Free Body Diagram:  determine the vertical reaction at the left hand beam support. 

You are probably using incorrect signs in your calculations!

Sum of clockwise moments = Sum of counterclockwise moments (for equilibrium)

You can assign clockwise moments as positive (you could reverse this as long as you're consistent).

$P_1 = 539\,\text{N}$ (downward force, $3\,\text{m}$ left of roller support A)

$P_2 = 431\,\text{N}$ (downward force, $4\,\text{m}$ right of pin support B)

Distance between supports: $I_2 = 4\,\text{m}$

Roller at A (left), Pin at B (right)

Counterclockwise (CCW) moments are positive

Clockwise (CW) moments are negative

Moment Equilibrium About B ($\sum M_B = 0$):

$R_A$ creates a counterclockwise moment about B.

$+M_{P_1} - M_{P_2} - M_{R_A} = 0$


Where:

$M_{P_1} = +P_1 \times (I_1 + I_2) = +539 \times 7 = +3773\text(N·m) \text(CCW)$

$M_{P_2} = -P_2 \times I_3 = -431 \times 4 = -1724\text(N·m) \text(CW)$

$M_{R_A} = -R_A \times 4\quad\text{(CW)}$

Substituting into equilibrium equation:

$3773 - 1724 - 4R_A = 0$

$2049 - 4R_A = 0$

$R_A = \frac{2049}{4} = 512.25\text{N}$

Vertical Force Equilibrium ($\sum F_y = 0$):

$R_A + R_B = P_1 + P_2$

$512.25 + R_B = 539 + 431$

$R_B = 970 - 512.25 = 457.75\text{N}$

Final Reactions:

$R_A = 512N$

$R_B = 458N$

Verification:

$512 + 458 = 970\,\text{N} \approx 539 + 431 = 970\,\text{N}\quad\text{\checkmark}$