Solve the following system of equations: 2x − y = 0 and 4x2 + 3y2 − 2x + y = 16

The given equations are:

1. $2x - y = 0$
2. $4x^2 + 3y^2 - 2x + y = 16$

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

$$y = 2x$$

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

$$4x^2 + 3(2x)^2 - 2x + (2x) = 16$$

$$4x^2 + 3(4x^2) - 2x + 2x = 16$$ $$4x^2 + 12x^2 = 16$$ $$16x^2 = 16$$

Divide both sides by 16:

$$x^2 = \frac{16}{16}$$ $$x^2 = 1$$

Take the square root of both sides:

$$x = \pm\sqrt{1}$$ $$x = \pm 1$$

So, we have two possible values for $x$: $x_1 = 1$ and $x_2 = -1$.

For $x_1 = 1$:

$$y_1 = 2(1)$$ $$y_1 = 2$$

This gives the solution pair $(1, 2)$.

For $x_2 = -1$:

$$y_2 = 2(-1)$$ $$y_2 = -2$$

This gives the solution pair $(-1, -2)$.

The solutions to the pair of equations are \boxed{\text{(x, y) = (1, 2) and (x, y) = (-1, -2)}}.