{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: \( (-2, 5) \)