We are solving the system of equations:

1. $\frac{x}{3} - \frac{y}{4} = 1$
2. $y^2 - 2y = 4x + 5$

Step 1: Simplify the equations

1. Multiply through by the common denominator of 12 in the first equation: 4x - 3y = 12 \tag{1'}

2. Rewrite the second equation: y^2 - 2y - 4x - 5 = 0 \tag{2'}

Now the system becomes: 4x - 3y = 12 \tag{1'} y^2 - 2y - 4x - 5 = 0 \tag{2'}

Step 2: Express $x$ in terms of $y$ from (1')

From $(1')$: 4x = 3y + 12 \implies x = \frac{3y + 12}{4} \tag{3}

Step 3: Substitute $x$ into (2')

Substitute $x = \frac{3y + 12}{4}$ into $(2')$: $y^2 - 2y - 4\left(\frac{3y + 12}{4}\right) - 5 = 0$ Simplify: $y^2 - 2y - (3y + 12) - 5 = 0$ $y^2 - 2y - 3y - 12 - 5 = 0$ y^2 - 5y - 17 = 0 \tag{4}

Step 4: Solve the quadratic equation for $y$

The quadratic equation is: $y^2 - 5y - 17 = 0$ Solve using the quadratic formula: $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \quad a = 1, \, b = -5, \, c = -17$ $y = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(-17)}}{2(1)}$ $y = \frac{5 \pm \sqrt{25 + 68}}{2}$ $y = \frac{5 \pm \sqrt{93}}{2}$ $y = \frac{5 \pm \sqrt{93}}{2}$

Thus: $y_1 = \frac{5 + \sqrt{93}}{2}, \quad y_2 = \frac{5 - \sqrt{93}}{2}$

Step 5: Find corresponding $x$ values

Using $x = \frac{3y + 12}{4}$, substitute each $y$ value:

For $y_1 = \frac{5 + \sqrt{93}}{2}$: $x_1 = \frac{3\left(\frac{5 + \sqrt{93}}{2}\right) + 12}{4}$ $x_1 = \frac{\frac{15 + 3\sqrt{93}}{2} + 12}{4}$ $x_1 = \frac{\frac{15 + 3\sqrt{93} + 24}{2}}{4}$ $x_1 = \frac{39 + 3\sqrt{93}}{8}$

For $y_2 = \frac{5 - \sqrt{93}}{2}$: $x_2 = \frac{3\left(\frac{5 - \sqrt{93}}{2}\right) + 12}{4}$ $x_2 = \frac{\frac{15 - 3\sqrt{93}}{2} + 12}{4}$ $x_2 = \frac{\frac{15 - 3\sqrt{93} + 24}{2}}{4}$ $x_2 = \frac{39 - 3\sqrt{93}}{8}$

Final Solution:

The solutions are: $(x_1, y_1) = \left(\frac{39 + 3\sqrt{93}}{8}, \frac{5 + \sqrt{93}}{2}\right)$ $(x_2, y_2) = \left(\frac{39 - 3\sqrt{93}}{8}, \frac{5 - \sqrt{93}}{2}\right)$

Let me know if you need detailed numeric approximations or further clarifications!

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Here are five related questions:

1. How is the quadratic formula derived?
2. What are other methods to solve simultaneous equations?
3. Can this problem be solved graphically?
4. How does the discriminant of a quadratic equation affect the solution?
5. How can we interpret these solutions geometrically?

Tip: Always verify solutions by substituting them back into the original equations!