$$ X := \left\{ (x,y) \in \Bbb R^2\ : 3\le 2x+3y\le 8 \right\} $$
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 $\alpha=1$ $2x_1+3y_1$
On
Let $$ X := \left\{ (x,y) \in \Bbb R^2\ : 3\le 2x+3y\le 8 \right\} $$
Let $\mathbf{x} = (x_1, y_1) \in X$.
Let $\mathbf{y}=(x_2, y_2) \in X$.
By definition, $$ 3\le 2x_1+3 y_1\le 8 \tag{1} $$ $$ 3 \le 2x_2 + 3 y_2 \le 8 \tag{2} $$
Take any convex combination of $\mathbf{x}, \mathbf{y}$, viz. $$ \mathbf{z} = \alpha \mathbf{x} + (1 - \alpha) \mathbf{y}, $$ where $0 \le \alpha \le 1$.
If we define $\mathbf{z} = (\xi, \eta)$, then $$ \xi = \alpha x_1 + (1 - \alpha) x_2 $$ $$ \eta = \alpha y_1 + (1 - \alpha) y_2 $$
From (1) and (2), we get $$ 3 \alpha \le \alpha (2 x_1 + 3 y_1) \leq 8 \alpha \tag{3} $$ $$ 3 (1-\alpha) \le (1-\alpha) (2 x_2 + 3 y_2) \leq 8 (1-\alpha) \tag{4} $$
Adding (3) and (4), we get $$ 3 \leq 2 \xi + 3 \eta \leq 8 $$ which shows that $\mathbf{z} \in X$.
Hence, $X$ is a convex set.
$X$ 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 $(x,y)\rightarrow2x+3y$.