Solve this problem using branch and bound algorithm.

First branch on $x_{1}$ by creating two new subproblems; one with the contraint $x_{1} \le 3$ , and another with $x_{1} \ge 4$. (Hence we've "branched" on x1, and "bounded" it to the nearest integer values 3 or 4). Solve the LP relaxation for each:

Subproblem 1
maximize $z=5x_{1}+4x_{2}$
subject to    $x_{1}+x_{2}\le5$
                 $10x_1+6x_2 \le 45$ 
                                        $x_1 \le 3$ 
LP Relaxation: {$x_1$,$x_2$} = {3,2}, z = 23

Subproblem 2
maximize $z=5x_{1}+4x_{2}$
subject to    $x_{1}+x_{2}\le5$
                 $10x_1+6x_2 \le 45$
                                        $x_1 \ge 4$
LP Relaxation: {$x_1$,$x_2$} = {4,5/6}, z = 23.333

Next, we branch the Subproblem 2 on x2, since x1 is an integer now. Bound $x_2$ to the closest integer values 0 or 1. Also keep the previous bounding constraint on $x_1$:

Subproblem 3
maximize $z=5x_{1}+4x_{2}$
subject to    $x_{1}+x_{2}\le5$
                 $10x_1+6x_2 \le 45$
                                        $x_1 \ge 4$
                                        $x_2 \le 0$
LP Relaxation: {x1,x2} = {4.5,0}, z = 22.5

Subproblem 4
maximize $z=5x_{1}+4x_{2}$
subject to    $x_{1}+x_{2}\le5$
                 $10x_1+6x_2 \le 45$
                                        $x_1 \ge 4$
                                        $x_2 \ge 1$
Infeasible

Of the four subproblems, the only feasible solution is Subproblem 1, and so the optimal solution is {$x_1$, $x_2$} = {3,2}, with an objective value of 23.