Question

Samantha took out two loans totaling $6,000 to pay her first year of college. She borrowed the maximum amount she could at 3.5% simple annual interest and the remainder at 7% simple annual interest. At the end of the first year, she owed $259 in interest. How much was borrowed at each rate?

Answer 1

Our simple interest formula is I = prt, where I is the amount of interest, p is the amount of principal, r is the percentage written as a decimal, and t is the amount of time (in this case in years).  We will define our variable x as the amount borrowed at the lower percentage rate.  Our formula would then look like
$I=x(0.035)(1) \\ =0.035x$.  (Remember that when we convert percentages to decimals, we divide by 100; 3.5/100 = 0.035.)
The remaining money borrowed was invested at 7% interest.  The expression to represent the remaining money would be 6000 - x, as it is what was left over to borrow.  The interest formula for this loan would be
$I=(6000-x)(0.07)(1)$. (Again, we must divide 7 by 100 to convert the percentage; 7/100=0.07.)
Using the distributive property we have:
$I=6000*0.07-x*0.07 \\ =420-0.07x$ (t in this case is 1, since it is 1 year.)
The total amount of interest for both loans for one year was $259, so we have:
$259=0.035x+420-0.07x$
Combine our like terms:
$259=-0.035x+420$
Cancel 420 by subtracting:
$259-420=-0.035x+420-420 \\ -161=-0.035x$
Cancel -0.035 by dividing:
$\frac{-161}{-0.035}=\frac{-0.035x}{-0.035} \\ 4600=x$
This means she borrowed $4600 at the lower interest rate.  The remainder would be $6000-$4600=$1400 at the higher interest rate.