Creating Horizontal Lines With A Specific Distance

Question

I'm relatively new at LaTeX. I've read material online and have been practicing. However, I have not been able to find a solution to two of my questions. I'm hoping someone may be of assistance. This is what I have...

\begin{align}
-4(-5+6x) & = \hspace{1mm} 188  \\ 
(-4 \times -5) + (-4\times6x) & = \hspace{1mm} 188 \\
20-24x & = \hspace{1mm} 188 \\
-20\hspace{11.8mm}  & \hspace{3.5mm} -20 \\[-11pt]
\cline{1-2}
-24x & = \hspace{1mm} 168 \\
\dfrac{\cancel{-24} x}{\cancel{-24}} & =\dfrac{168}{-24}\\
x & = -7
\end{align}

My question is regarding the horizontal line. How can I minimize the distance? I would like for the horizontal line to line up, so to speak, directly under the -20 from line (4), that being the -20 on the left hand side of the equality. Also, I would like for it to end directly under the 0 from the -20 on line (4), that being the -20 on the right hand side. Furthermore, in line (6), how can I align the 4 from the right hand side of the denominator so that it is directly under the 8 of the numerator, that being the right hand side as well.

Thanks,

Miguel

Answer 1

Alignments are possible (even fractions) using array:

\documentclass{article}
\usepackage{mathtools,cancel}% http://ctan.org/pkg/{mathtools,cancel}
\newcommand{\DFrac}[3][c]{\begin{array}{#1}#2\\\hline#3\end{array}}
\begin{document}
[
  \setlength{\arraycolsep}{0pt}
  \renewcommand{\arraystretch}{1.2}
  \begin{array}{rr}
    -4(-5 + 6x) ={} & 188 \
    (-4 \times -5) + (-4 \times 6x) ={} & 188 \
    20 - 24x ={} & 188 \
    \multicolumn{2}{r}{\begin{array}{r}
      -20 \phantom{- 24x ={}} -20 \[0pt] \hline
    \end{array}} \
    -24x ={} & 168 \
    \DFrac{\cancel{-24} x}{\cancel{-24}} ={} & \DFrac[r]{168}{-24} \
    x ={} & -7
  \end{array}
]
\end{document}

Let's consider the code line-by-line

• \setlength{\arraycolsep}{0pt}

  Remove the gap (by setting it to 0pt) between columns in an array environment.

• \renewcommand{\arraystretch}{1.2}

  Increase the vertical spacing between rows inside a tabular/array. As reference, see Column and row padding in tables.

• \begin{array}{rr}

  Start an array (tabular-like structure in math mode) that will have 2 columns, both of which are right-aligned.

• -4(-5 + 6x) ={} & 188 \\

  First column content stretches up to &, while the second column stretches up to \\ - the line-breaking control. We use ={} since = is a binary relation. As such, it sets a specific space around it which differs from the space around a regular symbol, or a binary operator like + or -. As reference, see What is the difference between \mathbin vs. \mathrel? By providing an empty right-hand side {} for =, the appropriate spacing is established. Note that we set the column spacing to 0pt, so this spacing actually helps for making it work with 188.

• (-4 \times -5) + (-4 \times 6x) ={} & 188 \\

  Similar to the above.

• 20 - 24x ={} & 188 \\

  Similar to the above.

• \multicolumn{2}{r}{\begin{array}{r}

  Here we start a \multicolumn that will span 2 columns and will be right-aligned as well. Inside of it we will create another array that now only contains a single column, also right-aligned. Why is this? Well, we eventually want to draw a horizontal rule that spans only the with of the contents of this structure, so we contain it in its own array.

• -20 \phantom{- 24x ={}} -20 \\[0pt] \hline

  This sets -20, followed by a \phantom setting of -24x ={}. The \phantom setting creates a box (with width and height) that matches it's content but without printing it. Another way of viewing it would be that -24x ={} is set using white font colour. We also end the line with a [0pt] vertical gap, and insert a horizontal rule.

• \end{array}} \\

  The \multicolumn is ended (it held the nested array).

• -24x ={} & 168 \\

  Similar to what is mentioned above.

• \DFrac{\cancel{-24} x}{\cancel{-24}} ={} & \DFrac[r]{168}{-24} \\

  Similar to what is mentioned above, yet we now insert a \DFrac. \DFrac is defined as an array itself containing two elements stacked on top of one another and an \hline inbetween; similar to a rudimentary \frac. The optional argument [r] is used as a replacement for the default centred view of the \DFrac. That is, it aligns the numerator/denominator of the fraction to the right, rather than centred.

• x ={} & -7

  Similar to what was mentioned above.

• \end{array}

  End the outer array structure.

Answer 2

One point about position of -20 is unclear for me, but the rest should be as you are expecting, I hope. Rather unstardard usage of align, but works.

\documentclass{article}
\usepackage{amsmath}
\usepackage{cancel}
\begin{document}

\newlength\mxx
\settowidth{\mxx}{$-24$}
\begin{align}
-4(-5+6x)  = \hspace{1mm} 188 & \\ 
(-4 \times -5) + (-4\times6x)  = \hspace{1mm} 188 &\\
20-24x  = \hspace{1mm} 188 &\\
%-20\hspace{11.8mm}  & \hspace{3.5mm} -20 \\[-11pt]
-20\hspace{11.8mm}    -20 &\\[-14pt]
\cline{1-1}
-24x  = \hspace{1mm} 168 &\\
 \dfrac{\cancel{-24} x}{\cancel{-24}}  = \dfrac{\makebox[\mxx][r]{168}}{\makebox[\mxx][r]{$-24$}}\!&\\
x  = -7 &
\end{align}

\end{document}