systems of two equals

Question

A self-employed contractor nearing retirement made two investments totaling $ 15,000. In one year, these investments yielded $573 in simple interest. Part of the money was invested at 3% and the rest at 4.5%. How much was invested at each rate?

Answer 1

Equations:
Quantity: x + y = 15000
Interest: 0.03x + 0.045y = 573

 

Multiplying quantity by .03:

.03x + .03y = 450

.03x + .045y = 573

 

Subtract first from second equation:

.015y=123

y=8200

x=6800

 

Hope that helps!

Deanna

Answer 2

573 / 15,000 = .0382 = 3.82% is the average interest rate

4.5% - 3.82% = .68 is the difference between 4.5% and the average

3.82% - 3% = .82 is the difference between 3% and the average

.68/1.5 = .45333 = 45 1/3% of 15,000 at 3% = $6,800

.82/1.5 = .54666 = 54 2/3% of 15,000 at 4.5% = $8,200

 

 

Another method is dividing the investments into x and (15,000 - x)

.045x + .03(15000-x) = 573

.045x + 450 - .03x = 573

.015x = 123

x = 8,200 at 4.5%

15,000 - x = 6,800 at 3%