How to fix overfull \hbox in multiline equation

The equation is simply longer than the allowable horizontal space, split it over five lines for readability. When splitting lines, you should have the same virtual space as if the = sign is present in all lines. To do this, insert \phantom{{}={}} at the beginning of lines without =. Moreover, the two braces {}={} are added because = is a binary operator just like + or - and these braces act as math atoms for correct spacing around the = sign.

\documentclass[letterpaper,12pt]{article}
\addtolength{\oddsidemargin}{-.5in}
\addtolength{\evensidemargin}{1in}
\addtolength{\textwidth}{1.15in}
\addtolength{\topmargin}{-.75in}
\addtolength{\textheight}{1in}
\usepackage{amsmath}
\begin{document}
\begin{align}
    \nonumber E_{AH}^* &= (S_b^*, I_b^*, S_h^*, I_a^*, R_h^*) \\ \nonumber
    &= \Big(\frac{\mu_b + \delta_b}{\beta_B}, \frac{\Lambda_b\beta_B - \mu_b(\mu_b+\delta_b)}{\beta_B(\mu_b + \delta_b)},\\ \nonumber 
    &\phantom{{}={}} \frac{\Lambda_h\beta_B(\mu_b+\delta_b)}{\mu_h\beta_B(\mu_b+\delta_b)+\beta_{BH}\lbrack\Lambda_b\beta_B-\mu_b(\mu_b+\delta_b)\rbrack}, \\ \nonumber
    &\phantom{{}={}} \frac{\Lambda_h\beta_{BH}\lbrack\Lambda_b\beta_B-\mu_b(\mu_b+\delta_b)\rbrack}{(\mu_h+d+\gamma_a)\big\lbrack\mu_h\beta_B(\mu_b+\delta_b)+\beta_{BH}\big(\Lambda_b\beta_B-\mu_b(\mu_b+\delta_b)\big)\big\rbrack},\\
    &\phantom{{}={}} \frac{\gamma_a\Lambda_h\beta_{BH}\lbrack\Lambda_b\beta_B-\mu_b(\mu_b+\delta_b)\rbrack}{\mu_h(\mu_h+d+\gamma_a)\big\lbrack\mu_h\beta_B(\mu_b+\delta_b)+\beta_{BH}\big(\Lambda_b\beta_B-\mu_b(\mu_b+\delta_b)\big)\big\rbrack}\Big). \label{Eqn: EEAH}
\end{align}
\end{document}

enter image description here


Your readers will have a hard time in finding the correspondence between symbols and their expansion. I'd suggest centering the main equation and explain the symbols below it.

\documentclass[letterpaper,12pt]{article}
\usepackage{amsmath}

\begin{document}

\begin{gather}
E_{AH}^* = (S_b^*, I_b^*, S_h^*, I_a^*, R_h^*) \label{Eqn: EEAH} \\
\begin{align*}
S_b^* &= \frac{\mu_b + \delta_b}{\beta_B} \\
I_b^* &= \frac{\Lambda_b\beta_B - \mu_b(\mu_b+\delta_b)}{\beta_B(\mu_b + \delta_b)} \\
S_h^* &=
  \frac{\Lambda_h\beta_B(\mu_b+\delta_b)}
       {\mu_h\beta_B(\mu_b+\delta_b)+\beta_{BH}[\Lambda_b\beta_B-\mu_b(\mu_b+\delta_b)]}
  \\
I_a^* &=
  \frac{\Lambda_h\beta_{BH}[\Lambda_b\beta_B-\mu_b(\mu_b+\delta_b)]}
       {(\mu_h+d+\gamma_a)\bigl[\mu_h\beta_B(\mu_b+\delta_b)+\beta_{BH}
        \bigl(\Lambda_b\beta_B-\mu_b(\mu_b+\delta_b)\bigr)\bigr]}
  \\
R_h^* &=
  \frac{\gamma_a\Lambda_h\beta_{BH}[\Lambda_b\beta_B-\mu_b(\mu_b+\delta_b)]}
       {\mu_h(\mu_h+d+\gamma_a)\bigl[\mu_h\beta_B(\mu_b+\delta_b)+
        \beta_{BH}\bigl(\Lambda_b\beta_B-\mu_b(\mu_b+\delta_b)\bigr)\bigr]}. 
\end{align*}
\end{gather}

\end{document} 

enter image description here


I would (a) introduce one more line break and (b) align all rows vertically (with offsets) relative to the first = symbol. I would also increase the space between successive rows a bit and increase the size of the outermost parenthesis from \Big to \bigg.

enter image description here

\documentclass[letterpaper,12pt]{article}
\addtolength{\oddsidemargin}{-.5in}
\addtolength{\evensidemargin}{1in}
\addtolength{\textwidth}{1.15in}
\addtolength{\topmargin}{-.75in}
\addtolength{\textheight}{1in}
\usepackage{amsmath}
\begin{document}
\begin{align}
E_{AH}^* &= (S_b^*, I_b^*, S_h^*, I_a^*, R_h^*) \notag \\[1ex]
&= \biggl(\frac{\mu_b + \delta_b}{\beta_B}, 
   \frac{\Lambda_b\beta_B - \mu_b(\mu_b+\delta_b)}{\beta_B(\mu_b + \delta_b)}, \notag \\[1.5ex]
&\qquad \frac{\Lambda_h\beta_B(\mu_b+\delta_b)}{\mu_h\beta_B(\mu_b+\delta_b)+\beta_{BH}\lbrack\Lambda_b\beta_B-\mu_b(\mu_b+\delta_b)\rbrack}, \notag \\[1.5ex]
&\qquad \frac{\Lambda_h\beta_{BH}\lbrack\Lambda_b\beta_B-\mu_b(\mu_b+\delta_b)\rbrack}{(\mu_h+d+\gamma_a)\big\lbrack\mu_h\beta_B(\mu_b+\delta_b)+\beta_{BH}\bigl(\Lambda_b\beta_B-\mu_b(\mu_b+\delta_b)\bigr)\big\rbrack}, \notag \\[1.5ex]
&\qquad\frac{\gamma_a\Lambda_h\beta_{BH}\lbrack\Lambda_b\beta_B-\mu_b(\mu_b+\delta_b)\rbrack}{\mu_h(\mu_h+d+\gamma_a)\big\lbrack\mu_h\beta_B(\mu_b+\delta_b)+\beta_{BH}\bigl(\Lambda_b\beta_B-\mu_b(\mu_b+\delta_b)\bigr)\big\rbrack}\biggr). \label{Eqn: EEAH}
\end{align}
\end{document}