Two investments totaling $ 18,000 produce an annual income of $1370. one investment yields 7% per year, while the other yields 8% per year. how much is invested at each rate?

Let amount invested at 6% be S
Then amount invested at 8% = 16,000 - S 
We therefore have: .06(S) + .08(16,000 - S) = 1,140 .06S + 1,280 - .08S = 1,140 .06S - .08S = 1,140 - 1,280 - .02S = - 140 S, or amount invested at 6% = $ 7,000Amount invested at 8% = $16,000 - 7,000, or $ 9,000

)because it's going in order of the math sign