We have two equations to solve. The first equation is: $(x+2)^2 - (x-y)(x+y) - 3(y+5) = 0$ The second equation is: $(2y-3)^2 - y(4y-y) + 4(3x-\frac{3}{4}) = 0$

1. Problem Description

We have two equations to solve.

The first equation is:

$(x+2)^2 - (x-y)(x+y) - 3(y+5) = 0$

The second equation is:

$(2y-3)^2 - y(4y-y) + 4(3x-\frac{3}{4}) = 0$

2. Solution Steps

First, let's solve the first equation:

$(x+2)^2 - (x-y)(x+y) - 3(y+5) = 0$

Expand the terms:

$x^2 + 4x + 4 - (x^2 - y^2) - 3y - 15 = 0$

$x^2 + 4x + 4 - x^2 + y^2 - 3y - 15 = 0$

$4x + y^2 - 3y - 11 = 0$

$4x = -y^2 + 3y + 11$

$x = \frac{-y^2 + 3y + 11}{4}$

Now, let's solve the second equation:

$(2y-3)^2 - y(4y-y) + 4(3x-\frac{3}{4}) = 0$

Expand the terms:

$4y^2 - 12y + 9 - y(3y) + 12x - 3 = 0$

$4y^2 - 12y + 9 - 3y^2 + 12x - 3 = 0$

$y^2 - 12y + 6 + 12x = 0$

Now substitute the value of $x$ from the first equation into the second equation:

$y^2 - 12y + 6 + 12(\frac{-y^2 + 3y + 11}{4}) = 0$

$y^2 - 12y + 6 + 3(-y^2 + 3y + 11) = 0$

$y^2 - 12y + 6 - 3y^2 + 9y + 33 = 0$

$-2y^2 - 3y + 39 = 0$

$2y^2 + 3y - 39 = 0$

Now solve this quadratic equation for $y$:

$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$y = \frac{-3 \pm \sqrt{3^2 - 4(2)(-39)}}{2(2)}$

$y = \frac{-3 \pm \sqrt{9 + 312}}{4}$

$y = \frac{-3 \pm \sqrt{321}}{4}$

So, $y_1 = \frac{-3 + \sqrt{321}}{4}$ and $y_2 = \frac{-3 - \sqrt{321}}{4}$

Now, substitute these values of $y$ into the equation $x = \frac{-y^2 + 3y + 11}{4}$ to find the corresponding $x$ values.

$x_1 = \frac{-(\frac{-3 + \sqrt{321}}{4})^2 + 3(\frac{-3 + \sqrt{321}}{4}) + 11}{4}$

$x_2 = \frac{-(\frac{-3 - \sqrt{321}}{4})^2 + 3(\frac{-3 - \sqrt{321}}{4}) + 11}{4}$

We can approximate $\sqrt{321} \approx 17.92$

$y_1 \approx \frac{-3 + 17.92}{4} \approx \frac{14.92}{4} \approx 3.73$

$y_2 \approx \frac{-3 - 17.92}{4} \approx \frac{-20.92}{4} \approx -5.23$

$x_1 \approx \frac{-(3.73)^2 + 3(3.73) + 11}{4} \approx \frac{-13.91 + 11.19 + 11}{4} \approx \frac{8.28}{4} \approx 2.07$

$x_2 \approx \frac{-(-5.23)^2 + 3(-5.23) + 11}{4} \approx \frac{-27.35 - 15.69 + 11}{4} \approx \frac{-32.04}{4} \approx -8.01$

3. Final Answer

The approximate solutions are:

$x_1 \approx 2.07$, $y_1 \approx 3.73$

$x_2 \approx -8.01$, $y_2 \approx -5.23$

The exact solutions are:

$y = \frac{-3 \pm \sqrt{321}}{4}$

$x = \frac{-y^2 + 3y + 11}{4}$

So, the solution pairs are:

$(x_1, y_1) = (\frac{-(\frac{-3 + \sqrt{321}}{4})^2 + 3(\frac{-3 + \sqrt{321}}{4}) + 11}{4}, \frac{-3 + \sqrt{321}}{4})$

$(x_2, y_2) = (\frac{-(\frac{-3 - \sqrt{321}}{4})^2 + 3(\frac{-3 - \sqrt{321}}{4}) + 11}{4}, \frac{-3 - \sqrt{321}}{4})$