An investor decides to invest some cash in an account paying

An investor decides to invest some cash in an account paying 6% annual interest, and to put the rest in a stock fund that ends up earning 9% over the course of a year. The investor puts $1800 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 $1890. How much money was invested at each of the two rates? Round to the nearest integer.

Solution

Let x , x+1800 be the amounts invested in the 2nd  and 1 st accounts

interest in the 1st account = ( x+1800)6 /100

interest in the 2nd account = (x) 9/100

sum of the interest = (x+1800) 6/100 + 9x/100 = 1890 dividing by 3

2x+3600+3x= 63000

5x= 63000-3600= 594 00

x= 11880 investment in the 2nd account and

x+1800 = 13680 is the amount in the 1 st account