1. Let's start with a problem on quadratic equations: Solve for $x$ in the equation $$x^2 - 5x + 6 = 0.$$ 2. The formula to solve quadratic equations is the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.$$ Here, $a$, $b$, and $c$ are coefficients from the quadratic equation $ax^2 + bx + c = 0$. 3. For our equation, $a=1$, $b=-5$, and $c=6$. Plugging these into the formula: $$x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \times 1 \times 6}}{2 \times 1} = \frac{5 \pm \sqrt{25 - 24}}{2} = \frac{5 \pm \sqrt{1}}{2}.$$ 4. Simplify the square root: $$\sqrt{1} = 1.$$ So, $$x = \frac{5 \pm 1}{2}.$$ 5. This gives two solutions: $$x_1 = \frac{5 + 1}{2} = 3,$$ $$x_2 = \frac{5 - 1}{2} = 2.$$ 6. Therefore, the solutions to the quadratic equation are $x=3$ and $x=2$. --- 7. Next, let's solve a system of linear equations: $$\begin{cases} 2x + 3y = 6 \\ x - y = 1 \end{cases}$$ 8. From the second equation, express $x$ in terms of $y$: $$x = y + 1.$$ 9. Substitute $x = y + 1$ into the first equation: $$2(y + 1) + 3y = 6.$$ 10. Simplify and solve for $y$: $$2y + 2 + 3y = 6 \Rightarrow 5y + 2 = 6 \Rightarrow 5y = 4 \Rightarrow y = \frac{4}{5}.$$ 11. Substitute $y = \frac{4}{5}$ back into $x = y + 1$: $$x = \frac{4}{5} + 1 = \frac{9}{5}.$$ 12. The solution to the system is $x=\frac{9}{5}$ and $y=\frac{4}{5}$.