# Vertical alignment of different systems of equations

I would like to have the following systems horizontally aligned (the misalignment is due to some of the equations having a fraction, others not). Basically, I want (for example) all the $q_1$ on the same line. Thanks in advance for your help!

Here it is the code

\begin{tcolorbox}[title={2. Calcolo $0+X\cdot a=b$}, colback=white,colframe=black!20,coltitle=black]
\eqal{&\hs{1}0&+X\cdot &\hs{1}a&=&\hs{1}b&\hs{.2}\implies&\hs{1}\delta\mathcal{L}^{ab}_{p/c,i}\\
&\left\{
\begin{aligned}
q_1^0&=0\\
q_2^0&=0\\
q_3^0&=0\\
q_4^0&=-\frac{T}{h}\\
P_1^0&=-\frac{TL}{h}\\
P_2^0&=0\\
P_3^0&=0\\
P_4^0&=\frac{TL}{h}
\end{aligned}
\right.&+X\cdot\hs{.2}
&\left\{
\begin{aligned}
q_1^a&=-\frac{1}{2L}\\
q_2^a&=\frac{1}{2L}\\
q_3^a&=-\frac{1}{2L}\\
q_4^a&=\frac{1}{2L}\\
P_1^a&=1\\
P_2^a&=-1\\
P_3^a&=1\\
P_4^a&=-1
\end{aligned}
\right.&=\hs{.2}
&\left\{
\begin{aligned}
q_1^b&=-\frac{X}{2L}\\
q_2^b&=\frac{X}{2L}\\
q_3^b&=-\frac{X}{2L}\\
q_4^b&=-\frac{T}{h}+\frac{X}{2L}\\
P_1^b&=-\frac{TL}{h}+X\\
P_2^b&=-X\\
P_3^b&=X\\
P_4^b&=\frac{TL}{h}-X
\end{aligned}
\right.\notag
&\hs{.2}&\hs{.2}\left\{
\begin{aligned}
\delta\mathcal{L}_{c1}^{ab}&=\frac{Lb}{Gs}\left(-\frac{1}{2L}\right)\left(-\frac{X}{2L}\right)\\
\delta\mathcal{L}_{c2}^{ab}&=\frac{Lh}{Gs}\left(\frac{1}{2L}\frac{X}{2L}\right)\\
\delta\mathcal{L}_{c3}^{ab}&=\frac{Lb}{Gs}\left(-\frac{1}{2L}\right)\left(-\frac{X}{2L}\right)\\
\delta\mathcal{L}_{c4}^{ab}&=\frac{Lh}{Gs}\left(\frac{1}{2L}\right)\left(-\frac{T}{h}+\frac{X}{2L}\right)\\
\delta\mathcal{L}_{p1}^{ab}&=\frac{L}{3EA}\left[1\left(-\frac{TL}{h}+X\right)\right]\\
\delta\mathcal{L}_{p2}^{ab}&=\frac{L}{3EA}\left(-1\right)\left(-X\right)\\
\delta\mathcal{L}_{p3}^{ab}&=\frac{L}{3EA}\left(1\cdot X\right)\\
\delta\mathcal{L}_{p4}^{ab}&=\frac{L}{3EA}\left[-1\left(\frac{TL}{h}-X\right)\right]\\
\end{aligned}
\right.
}
\end{tcolorbox}


Please note \hs{X} is \hspace{Xcm} (a newcommand*) and \eqal{X} is \begin{eqnarray} \left{ \begin{aligned} X \end{aligned} \right. \end{eqnarray}

• welcome to tex.se! what you try so far? please show this! it is not fun to retype your equations from scratch. help us to help you! Aug 26, 2018 at 16:23
• Right, I'm sorry but I was not sure how to paste the code. Here it is Aug 26, 2018 at 16:28
• Thank you!!! (adding exclamation marks so I have enough characters :) ) Aug 26, 2018 at 16:51
• thank you for code snippet, but i expect complete document beginning with \documentclass ... and ending with \end{document} with preamble loaded with only necessary package for compiling your code snippet. it seems that it contain errors, please check again. Aug 26, 2018 at 17:37

\documentclass{article}
\usepackage{amsmath,array}

\newcommand{\phnf}{\vphantom{\left(\frac{A}{B}\right)}}

\begin{document}
\begin{array}{ @{}c @{}>{{}}c<{{}} @{}c @{}>{{}}c<{{}} @{}c @{}>{{}}c<{{}} @{}c @{}>{{}}c<{{}} @{}c } 0 &+& X &\cdot& a&=& b &\implies&\delta\mathcal{L}^{ab}_{p/c,i} \\[4ex] \begin{cases} \begin{aligned} q_1^0&=0 \phnf \\ q_2^0&=0 \phnf \\ q_3^0&=0 \phnf \\ q_4^0&=-\frac{T}{h} \phnf \\ P_1^0&=-\frac{TL}{h} \phnf \\ P_2^0&=0 \phnf \\ P_3^0&=0 \phnf \\ P_4^0&=\frac{TL}{h} \phnf \end{aligned} \end{cases} &+& X &\cdot& \begin{cases} \begin{aligned} q_1^a&=-\frac{1}{2L} \phnf \\ q_2^a&=\frac{1}{2L} \phnf \\ q_3^a&=-\frac{1}{2L} \phnf \\ q_4^a&=\frac{1}{2L} \phnf \\ P_1^a&=1 \phnf \\ P_2^a&=-1 \phnf \\ P_3^a&=1 \phnf \\ P_4^a&=-1 \phnf \end{aligned} \end{cases} &=& \begin{cases} \begin{aligned} q_1^b&=-\frac{X}{2L} \phnf \\ q_2^b&=\frac{X}{2L} \phnf \\ q_3^b&=-\frac{X}{2L} \phnf \\ q_4^b&=-\frac{T}{h}+\frac{X}{2L} \phnf \\ P_1^b&=-\frac{TL}{h}+X \phnf \\ P_2^b&=-X \phnf \\ P_3^b&=X \phnf \\ P_4^b&=\frac{TL}{h}-X \phnf \end{aligned} \end{cases} && \begin{cases} \begin{aligned} \delta\mathcal{L}_{c1}^{ab}&=\frac{Lb}{Gs}\left(-\frac{1}{2L}\right)\left(-\frac{X}{2L}\right) \phnf \\ \delta\mathcal{L}_{c2}^{ab}&=\frac{Lh}{Gs}\left(\frac{1}{2L}\frac{X}{2L}\right) \phnf \\ \delta\mathcal{L}_{c3}^{ab}&=\frac{Lb}{Gs}\left(-\frac{1}{2L}\right)\left(-\frac{X}{2L}\right) \phnf \\ \delta\mathcal{L}_{c4}^{ab}&=\frac{Lh}{Gs}\left(\frac{1}{2L}\right)\left(-\frac{T}{h}+\frac{X}{2L}\right) \phnf \\ \delta\mathcal{L}_{p1}^{ab}&=\frac{L}{3EA}\left[1\left(-\frac{TL}{h}+X\right)\right] \phnf \\ \delta\mathcal{L}_{p2}^{ab}&=\frac{L}{3EA}\left(-1\right)\left(-X\right) \phnf \\ \delta\mathcal{L}_{p3}^{ab}&=\frac{L}{3EA}\left(1\cdot X\right) \phnf \\ \delta\mathcal{L}_{p4}^{ab}&=\frac{L}{3EA}\left[-1\left(\frac{TL}{h}-X\right)\right] \phnf \end{aligned} \end{cases} \end{array}

\end{document}


And never ever use eqnarray.

It's up to you having a suitable line width to contain the thing.

• Thank you very much!! I will fix my super-uber use of eqnarray asap :) Aug 26, 2018 at 17:57