Solution: Set $ f(2) = g(2) $. Compute $ f(2) = 4 - 6 + m = -2 + m $ and $ g(2) = 4 - 6 + 5m = -2 + 5m $. Equate: $ -2 + m = -2 + 5m $. Simplify: $ 0 = 4m $, so $ m = \boxed{0} $.

["Solution: Setting $ f(2) = g(2) $ to Solve for $ m $", "In many mathematical problems, determining the value of a parameter like $ m $ boils down to equating function outputs at a specific input. This exercise demonstrates a fundamental algebraic technique: setting $ f(2) = g(2) $ when both functions behave identically at $ x = 2 $.", "Let’s begin with the given expressions:", "$$\nf(x) = 4 - 6 + m \quad \ ext{and} \quad g(x) = 4 - 6 + 5m\n$$", "Simplify both expressions:", "- $ f(x) = (4 - 6) + m = -2 + m $\n- $ g(x) = (4 - 6) + 5m = -2 + 5m $", "Since we’re tasked with finding $ m $ such that $ f(2) = g(2) $, set the simplified expressions equal:", "$$\n-2 + m = -2 + 5m\n$$", "Subtract $ -2 $ from both sides:", "$$\nm = 5m\n$$", "Now subtract $ m $ from both sides:", "$$\n0 = 4m\n$$", "Divide both sides by 4:", "$$\nm = 0\n$$", "This elegant solution shows that the only value of $ m $ that makes $ f(2) = g(2) $ is $ \boxed{0} $.\nThis method is widely applicable—whether comparing quadratic functions, piecewise definitions, or system outputs—proving that equality at a point often reveals critical constraints on parameters.", "Why This Technique Matters\nSetting functions equal at a known input is a powerful tool in solving for unknowns, especially when functions differ by a constant coefficient (like $ m $ here). It avoids trial and error and grounds solutions in algebraic logic, making it essential for algebra, calculus, and applied mathematics.", "---", "Summary:\n- Start with $ f(2) = g(2) $\n- Simplify both sides\n- Equate and solve: $ -2 + m = -2 + 5m \Rightarrow m = 0 $\n- Final answer: $ \boxed{0} $"]