I'd say that since you're putting in x in the equations for x (i.e. f(x) = 3x -1 and g(x) = (x-2)^2 + 3, you "replaced" the x with another x), you would just add the two functions together to create a polynomial.(3x - 1) + [(x - 2)^2 + 3]Then you'd expand (x - 2)^2 and combine like terms.(3x - 1) + (x^2 - 4x + 4 + 3)(3x - 1) + (x^2 - 4x + 7)Then just add the two polynomials together.x^2 - x + 6This is your answer. Because you inserted a variable in the functions, you get an answer that still has a variable in it. That's what I think, anyway. Hopefully this explanation was correct and helpful. ~ Azaler