Solve the following inequations and give the region in the plane where they are satisfied:
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$$y-3x > 2$$
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$$2(x-y)-3(y+2) < 2x+1$$
- $$\dfrac{-x+4y}{3}-x \geqslant 2y+x$$
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.
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$$y-3x > 2 \Rightarrow y > 3x -2$$. The solution region is above the straight line (we have an inequality of the type $$>$$).
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$$2(x-y)-3(y+2) < 2x+1 \Rightarrow 2x-2y-3y-6 < 2x+1 \Rightarrow$$ $$\Rightarrow 2x-2x-6-1 < 5y \Rightarrow y > -\dfrac{7}{5}$$. The solution region is above the straight line.
- $$\dfrac{-x+4y}{3}-x\geqslant 2y+x \Rightarrow \dfrac{-x+4y-3x}{3} \geqslant 2y+x \Rightarrow $$ $$\Rightarrow -x+4y-3x \geqslant 6y+3x \Rightarrow -x-3x-3x \geqslant 6y-4y \Rightarrow y \leqslant \dfrac{-7}{2}x$$. The solution region is below the straight line, taking also the line points.
Solution:
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$$y > 3x -2 \Rightarrow $$ Points above de line $$y-3x = 2$$ without the points in the line.
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$$y > -\dfrac{7}{5} \Rightarrow $$ 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.