xy+x=-4, x-y=6

To solve the system of equations:

1. \( xy + x = -4 \) (Equation 1)
2. \( x - y = 6 \) (Equation 2)

We can first express \( y \) in terms of \( x \) using Equation 2:

From Equation 2: \[ y = x - 6 \]

Now, substitute \( y \) into Equation 1:

\[ x(x - 6) + x = -4 \]

Expanding this we get: \[ x^2 - 6x + x = -4 \]

This simplifies to: \[ x^2 - 5x + 4 = -4 \]

Now, add 4 to both sides: \[ x^2 - 5x + 8 = 0 \]

Next, we will use the quadratic formula to solve for \( x \). The quadratic formula is given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 1 \), \( b = -5 \), and \( c = 8 \).

Calculating the discriminant \( b^2 - 4ac \): \[ (-5)^2 - 4(1)(8) = 25 - 32 = -7 \]

Since the discriminant is negative, there are no real solutions for \( x \). This means that the solution for this system of equations involves complex numbers.

We can find the complex solutions as follows:

\[ x = \frac{-(-5) \pm \sqrt{-7}}{2(1)} = \frac{5 \pm i\sqrt{7}}{2} \]

So the two values of \( x \) are: \[ x_1 = \frac{5 + i\sqrt{7}}{2}, \quad x_2 = \frac{5 - i\sqrt{7}}{2} \]

Now we can substitute these back into \( y = x - 6 \) to find the corresponding \( y \) values.

For \( x_1 \): \[ y_1 = \frac{5 + i\sqrt{7}}{2} - 6 = \frac{5 + i\sqrt{7}}{2} - \frac{12}{2} = \frac{-7 + i\sqrt{7}}{2} \]

For \( x_2 \): \[ y_2 = \frac{5 - i\sqrt{7}}{2} - 6 = \frac{5 - i\sqrt{7}}{2} - \frac{12}{2} = \frac{-7 - i\sqrt{7}}{2} \]

Thus, the complex solutions of the system of equations are: \[ (x_1, y_1) = \left( \frac{5 + i\sqrt{7}}{2}, \frac{-7 + i\sqrt{7}}{2} \right) \] \[ (x_2, y_2) = \left( \frac{5 - i\sqrt{7}}{2}, \frac{-7 - i\sqrt{7}}{2} \right) \]