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$X=\{(x,y)\in R^2\ :\ 3\le 2x+3y\le 8\}$

i tried to solve it as:

Let set $X$ is convex for $x_2,y_2\in X$ such that $\alpha x_1+(1-\alpha)x_2$,$\alpha y_1+(1-\alpha)y_2\in X$

Now, $3\le 2x+3y\le 8$ $2{\alpha x_1+(1-\alpha)x_2}+{3{\alpha y_1+(1-\alpha)y_2}}$ implies ${2\alpha x_1+2x_2-2\alpha x_2}+{3\alpha y_1+3y_2-3\alpha y_2}$; ${2\alpha(x_1-x_2)+2x_2}+{3\alpha(y_1-y_2)+3y_2}$

when $\alpha =0$, $2x_2+3y_2$ and when $\alpha=1$ $2x_1+3y_1$

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  • $\begingroup$ Welcome to MathSE! You are more likely to get a good answer to your question if you follow a few guidelines. In particular, make your question clear. Just what are you asking? $\endgroup$ – Rory Daulton Mar 26 '16 at 22:03
  • $\begingroup$ Generally it is easy to apply the definition of convexity directly to sets that are defined by linear inequalities. Since your set $X$ is defined by a pair of inequalities, $3\le 2x+3y$ and $2x+3y \le 8$, this should be a particularly straightforward approach. $\endgroup$ – hardmath Mar 26 '16 at 22:26
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$X$ is a convex because it is the intersection of 2 half-planes

$$\{(x,y)|2x+3y\leq8\} \ \ \text{and} \ \ \{(x,y)|2x+3y\geq3\} $$

and it is known that a half plane is a convex set.

The reason is that a half-plane is the pre-image by a continuous function of a convex set (Convex Sets Pre-image); for example, in the first case, the pre-image of $(-\infty,8]$ by the continuous function $(x,y)\rightarrow2x+3y$.

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