$M_1L_1 + 2M_2(L_1 + L_2) + M_3L_2 + \dfrac{6A_1\bar{a}_1}{L_1} + \dfrac{6A_2\bar{b}_2}{L_2} = 0$
Where,
$L_1 = 6 ~\text{ft}$
$L_2 = 0$
$M_1 = -\frac{1}{2}(4)(80)] \, \frac{1}{3}(4) \, ] = 213.33 ~ \text{lb}\cdot\text{ft}$
$M_3 = 0$
$\dfrac{6A_1\bar{a}_1}{L_1} = \frac{7}{60}(80)(6^3) + \frac{8}{60}(200)(6^3) = 7776 ~ \text{lb}\cdot\text{ft}^2$
$\dfrac{6A_2\bar{b}_2}{L_2} = 0$
Thus,
$-213.33(6) + 2M_{wall}(6 + 0) + 0 + 7776 + 0 = 0$
$M_{wall} = -541.33 ~ \text{lb}\cdot\text{ft}$ answer
$M_{wall} = \Sigma M_{\text{to the left of wall}}$
$-541.33 = 6R - \frac{1}{2}(10)(200)(\frac{10}{3})$
$6R = 2792$
$R = 465.33 ~ \text{lb}$ answer
Check by Area-Moment Method
$A_1 = \frac{1}{2}(6)(6R) = 18R$
$x_1 = \frac{1}{3}(6) = 2 ~ \text{ft}$
$A_2 = \frac{1}{4}(10)(3333.33) - \frac{1}{4}(4)(213.33) = 8120 ~ \text{lb}\cdot\text{ft}^2$
$x_2 = \dfrac{\frac{1}{4}(10)(3333.33)(\frac{10}{5}) - \frac{1}{4}(4)(213.33)(6 + \frac{4}{5})}{8120} = 1.8739 ~ \text{ft}$
$EI \, t_{B/A} = 0$
$(\text{Area}_AB) \cdot \bar{X}_B = 0$
$A_1(6 - x_1) - A_2(6 - x_2) = 0$
$18R(6 - 2) - 8120(6 - 1.8739) = 0$
$R = 465.33 ~ \text{lb}$ okay
$M_{wall} = 6(465.33) - 3333.33$
$M_{wall} = 541.33 ~\text{lb}\cdot\text{ft}$ okay
