I have the code:

```
\begin{flalign} 
&f_1(y_i; \boldsymbol{\theta}_1) = \lambda e^{-\lambda y_i} \\
&l_1(\boldsymbol{\theta}_1; \textbf{y}, \textbf{x}) = `r num.not.censored` 
  \log{(\lambda)} - \lambda \sum_{i=1}^{`r num.data`} y_i \\
&f_2(y_i; x_i, \boldsymbol{\theta}_2) = e^{\beta_1 + \beta_2 x_i} 
  \exp{(-e^{\beta_1 + \beta_2 x_i}y_i)} \\
&l_2(\boldsymbol{\theta}_2; \textbf{y}, \textbf{x}) = 
  \sum_{i=1}^{`r num.not.censored`} (\beta_1 + \beta_2 x_i) 
 -\sum_{i=1}^{`r num.data`} (e^{\beta_1 + \beta_2 x_i}y_i) \\
&f_3(y_i; x_i, \boldsymbol{\theta}_3) = \frac{1}{\sqrt{2\pi\sigma^2}} 
   \exp{\bigg(\frac{-(y_i - (\gamma_1 + \gamma_2 x_i))^2}{2\sigma^2}\bigg)} \\
&l_3(\boldsymbol{\theta}_3; \textbf{y}, \textbf{x}) = - 28 
   \log(2 \pi \sigma^2) - \sum_{i=1}^{`r num.not.censored`} 
   \bigg(\frac{(y_i - (\gamma_1 + \gamma_2 x_i))^2}{2 \sigma^2}\bigg)
   \space + \\ 
&\sum_{i=`r num.not.censored + 1`}^{`r num.data`} \log 
   \bigg(\frac{1}{2} - \frac{1}{2}\text{erf}\Big(\frac{y_i - 
   (\gamma_1 + \gamma_2 x_i)}{\sigma \sqrt{2}}\Big)\bigg)
\end{flalign}

```

It renders as:

[![enter image description here](https://i.stack.imgur.com/sSBrG.png)](https://i.stack.imgur.com/sSBrG.png)

How can I shift line 7 to the right, so that it starts at equals sign of line 6?

I also want to remove the equation (7) label, since it is still part of equation (6).

Thanks,

Jack