Answer :
y + 4 = x^2 . . . (1)
y - x = 2 . . . (2)
From (2), y = 2 + x . . . (3)
Putting (3) into (1), gives
2 + x + 4 = x^2
x^2 - x - 6 = 0
(x + 2)(x - 3) = 0
x = -2 or x = 3
y = 2 - 2 = 0 or y = 2 + 3 = 5
Solution set is (-2, 0) and (3, 5)
y - x = 2 . . . (2)
From (2), y = 2 + x . . . (3)
Putting (3) into (1), gives
2 + x + 4 = x^2
x^2 - x - 6 = 0
(x + 2)(x - 3) = 0
x = -2 or x = 3
y = 2 - 2 = 0 or y = 2 + 3 = 5
Solution set is (-2, 0) and (3, 5)
Answer:
[tex](-2,0)[/tex] and [tex](3,5)[/tex]
Step-by-step explanation:
we have
[tex]y+4=x^{2}[/tex] -----> equation A
Is a vertical parabola open upward
[tex]y-x=2[/tex] ----> equation B
Is the equation of a line
we know that
The solution of the system of equations is the intersection point both graphs
Using a graphing tool
The intersection points are [tex](-2,0)[/tex] and [tex](3,5)[/tex]
see the attached figure
