Algebra

Connie invested all of her $3000 savings, part at an annual rate of 6% and the rest at an annual rate of 9%. Her annual income from both the investments was 8% of her total investment. How much did she invest at 6%?

I don't know how to set up the problem? What are the two starting equations?

kelprofessor said:

Connie invested all of her $3000 savings, part at an annual rate of 6% and the rest at an annual rate of 9%. Her annual income from both the investments was 8% of her total investment. How much did she invest at 6%?

I don't know how to set up the problem? What are the two starting equations?

Click to expand...

Start by naming things.

Let x = amount invested at 6%
Let y = amount invested at 9%

One equation will be based on the fact that the total invested is $3000:

x + y = 3000

One equation will be based on the amount of interest earned. We know that the total of the interest for one year on the two investments is 8% of her total investment of $3000.

.06x + .09y = .08(3000)

There are your two equations.

What's a "kel" professor?

kel are my initials.

Let x = amount invested at 6%
Let 3000 - x = amount invested at 9%

0.06x + 0.09(3000 - x) = 0.08(3000)

Nice one, BJ; you'll soon be a full time tutor here, with full employment benefits :roll: