Need help setting up an equation to answer a word problem involving a loan with 2 interest rates...?

simple interest: I = P·R·T, where P is the principal, R is the rate (expressed as a decimal instead of as a percentage), and T is the time (usually measured in years).

Here we have two rates (R₁ & R₂), with two amounts of interest (I₁ & I₂) on two portions of the principal (P₁ & P₂),

so we have I₁ = P₁R₁T and I₂ = P₂R₂T.

We are not given the amount of time, so let's suppose it to be 1 year for simplicity.

Rewrite the interest equations as I₁ = P₁R₁ and I₂ = P₂R₂, leaving the time implicit.

I₁ + I₂ = 1000, so P₁R₁ + P₂R₂ = 1000. Substitute the values for R₁ & R₂:

P₁(0.08) + P₂(0.18) = 1000.

We are given that P = P₁ + P₂ = 12,000, so P₁ = 12,000 - P₂. Substitute this in the previous equation:

(12,000 - P₂)(0.08) + P₂(0.18) = 1000, or

960 - 0.08P₂ + 0.18P₂ = 1000, or

0.10P₂ = 40, so

P₂ = $400, and

P₁ = 12,000 - P₂ = $11,600.

check:

(0.08)($11,600) + (0.18)($400) = $1000.

You are asked to find the portions of the loan at each rate. Choose the variables to represent those portions. Let x and y represent the parts of the loan at 8% and 18%, respectively.

You are given the total amount of the loan, and the total amount of interest. You can write an equation for each of these.

.. x + y = 12000 . . . . . .. the sum of both parts of the loan is $12k

.. .08x +.18y = 1000 . . . the total interest for the two parts is $1000

Subtract .08 times the first equation from the second to eliminate x as a variable.

.. (.08x +.18y) -.08(x + y) = 1000 -.08*(12000)

.. 0.10y = 40

.. y = 400 . . . . . . . . . . . . . . . . .. $400 of the loan was charged 18%

.. x = 12000 -400 = 11,600 . . . . $11,600 of the loan was charged 8%

_____

If you wish, you can just use one equation for y. Since x = 12000 -y, you can just write the second equation in terms of y as

.. 0.08(12000 -y) +0.18y = 1000

This gives you essentially the same steps:

.. 960 +0.10y = 1000

.. 0.10y = 40

.. y = 400

You don't actually need to write an equation involving x, since you know that the amount at the lower rate will be 12000 minus the amount at the higher rate.

It usually works well to let the variable in a mixture problem like this represent the amount associated with the highest cost or largest contributor. That way, the numbers remain positive. If you use a variable for the amount at 8% (x), then you will find you are doing arithmetic with negative numbers. The result is the same, but most people prefer working with positive numbers.