[tex]y+2 < -\dfrac{2}{3}(x-6)\\\\y+2 < -\dfrac{2}{3}x+4\ \ \ |-2\\\\y < -\dfrac{2}{3}x+2\\--------------------------\\y=-\frac{2}{3}x+2\\\\for\ x=0\to y=-\dfrac{2}{3}\cdot0+2=2\to(0;\ 2)\\\\for\ x=3\to y=-\dfrac{2}{3}\cdot3+2=0\to(3;\ 0)[/tex]
Answer in the attachment.
