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
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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
