Chapter 10 - Section 10.8 - Systems of Nonlinear Equations - 10.8 Exercises - Page 754: 23

$(6, 2)$ $(-2, -6)$

Given the system of equations: 1. $x - y = 4$ 2. $xy = 12$ It asks to find all solutions. Substitute $y = x-4$ $x (x-4) = 12$ $x^2 - 4x - 12 = 0$ $(x-6)(x+2) = 0$ $x = 6, -2$ Substitute the values into the equation $6y = 12$ $y = 2$ $-2y = 12$ $y = -6$ All solutions: $(6, 2)$ $(-2, -6)$