To determine the reactions at supports A and B, we will use the equations of static equilibrium. Let $R_A$ be the reaction force at support A. Let $R_B$ be the reaction force at support B. The downward point load is 10 units. The distance from A to the load is 3 units. The distance from the load to B is 3 units. The total length of the beam is $3 + 3 = 6$ units. Step 1: Apply the equilibrium equation for the sum of vertical forces. Assume upward forces are positive and downward forces are negative. $$ \sum F_y = 0 $$ $$ R_A + R_B - 10 = 0 $$ $$ R_A + R_B = 10 \quad (*)$$ Step 2: Apply the equilibrium equation for the sum of moments about point A. Assume counter-clockwise moments are positive. $$ \sum M_A = 0 $$ The load of 10 units creates a clockwise moment about A. The reaction $R_B$ creates a counter-clockwise moment about A. $$ (R_B \times 6) - (10 \times 3) = 0 $$ $$ 6 R_B - 30 = 0 $$ $$ 6 R_B = 30 $$ $$ R_B = \frac{30}{6} $$ $$ R_B = 5 $$ Step 3: Substitute the value of $R_B$ into equation $(*)$ to find $R_A$. $$ R_A + 5 = 10 $$ $$ R_A = 10 - 5 $$ $$ R_A = 5 $$ The reactions at the supports are: $R_A = 5$ units $R_B = 5$ units The final answer is $\boxed{R_A = 5, R_B = 5}$.