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ok do you have an example?

These are generally specific to the word problem, and probably best learned by example

that should be $120

hmmm, is that the exact problem?

yes

what should be $120?

and is it after one year?

that is how much less the interest earned is on $6000

yes annual interest

ok got it

so the amount of interest earned in a year is .% x the investment

so for the first part of the word problem we know this:
1. 6000 * x = 10000 * y - 120

now, they also give you the second interest rate - for the 10000 value - is 1%

which is .01

no $10000 is one percent less than the rate for $6000

so now you need to substitute one equation into the other

now given that x = .5%, and y = x - .01, we can solve for y

ack sorry

I messed up in step 4

120 = 4000x - 100 means 220 = 4000x

I got confused at step 2 lol

so x = 220/4000 = .055 = 5.5%

1% should equal the $120 dollars is what I was thinking

y = x - .01 means y = .055 - .01 = .045 = 4.5%

I guess I have more of a problem setting up the equation. Once I seen it I could solve it

well, practice is key with these. and honestly - that was a tricky one

took me a long time to understand what they wanted

Ok thanks a lot! I haven't had math in over 15 years.

np =)