Solve the system of equations: 2x + y = 9 and 3x2 + 2y2 + x − y = 45

The given equations are:

1. $2x + y = 9$
2. $3x^2 + 2y^2 + x - y = 45$

From equation (1), we can express $y$ in terms of $x$:

$$y = 9 - 2x$$

Substitute $y = 9 - 2x$ into equation (2):

$$3x^2 + 2(9 - 2x)^2 + x - (9 - 2x) = 45$$

First, expand $(9 - 2x)^2$:

$$(9 - 2x)^2 = 9^2 - 2(9)(2x) + (2x)^2 = 81 - 36x + 4x^2$$

Now substitute this back into the equation:

$$3x^2 + 2(81 - 36x + 4x^2) + x - 9 + 2x = 45$$ $$3x^2 + 162 - 72x + 8x^2 + x - 9 + 2x = 45$$

Combine like terms:

$$(3x^2 + 8x^2) + (-72x + x + 2x) + (162 - 9) = 45$$ $$11x^2 - 69x + 153 = 45$$

Subtract 45 from both sides to set the equation to 0:

$$11x^2 - 69x + 153 - 45 = 0$$ $$11x^2 - 69x + 108 = 0$$

For the equation $ax^2 + bx + c = 0$, the quadratic formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
Here, $a = 11$, $b = -69$, $c = 108$.

Calculate the discriminant $\Delta = b^2 - 4ac$:

$$\Delta = (-69)^2 - 4(11)(108)$$ $$\Delta = 4761 - 4752$$ $$\Delta = 9$$

Now, substitute the values into the quadratic formula:

$$x = \frac{-(-69) \pm \sqrt{9}}{2(11)}$$ $$x = \frac{69 \pm 3}{22}$$

This gives two possible values for $x$:

$$x_1 = \frac{69 + 3}{22} = \frac{72}{22} = \frac{36}{11}$$ $$x_2 = \frac{69 - 3}{22} = \frac{66}{22} = 3$$

For $x_1 = \frac{36}{11}$:

$$y_1 = 9 - 2\left(\frac{36}{11}\right)$$ $$y_1 = 9 - \frac{72}{11}$$ $$y_1 = \frac{9 \times 11}{11} - \frac{72}{11}$$ $$y_1 = \frac{99 - 72}{11}$$ $$y_1 = \frac{27}{11}$$

This gives the solution pair $\left(\frac{36}{11}, \frac{27}{11}\right)$.

For $x_2 = 3$:

$$y_2 = 9 - 2(3)$$ $$y_2 = 9 - 6$$ $$y_2 = 3$$

This gives the solution pair $(3, 3)$.

The solutions to the pair of equations are \boxed{\text{(x, y) = \left(\frac{36}{11}, \frac{27}{11}\right) and (x, y) = (3, 3)}}.