Problem: if x dollars is invested at 10 percent for one year and y dollars is invested at 8 percent for one year , the annual income from the 10 percent investment will exceed the annual income from the 8 percent investment by $ 38 . if $ 2,000 is the total amount invested , how much is invested at 8 percent ?
Rationale: "0.1 x = 0.08 ( 2000 - x ) + 38 0.18 x = 198 x = 1100 then the amount invested at 8 % is $ 2000 - $ 1100 = $ 900 the answer is c ."
annotated_formula: subtract(multiply(multiply(const_100, 10), const_2), divide(add(multiply(multiply(10, 8), const_2), 38), add(divide(10, const_100), divide(8, const_100))))