Use a system of equations to solve the following problem.

Question

An investor decides to invest some cash in an account paying 9% annual interest, and to put the rest in a stock fund that ends up earning 11% over the course of a year. The investor puts $⁢1400 more in the first account than in the stock fund, and at the end of the year finds the total interest from the two investments was $⁢1060. How much money was invested at each of the two rates? Round to the nearest integer.

Answer 1

Let "x" represent the amount of cash they put in the 9% account and let "y" represent the amount they put in the 11% stock fund.

Because "the investor puts $⁢1400 more in the first account than in the stock fund" x = 1400 + y

To calculate the interest in one year we would multiply the interest rate by the amount we put in each account so:

(x)(0.09) + (y)(0.11)

and we are told the total interest was $1060 so:

0.09x + 0.11y = 1060

Because "x" is the same as "1400 + y" (from our first equation), we can plug in "1400 + y" where we see "x":

0.09(1400 + y) + 0.11y = 1060

126 + 0.09y + 0.11y = 1060

0.2y = 934

y = 934/0.2 = 4670

If x = 1400 + y then x = 1400 + 4670 = 6070

So they placed $6070 in the first account (9% account) and $4670 in the second account (11% stick fund)