Question

S=align="absmiddle" class="latex-formula">
(a+l) solve for a

Answer 1

$\bf S=\cfrac{n}{2}(a+I)\implies S=\cfrac{n(a+I)}{2}\implies 2S=n(a+I)\implies \stackrel{\textit{distributing}}{2S=na+nI} \\\\\\ 2S-nI=na\implies \cfrac{2S-nI}{n}=a$