Can I please get help with this question?

Question

Susan placed $5,500 into three investments: the first paid 10% simple interest, the second paid 12% simple interest, and the third paid 18% simple interest. The total interest earned after one year is $764. If the amount of money placed in the third investment is $300 more than the amount placed in the second, find the amount of each investment in dollars.

amount invested at 10%$ =

amount invested at 12%$=

 amount invested at 18%=

Answer 1

let T = amount invested for the investment earning 18%, S= amount for 12% and F= 10%

T = 300 + S

F + S + T = 5500

.1F + .12S + .18T = 764

.1F + .1S + .1T = 550

subtract to eliminate one variable

.02S + .08T = 214

.02S + .08(300+S) = 214

.02S + .08S = 214 - .08(300) = 214 - 24

.1S = 190

S = $1,900 invested at 12%

T = $2,200 invested at 18%

F = 5500-1900-2200 = 5500-4100 = $1,400 invested at 10%

.1(1400) + .12(1900) + .18(2200)

= 140 + 228 + 396 = 764