Solve the following system of simultaneous equations: 2x − y = 3 x2 - xy = -4

Given the first equation:

$$2x - y = 3$$

Rearrange to solve for $y$:

$$-y = 3 - 2x$$ $$y = 2x - 3$$

Given the second equation:

$$x^2 - xy = -4$$

Substitute $y = 2x - 3$:

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

Expand the equation:

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

Combine like terms:

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

Move all terms to one side to form a standard quadratic equation $ax^2 + bx + c = 0$:

$$x^2 - 3x - 4 = 0$$

Factor the quadratic equation:

$$(x - 4)(x + 1) = 0$$

This gives two possible values for $x$:

$$x - 4 = 0 \implies x = 4$$ $$x + 1 = 0 \implies x = -1$$

For $x = 4$:

$$y = 2(4) - 3$$ $$y = 8 - 3$$ $$y = 5$$

One solution is $(x, y) = (4, 5)$.

For $x = -1$:

$$y = 2(-1) - 3$$ $$y = -2 - 3$$ $$y = -5$$

The second solution is $(x, y) = (-1, -5)$.

The solutions to the simultaneous equations are:

$$x = 4, y = 5 \quad \text{and} \quad x = -1, y = -5$$