This problem can be concluded as colorbox inside align, so we can use

\documentclass{article} \usepackage{xcolor} \usepackage{mathtools} % or \usepackage{amsmath} % Reference: https://tex.stackexchange.com/a/13693/234654 \newlength\dlf \newcommand\alignedbox[2]{ % #1 = before alignment % #2 = after alignment & \begingroup \settowidth\dlf{$\displaystyle #1$} \addtolength\dlf{\fboxsep+\fboxrule} \hspace{-\dlf} \fcolorbox{yellow}{yellow}{$\displaystyle #1 #2$} \endgroup } \begin{document} \begin{align*} %-- i^7 &= 1 \times i^3 \\[5mm] %-- i^7 &= i^3 \\[5mm] %-- i^7 &= i^2 \times i \\[5mm] %-- i^7 &= -1 \times i \\[5mm] %-- i^7 &= -i \\[5mm] %-- \alignedbox{i^7}{=-i} \end{align*} \end{document} 

This genius idea is borrowed from here.

This problem can be concluded as colorbox inside align, so we can use

\documentclass{article} \usepackage{xcolor} \usepackage{mathtools} % Reference: https://tex.stackexchange.com/a/13693/234654 \newlength\dlf \newcommand\alignedbox[2]{ % #1 = before alignment % #2 = after alignment & \begingroup \settowidth\dlf{$\displaystyle #1$} \addtolength\dlf{\fboxsep+\fboxrule} \hspace{-\dlf} \fcolorbox{yellow}{yellow}{$\displaystyle #1 #2$} \endgroup } \begin{document} \begin{align*} %-- i^7 &= 1 \times i^3 \\[5mm] %-- i^7 &= i^3 \\[5mm] %-- i^7 &= i^2 \times i \\[5mm] %-- i^7 &= -1 \times i \\[5mm] %-- i^7 &= -i \\[5mm] %-- \alignedbox{i^7}{=-i} \end{align*} \end{document} 

This genius idea is borrowed from here.

This method calculates the length before the aligned mark & into variable \dlf. Then \hspace is used to left shift the colorbox. Now the alignment is achieved.