2

I would like to align equations on the right side to this imaginary red line (to the left). Also, it would be really nice to have just one number of the equation (just like in equation environment) instead of three.

I tried to use cases but then I couldn't use align.

If you have better ideas how to clearly present following system of equations, then I would be grateful for the suggestions.

\begin{align} 
\sum F_x: & & -Q \cdot \cos \gamma - L \cdot \sin \beta + R_{B_x} = 0\\
\sum F_y: & & Q \sin \gamma + L \cos \beta + R_{B_y} - mg = 0 \\ 
\sum (M_z)_B: & & mgh \cdot \cos \beta - Lh - Qa \cdot \cos(90 - \beta - \gamma) = 0
\end{align}

Current status

  • A single & should be sufficient for aligning, this will bring the RHS next to the colon. You can then manually add the same spacing to each line – Dai Bowen Jun 18 '16 at 17:08
1

To typeset a multi-line system of equations with a single, vertically centered equation number, you might use a split environment inside an equation environment:

enter image description here

\documentclass{article}
\usepackage{amsmath}
\begin{document}
\begin{equation} \begin{split}
\sum F_x &\colon {-}Q  \cos \gamma - L  \sin \beta + R_{B_x} = 0\\
\sum F_y &\colon Q \sin \gamma + L \cos \beta + R_{B_y} - mg = 0 \\ 
\sum (M_z)_B &\colon mgh  \cos \beta - Lh - Qa  \cos(90 - \beta - \gamma) = 0
\end{split} \end{equation}
\end{document}
1

OK, I came up with following solution:

\begin{subequations} \label{eq:equilibrium}
\begin{align} 
&\sum F_x: &&  -Q \cdot \cos \gamma - L \cdot \sin \beta + R_{B_x} = 0\\
&\sum F_y: &&    Q \sin \gamma + L \cos \beta + R_{B_y} - mg = 0 \\ 
&\sum (M_z)_B: &&  mgh \cdot \cos \beta - Lh - Qa \cdot \cos(90 - \beta - \gamma) = 0
\end{align}
\end{subequations}

that results with following: enter image description here

This satisfies my needs, but if anyone has better solutions, then please share it :)

1

You can do that with alignedat:

\documentclass{article}
\usepackage{mathtools}

\begin{document}

\begin{equation}
    \begin{alignedat}{2}
      ∑ F_x: & \qquad & & \mathllap{ -}Q · \cos γ- L · \sin β+ R_{B_x} = 0 \\
      ∑ F_y: & & & Q · \sin γ+ L · \cos β+ R_{B_y} - mg = 0 \\
      ∑ (M_z)_B: & & & mgh · \cos β- Lh - Qa · \cos(90 - β- γ) = 0
    \end{alignedat}
\end{equation}

Current status

\end{document} 

enter image description here

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