Answer
$(3,6),(1,-2)$
Work Step by Step
$y = x^{2} - 3$ Equation $(1)$
$4x-y = 6$ Equation $(2)$
From Equation $(2),$
$4x-y = 6$
$4x-6 = y$
$y = 4x - 6$ Equation $(3)$
From Equation $(1)$ and Equation $(3)$
$x^{2} - 3 = 4x-6$
$x^{2} - 3 -4x+6 = 0$
$x^{2} -4x+3 = 0$
By factoring,
$(x-3)(x-1)=0$
$x=3$ or $x=1$
Substitute $ x$ values in Equation $(1)$
Let $x = 3$
$y = x^{2} - 3$
$y = 3^{2} - 3$
$y = 9 - 3$
$y = 6$
Let $x = 1$
$y = x^{2} - 3$
$y = 1^{2} - 3$
$y = 1 - 3$
$y = -2$
$(3,6),(1,-2)$ satisfy the given equations. So, solutions are $(3,6),(1,-2)$
