Chapter 10 - Section 10.3 - Solving Nonlinear Systems of Equations - Exercise Set - Page 624: 17

$(1,-3)$

$y= x^{2}-4$ Equation $(1)$ $y=x^{2}-4x$ Equation $(2)$ From Equation $(1)$ and Equation $(2)$ $x^{2}-4=x^{2}-4x$ $x^{2}-4-x^{2}+4x=0$ $4x-4 = 0$ $4(x-1) = 0$ $x= 1$ Substituting $x$ value in Equation $(1)$ $y= x^{2}-4$ $y= 1^{2}-4$ $y= 1-4$ $y= -3$ $(1,-3) $ satisfy both the equations. So, $(1,-3) $ is the solution.