.5% interest is "y" (in dollars).
Now, we can set up the following equations based on the given information:
Equation 1: x + y = 31,000 (since the total amount of the two loans is $31,000)
Equation 2: (2.8/100) * x + (4.5/100) * y = 1024.40 (since the interest on the first loan is calculated as a simple interest rate of 2.8% on "x" dollars, and the interest on the second loan is calculated as a simple interest rate of 4.5% on "y" dollars, which adds up to $1024.40 after one year)
To solve this system of equations, we can use the substitution or elimination method. Let's solve it using the substitution method:
From Equation 1, we can isolate one variable. Let's solve for "x":
x = 31,000 - y
Now, let's substitute this value of "x" into Equation 2:
(2.8/100) * (31,000 - y) + (4.5/100) * y = 1024.40
Simplifying the equation:
(0.028 * 31,000 - 0.028 * y) + (0.045 * y) = 1024.40
(868 - 0.028y) + 0.045y = 1024.40
Combine like terms:
868 + 0.017y = 1024.40
0.017y = 1024.40 - 868
0.017y = 156.40
Divide by 0.017:
y = 156.40 / 0.017
y ≈ $9,200
Now, substitute this value of "y" into Equation 1 to find the value of "x":
x + 9,200 = 31,000
Subtract 9,200 from both sides:
x = 31,000 - 9,200
x ≈ $21,800
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.