1. **State the problem:** Given the equation $kx + 2y + 18 = 0$, where $k$ is a constant, find $x$ when $y=6$ given that $x=3$ when $y=4$. 2. **Write the equation:** $$kx + 2y + 18 = 0$$ 3. **Use the known values to find $k$:** Substitute $x=3$ and $y=4$: $$k(3) + 2(4) + 18 = 0$$ $$3k + 8 + 18 = 0$$ $$3k + 26 = 0$$ 4. **Solve for $k$:** $$3k = -26$$ $$k = \frac{-26}{3}$$ 5. **Find $x$ when $y=6$:** Substitute $k = \frac{-26}{3}$ and $y=6$ into the original equation: $$\frac{-26}{3} x + 2(6) + 18 = 0$$ $$\frac{-26}{3} x + 12 + 18 = 0$$ $$\frac{-26}{3} x + 30 = 0$$ 6. **Isolate $x$:** $$\frac{-26}{3} x = -30$$ 7. **Solve for $x$:** $$x = \frac{-30}{\cancel{1}} \times \frac{\cancel{3}}{-26} = \frac{-30 \times 3}{-26} = \frac{90}{26}$$ 8. **Simplify the fraction:** $$x = \frac{45}{13}$$ **Final answer:** $$x = \frac{45}{13}$$