The solution is:
Point P ( -2 ; 0)
Point Q ( 1 ; -6)
2) If the linear equation never touches the quadratic function the system has no solution
The system:
2×x + y = -4 or y = -4 - 2×x
y = x² - x - 6 (1)
Solving ( by substitution of y value in equation (1) )
- 4 - 2× x = x² - x - 6
x² - x - 6 + 4 + 2×x = 0
x² + x -2 = 0
Solving for x
x₁,₂ = ( - 1 ± √1 +8 ) / 2
x₁,₂ = (-1 ± 3 )/2
x₁ = - 2
x₂ = 1
Now for x₁ = -2 y = (-2)² - ( -2) - 6 y = 0
Point P ( -2 ; 0)
For x₂ = 1 y = (1)² - ( 1) - 6 y = - 6
Point Q ( 1 ; -6)
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