Katy invests a total of $19,000 in two accounts paying 7% and 5% annual interest, respectively. how much was invested in each account if, after one year, the total interest was $1,210.00

Let x represent the dollar amount invested at 7% (the higher of the two rates). Then (19000-x) is the amount invested at 5%. The total interest earned in one year is the sum of the products of amount and rate. That sum is ...

... 1210 = (x)(.07) + (19000-x)(.05)

... 1210 = 0.02x + 950 . . . . . . collect terms

... 260 = 0.02x . . . . . . . . . . . . subtract 950

... 13000 = x . . . . . . . . . . . . . . divide by 0.02; amount at 7%

Then the amount invested at 5% is

... 19000-x = 19000-13000 = 6000

Comment on solution

Such a problem can be written with two variables, one for each amount. Then the two equations are equivalent to

Choosing one variable to represent the amount at the higher rate does two things:

If you choose x = investment at 5%, then the final arithmetic in the above solution becomes ...

... -120 = -0.02x ⇒ x = 6000 . . . . . same answer, but negative numbers to work with

you didn't post any answer choices, so its not 100% clear what your teacher is looking for

however, log(15/7) is equivalent to log(15) - log(7)

i'm using the rule log(x/y) = log(x) - log(y)

isaiah walks 3 miles due north, turns and then 5 miles due east. how far is he from his starting point?

a^2 + b^2 = c^2

let c = distance from starting point.

(3)^2 + (5)^2 = c^2

9 + 25 = c^2

34 = c^2

sqrt{34} = sqrt{c^2}

sqrt{34} = c