{3x+y=-1, {x-xy=8.

To solve the system of equations:

1. \( 3x + y = -1 \) (Equation 1)
2. \( x - xy = 8 \) (Equation 2)

We can express \( y \) from Equation 1 in terms of \( x \):

\[ y = -1 - 3x \]

Now, we substitute this expression for \( y \) into Equation 2:

\[ x - x(-1 - 3x) = 8 \]

Simplifying this:

\[ x + x(1 + 3x) = 8 \] \[ x + x + 3x^2 = 8 \] \[ 2x + 3x^2 = 8 \]

Rearranging gives us a quadratic equation:

\[ 3x^2 + 2x - 8 = 0 \]

We can solve this quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 3 \), \( b = 2 \), and \( c = -8 \):

\[ b^2 - 4ac = 2^2 - 4(3)(-8) = 4 + 96 = 100 \]

Now substituting into the formula:

\[ x = \frac{-2 \pm \sqrt{100}}{2 \cdot 3} = \frac{-2 \pm 10}{6} \]

This gives us two potential solutions for \( x \):

1. \( x = \frac{8}{6} = \frac{4}{3} \)
2. \( x = \frac{-12}{6} = -2 \)

Now we will find corresponding values of \( y \) for both \( x \) values:

Case 1: \( x = \frac{4}{3} \)

Substituting back into \( y = -1 - 3x \):

\[ y = -1 - 3\left(\frac{4}{3}\right) = -1 - 4 = -5 \]

Case 2: \( x = -2 \)

Substituting back into \( y = -1 - 3x \):

\[ y = -1 - 3(-2) = -1 + 6 = 5 \]

Thus, the solutions to the system of equations are:

1. \( (x, y) = \left( \frac{4}{3}, -5 \right) \)
2. \( (x, y) = (-2, 5) \)

These can be written as:

• Solution 1: \( \left( \frac{4}{3}, -5 \right) \)
• Solution 2: \( (-2, 5) \)