Question

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

Answer 1

• $13,000 at 7%
• $6,000 at 5%

Step-by-step explanation:

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

• x + y = 19000 . . . . . . total amount invested
• .07x +.05y = 1210 . . . total interest earned

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

1. It reduces the problem to one equation (equivalent to substituting y=19000-x in the above)
2. It ensures the numbers you're working with are positive numbers.

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