The equality constraint pins the solution to a line, so eliminate $x_1$ :

$5x_1+3x_2=15 \Longrightarrow x_1=3-\tfrac35x_2 .$

Now translate every remaining condition into a bound on $x_2$ :

$\begin{aligned} x_1\ge0:&\quad 3-\tfrac35x_2\ge0 &&\Longrightarrow x_2\le5,\\ -x_1+x_2\le1:&\quad -3+\tfrac35x_2+x_2\le1 &&\Longrightarrow \tfrac85x_2\le4 \Longrightarrow x_2\le2.5,\\ 2x_1+5x_2\le10:&\quad 6-\tfrac65x_2+5x_2\le10 &&\Longrightarrow \tfrac{19}{5}x_2\le4 \Longrightarrow x_2\le\tfrac{20}{19}, \end{aligned}$

together with $x_2\ge0$ . The feasible set is the non-empty segment $0\le x_2\le\tfrac{20}{19}$ — so (A) is out, and boundedness rules out (D).

On that segment the objective becomes a function of $x_2$ alone:

$x_1+\tfrac52x_2=\Bigl(3-\tfrac35x_2\Bigr)+\tfrac52x_2 =3+\tfrac{19}{10}x_2 ,$

which is strictly increasing. A strictly monotone objective on a segment attains its maximum at exactly one endpoint, namely $x_2=\tfrac{20}{19}$ , giving

$x_1=3-\tfrac35\cdot\tfrac{20}{19}=\tfrac{45}{19}, \qquad \text{optimal value}=3+\tfrac{19}{10}\cdot\tfrac{20}{19}=5 .$

So the optimum is unique.

A tempting trap. The objective gradient $(1,\tfrac52)$ is parallel to the third constraint's normal $(2,5)$ , which usually signals alternative optima along an edge. Here it does not, because the equality constraint has already reduced the feasible region to a line that meets $2x_1+5x_2=10$ in a single point.