1. **State the problem:** Solve the system of linear equations: $$\begin{cases} 2x = 2 \\ x + 3y = 6 \\ 4x - 5y = 7 \end{cases}$$ 2. **Solve the first equation:** $$2x = 2$$ Divide both sides by 2: $$\frac{\cancel{2}x}{\cancel{2}} = \frac{2}{2}$$ which simplifies to: $$x = 1$$ 3. **Substitute $x=1$ into the second and third equations:** Second equation: $$1 + 3y = 6$$ Subtract 1 from both sides: $$3y = 6 - 1$$ $$3y = 5$$ Divide both sides by 3: $$\frac{\cancel{3}y}{\cancel{3}} = \frac{5}{3}$$ $$y = \frac{5}{3}$$ Third equation: $$4(1) - 5y = 7$$ Simplify: $$4 - 5y = 7$$ Subtract 4 from both sides: $$-5y = 7 - 4$$ $$-5y = 3$$ Divide both sides by -5: $$\frac{\cancel{-5}y}{\cancel{-5}} = \frac{3}{-5}$$ $$y = -\frac{3}{5}$$ 4. **Check for consistency:** From the second equation, $y = \frac{5}{3}$, but from the third equation, $y = -\frac{3}{5}$. These values are not equal, so the system has no solution. 5. **Conclusion:** The system is inconsistent and has no solution. The initial solution $(1, -2)$ given does not satisfy both equations simultaneously. --- **Final answer:** The system has no solution because the two equations contradict each other.