Civil Engineering problem

Let the vertical reaction at $c$ be $R_c$ at $a$ be $R_a$ and at $b$ be $R_b$ .

Summation of moments at $b$ equals zero. We have

$8R_c+4R_a+4(4)=6(2)+2(4)(2)+2(4)(6)$

$8R_c+4R_a=60$

$R_a=15-2R_c$ $\color{#D61F06}(1)$

Summation of moments at $a$ equal zero. We have

$4R_c+4(4)+6(2)+2(4)(2)=4R_b+2(4)(2)$

$4R_c+16+12=4R_b$

$R_b=R_c+7$ $\color{#D61F06}(2)$

Summation of forces vertical. We have

$R_a+R_b+R_c=6+2(4)+2(4)$

$R_a+R_b+R_c=22$ $\color{#D61F06}(3)$

Substituting $\color{#D61F06}(1)$ and $\color{#D61F06}(2)$ in $\color{#D61F06}(3)$ , we have

$15-2R_c+R_c+7+R_c=22$

$0=0$