1. The problem asks to find the values of the compositions $(f \circ g)(-6)$ and $(g \circ f)(-6)$ given the functions $f(y) = 3y - 8$ and $g(y) = 6y + 2$. 2. Recall that the composition of functions $(f \circ g)(x)$ means $f(g(x))$, which is applying $g$ first, then $f$ to the result. Similarly, $(g \circ f)(x) = g(f(x))$. 3. First, calculate $g(-6)$: $$g(-6) = 6(-6) + 2 = -36 + 2 = -34$$ 4. Now, calculate $(f \circ g)(-6) = f(g(-6)) = f(-34)$: $$f(-34) = 3(-34) - 8 = -102 - 8 = -110$$ 5. Next, calculate $f(-6)$: $$f(-6) = 3(-6) - 8 = -18 - 8 = -26$$ 6. Then, calculate $(g \circ f)(-6) = g(f(-6)) = g(-26)$: $$g(-26) = 6(-26) + 2 = -156 + 2 = -154$$ 7. Final answers: $$(f \circ g)(-6) = -110$$ $$(g \circ f)(-6) = -154$$