How can I align equations belong to different { part? [closed] - latex

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I am a beginner in latex. I have the following piece of latex code. The code is working fine but I wish all the equality operators and the "if kflag=n" of each equation be aligned and written in one equation box with one equation counting number. How can it be done?
\begin{equation} %kflag=0
\left \{
\begin{array}{rl}
T =& (1-D)\sigma_{max,0}\times e^{1-\frac{\Delta_n}{\delta_n}}\times \frac{\Delta_n}{\delta_n}\\
K =& \frac{(1-D)\sigma_{max,0}}{\delta_n}\times e^{1-\frac{\Delta_n}{\delta_n}}\times (1-\frac{\Delta_n}{\delta_n})\\
\end{array}
\right.
\quad \text{if} \quad kflag=0
\end{equation}
\begin{equation} %kflag=1
\left \{
\begin{array}{rl}
T =& \alpha\sigma_{max,0}\times e^{1-\frac{\Delta_n}{\delta_n}}\times \frac{\Delta_n}{\delta_n}\\
K =& \frac{\alpha\sigma_{max,0}}{\delta_n}\times e^{1-\frac{\Delta_n}{\delta_n}}\times (1-\frac{\Delta_n}{\delta_n})\\
\end{array}
\right.
\quad \text{if} \quad kflag=1
\end{equation}
\begin{equation} %kflag=2
\left \{
\begin{array}{rl}
T =& (1-D)\sigma_{max,0}\times e^{1-\frac{\Delta_n}{\delta_n}}\times \frac{\Delta_n}{\delta_n}+T_{max}-\\
&(1-D)\sigma_{max,0}\times e\times\frac{\Delta_{max}}{\delta_n}+\\
&10\times \sigma_{max,0}\times e^{1-\frac{\Delta_n}{\delta_n}}\times \frac{\Delta_n}{\delta_n}\\
K =& 11\times\frac{(1-D)\sigma_{max,0}}{\delta_n}\times e^{1-\frac{\Delta_n}{\delta_n}}\times (1-\frac{\Delta_n}{\delta_n})\\
\end{array}
\right.
\quad \text{if} \quad kflag=2
\end{equation}
\begin{equation} %kflag=3
\left \{
\begin{array}{rl}
T =& T_{max}+K\times(\Delta_n-\Delta_{max})\\
K =& \frac{(1-D)\sigma_{max,0}*e}{\delta_n}\\
\end{array}
\right.
\quad \text{if} \quad kflag=3
\end{equation}
Right now the equations look like this

Answer 1

Here's an option using a multitude of nested structures - equation for the numbering, aligned for the horizontal alignment of structures and dcases (or cases) for the left-braced content.
\documentclass{article}
\usepackage{mathtools}
\begin{document}
\newcommand{\Ddn}{\frac{\Delta_n}{\delta_n}}
\newcommand{\smz}{\sigma_{\mathrm{max}, 0}}
\begin{equation}
\begin{aligned}
&\begin{dcases}
\phantom{K}\mathllap{T} = (1 - D) \smz \times e^{1 - \Ddn} \times \Ddn \\
K = \frac{(1 - D) \smz}{\delta_n} \times e^{1 - \Ddn} \times \bigl( 1 - \Ddn \bigr)
\end{dcases} & \text{if $k$-flag} = 0 \\ % k-flag = 0
&\begin{dcases}
\phantom{K}\mathllap{T} = \alpha \smz \times e^{1 - \Ddn} \times \Ddn \\
K = \frac{\alpha \smz}{\delta_n} \times e^{1 - \Ddn} \times \bigl( 1 - \Ddn \bigr)
\end{dcases} & \text{if $k$-flag} = 1 \\ % k-flag = 1
&\begin{dcases}
\phantom{K}\mathllap{T} = \begin{aligned}[t]
&(1 - D) \smz \times e^{1 - \Ddn} \times \Ddn + T_{\mathrm{max}} \\
&{} - (1 - D) \smz \times e \times \frac{\Delta_{\mathrm{max}}}{\delta_n} \\
&{} + 10 \times \smz \times e^{1 - \Ddn} \times \Ddn
\end{aligned} \\
K = 11 \times \frac{(1 - D) \smz}{\delta_n} \times e^{1 - \Ddn} \times \bigl( 1 - \Ddn \bigr)
\end{dcases} & \text{if $k$-flag} = 2 \\ % k-flag = 2
&\begin{dcases}
\phantom{K}\mathllap{T} = T_{\mathrm{max}} + K \times (\Delta_n - \Delta_{\mathrm{max}}) \\
K = \frac{(1 - D) \smz \times e}{\delta_n}
\end{dcases} & \text{if $k$-flag} = 3
\end{aligned}
\end{equation}
\end{document}