1. **State the problem:** We have the system of equations: $$4x - 6y = 10y + 2$$ $$ty = 12 + 2x$$ where $t$ is a constant. We need to find the value of $t$ such that the system has no solution. 2. **Rewrite the first equation:** Move all terms to one side: $$4x - 6y - 10y - 2 = 0$$ $$4x - 16y - 2 = 0$$ or equivalently $$4x - 16y = 2$$ 3. **Rewrite the second equation:** $$ty = 12 + 2x$$ Rearranged: $$2x - ty = -12$$ 4. **Express the system in standard form:** $$\begin{cases} 4x - 16y = 2 \\ 2x - ty = -12 \end{cases}$$ 5. **Condition for no solution:** For a system of two linear equations: $$a_1x + b_1y = c_1$$ $$a_2x + b_2y = c_2$$ The system has no solution if the lines are parallel but not coincident, i.e., $$\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$$ 6. **Identify coefficients:** $$a_1 = 4, b_1 = -16, c_1 = 2$$ $$a_2 = 2, b_2 = -t, c_2 = -12$$ 7. **Set the ratios for parallel lines:** $$\frac{4}{2} = \frac{-16}{-t}$$ Simplify left side: $$2 = \frac{16}{t}$$ Cross-multiplied: $$2t = 16$$ $$t = 8$$ 8. **Check the ratio of constants to ensure no solution:** $$\frac{c_1}{c_2} = \frac{2}{-12} = -\frac{1}{6}$$ Compare with $\frac{a_1}{a_2} = 2$: $$2 \neq -\frac{1}{6}$$ So the lines are parallel but not coincident, confirming no solution. **Final answer:** $$t = 8$$