1. **State the problem:** We want to understand why the linear system $$\begin{cases} 3x + y = 10 \\ 2x - 2y = 8 \end{cases}$$ is not easy to solve by comparison (substitution) method. 2. **Recall the comparison method:** This method involves solving each equation for the same variable and then setting the expressions equal to each other. 3. **Try to solve each equation for $y$:** From the first equation: $$y = 10 - 3x$$ From the second equation: $$2x - 2y = 8 \implies -2y = 8 - 2x \implies y = \frac{2x - 8}{2} = x - 4$$ 4. **Set the two expressions for $y$ equal:** $$10 - 3x = x - 4$$ 5. **Solve for $x$:** $$10 - 3x = x - 4$$ $$10 + 4 = x + 3x$$ $$14 = 4x$$ $$x = \frac{14}{4} = \frac{7}{2}$$ 6. **Find $y$ by substituting $x$ back:** $$y = 10 - 3 \times \frac{7}{2} = 10 - \frac{21}{2} = \frac{20}{2} - \frac{21}{2} = -\frac{1}{2}$$ 7. **Why is it not easy?** - The second equation requires rearranging and dividing by $-2$ to isolate $y$, which is a bit more involved than simple substitution. - The coefficients are not straightforward multiples, so the expressions for $y$ are different and require careful algebraic manipulation. **Final answer:** The system is not easy to solve by comparison because isolating the same variable in both equations involves different operations and fractions, making substitution less straightforward. **Solution:** $$x = \frac{7}{2}, \quad y = -\frac{1}{2}$$