For this case we have a system of two equations with two unknowns.
To solve, we equate both equations:
[tex]2x-2 = x ^ 2-x-6\\x ^ 2-3x-4 = 0[/tex]
We have a quadratic equation of the form:
[tex]x ^ 2 + bx + c = 0[/tex]
Where:
[tex]a = 1\\b = -3\\c = -4[/tex]
Its roots are given by:
[tex]x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2a}\\x = \frac {- (- 3) \pm \sqrt {(- 3) ^ 2-4 (1) (- 4)}} {2 (1)}\\x = \frac {3 \pm \sqrt {9 + 16}} {2}\\x = \frac {3 \pm \sqrt {25}} {2}\\x = \frac {3 \pm5} {2}\\[/tex]
We have two roots:
[tex]x_ {1} = \frac {3 + 5} {2} = 4\\x_ {2} = \frac {3-5} {2} = - 1[/tex]
Thus, we look for the values of y, by substituting any of the equations:
[tex]For\ x_ {1} = 4,\ y_ {1} = 2 (4) -2 = 6\\For\ x_ {2} = - 1,\ y_ {2} = 2 (-1) -2 = -4[/tex]
Answer:
[tex](x_ {1}, y_ {1}) = (4,6)\\(x_ {2}, y_ {2}) (- 1, -4)[/tex]
Option B
