-2
x^2 + 3x + 7 = 6x + 18 x^2 - 3x - 11 = 0
(-1, 0)
x2 + 3x + 7 = 5 ∴ x2 + 3x + 2 = 0 ∴ (x + 1)(x + 2) = 0 ∴ x ∈ {-2, -1}
-2
-2
x^2 + 3x + 7 = 6x + 18 x^2 - 3x - 11 = 0
(-1, 0)
(-3x+1)(x+2) = 0 x = -2 or x = 1/3
x2 + 3x + 7 = 5 ∴ x2 + 3x + 2 = 0 ∴ (x + 1)(x + 2) = 0 ∴ x ∈ {-2, -1}
It is linear.
-2
3x + 4y - 4 = 0 y = -2 3x + 4(-2) - 4 = 0 3x - 8 - 4 = 0 3x = 12 x = 4 (4,-2)
y = -1x/2 3x + 6y = 0 -3x -3x __________ 6y = 0 - 3x --- -------- 6 6 y= -3x/6 y= -1x/2
x^2+3x-18=0 (x+6)(x-3)=0
Yes.
(3x + 1)(4x - 1)(x - 2)