1. **State the problem:** Given the equation $mx + y = 12$, where $m$ is a constant, and the point $(x, y) = (3, 6)$ lies on the line, find the value of $x$ when $y = 4$. 2. **Use the equation and given point to find $m$:** Substitute $x=3$ and $y=6$ into the equation: $$m \cdot 3 + 6 = 12$$ Simplify: $$3m + 6 = 12$$ Subtract 6 from both sides: $$3m + \cancel{6} - \cancel{6} = 12 - 6$$ $$3m = 6$$ Divide both sides by 3: $$\frac{3m}{\cancel{3}} = \frac{6}{\cancel{3}}$$ $$m = 2$$ 3. **Rewrite the equation with $m=2$:** $$2x + y = 12$$ 4. **Find $x$ when $y=4$:** Substitute $y=4$: $$2x + 4 = 12$$ Subtract 4 from both sides: $$2x + \cancel{4} - \cancel{4} = 12 - 4$$ $$2x = 8$$ Divide both sides by 2: $$\frac{2x}{\cancel{2}} = \frac{8}{\cancel{2}}$$ $$x = 4$$ **Final answer:** When $y=4$, $x=4$.