ICw from the statics course

Draw the free body diagram $($FBD$)$ of the following figure then find the reaction forces at points A and point $B$

$2-$ Apply the equations of equilibrium & solve it

$\frac{{\displaystyle \underset{?}{\overset{?}{}}}{F}_y}{F_{Ay}+F_{By}-400-700}=0=0=0$

$F_{Bx}=0F_{Bx}=0$

$F_{Ay}+F_{BY}=400-7y{F}_{Ay}+F_{By}=1100$

$-544\mathrm{.}44+F_{By}=1100$

$+544\mathrm{.}44+544\mathrm{.}44$

$\frac{{\displaystyle \underset{?}{\overset{?}{}}}{M}_B}{F_{Ay}(4\mathrm{.}5)-400(3\mathrm{.}5)-700(1\mathrm{.}5)}=0=0=0$

$-$FAy$(4\mathrm{.}5)-1400-1050=0$

$\frac{F_{Ay}(4\mathrm{.}5)}{t^{4\mathrm{.}5}}=\frac{2450}{-4\mathrm{.}5}$

$F_{AY}=-544\mathrm{.}44$ Are the signs in the moment equation correct?