Answer:
The solution of the system of equations are;
(-2, -6) and (4, 6)
Step-by-step explanation:
-2·x + y = -2...............(1)
[tex]y = -\dfrac{1}{2} \cdot x^2 + 3 \cdot x + 2[/tex]........(2)
Equation (1) gives;
y = 2·x - 2
From which we have;
[tex]2 \cdot x - 2 = -\dfrac{1}{2} \cdot x^2 + 3 \cdot x + 2[/tex]
[tex]0= -\dfrac{1}{2} \cdot x^2 + x + 4[/tex]
x² -2·x -8 = 0
(x - 4)·(x + 2) = 0
x = 4 or x = -2
The y-coordinate values are;
y = 2×(-2) - 2 = -6 and y = 2×(4) - 2 = 6
The solution points are;
(-2, -6) and (4, 6).
The points where the equation, -2·x + y = -2 and the equation [tex]y = -\dfrac{1}{2} \cdot x^2 + 3 \cdot x + 2[/tex] intersect are (-2, -6) and (4, 6).
