Solve the following inequations and give the region in the plane where they are satisfied:

$$y3x > 2$$

$$2(xy)3(y+2) < 2x+1$$
 $$\dfrac{x+4y}{3}x \geqslant 2y+x$$
Development:
We are going to solve three inequations. We will give the expression of the inequation as $$y < ax+b$$ and will say what points of the plane we take.

$$y3x > 2 \Rightarrow y > 3x 2$$. The solution region is above the straight line (we have an inequality of the type $$>$$).

$$2(xy)3(y+2) < 2x+1 \Rightarrow 2x2y3y6 < 2x+1 \Rightarrow$$ $$\Rightarrow 2x2x61 < 5y \Rightarrow y > \dfrac{7}{5}$$. The solution region is above the straight line.
 $$\dfrac{x+4y}{3}x\geqslant 2y+x \Rightarrow \dfrac{x+4y3x}{3} \geqslant 2y+x \Rightarrow $$ $$\Rightarrow x+4y3x \geqslant 6y+3x \Rightarrow x3x3x \geqslant 6y4y \Rightarrow y \leqslant \dfrac{7}{2}x$$. The solution region is below the straight line, taking also the line points.
Solution:

$$y > 3x 2 \Rightarrow $$ Points above de line $$y3x = 2$$ without the points in the line.

$$y > \dfrac{7}{5} \Rightarrow $$ Points above de line $$y = \dfrac{7}{5}$$ without the points in the line.
 $$y \leqslant \dfrac{7}{2}x \Rightarrow $$ Points below de line $$y = \dfrac{7}{2}x$$ with the points in the line.