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% Document Info
\author{kira}
\pdfinfo{
  /Title (linear-relationships.pdf)
  /Creator (Cheatography)
  /Author (kira)
  /Subject (Linear Relationships Cheat Sheet)
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\noindent
\begin{multicols}{3}
\begin{tabulary}{5.8cm}{C}
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    \vspace{-7pt}
    {\parbox{\dimexpr\textwidth-2\fboxsep\relax}{\noindent
        \hspace*{-6pt}\includegraphics[width=5.8cm]{/web/www.cheatography.com/public/images/cheatography_logo.pdf}}
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\begin{tabulary}{11cm}{L}
    \vspace{-2pt}\large{\bf{\textcolor{DarkBackground}{\textrm{Linear Relationships Cheat Sheet}}}} \\
    \normalsize{by \textcolor{DarkBackground}{kira} via \textcolor{DarkBackground}{\uline{cheatography.com/46506/cs/13512/}}}
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\noindent
\begin{multicols}{3}
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  \mymulticolumn{2}{p{5.377cm}}{\bf\textcolor{white}{Cheatographer}}  \\
  \vspace{-2pt}kira \\
  \uline{cheatography.com/kira} \\
  \end{tabulary}
\vfill
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  \mymulticolumn{1}{p{5.377cm}}{\bf\textcolor{white}{Cheat Sheet}}  \\
   \vspace{-2pt}Published 12th November, 2017.\\
   Updated 12th November, 2017.\\
   Page {\thepage} of \pageref{LastPage}.
\end{tabulary}
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  \vspace{-5pt}
  %\includegraphics[width=48px,height=48px]{dave.jpeg}
  Measure your website readability!\\
  www.readability-score.com
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\begin{document}
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\begin{multicols*}{3}

\begin{tabularx}{5.377cm}{X}
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\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Solving Problems Using Linear Equations}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{{\bf{Worked Example 6:}} \newline % Row Count 1 (+ 1)
Mira is twice Sanjin's age. How old are Mira and Sanjin now if in 10 years the sum of their ages will be 89. Use {\emph{s}} to represent Sanjin's age.  \newline % Row Count 4 (+ 3)
	Sanjin's Age: s				Mira's present age: 2s \newline % Row Count 5 (+ 1)
		In 10 years: \newline % Row Count 6 (+ 1)
		Sanjin = s + 10		Mira's age = 2s + 10 \newline % Row Count 7 (+ 1)
				(s+10) + (2s+10) = 89 \newline % Row Count 8 (+ 1)
					      3s + 20 = 89 \newline % Row Count 9 (+ 1)
			  		   	      3s = 69 \newline % Row Count 10 (+ 1)
						       s = 23 \newline % Row Count 11 (+ 1)
		2s = 2 x 23 = 46 \newline % Row Count 12 (+ 1)
Sanjin is 23 years old and Mira is 46 years old. In 10 years they will be 33 and 56 old respectively; 33 + 56= 89% Row Count 15 (+ 3)
} \tn 
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\begin{tabularx}{5.377cm}{x{2.4885 cm} x{2.4885 cm} }
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\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Distance and Midpoint}}  \tn
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{\bf{ Distance Formula}} & {\bf{Midpoint Formula}} \tn 
% Row Count 2 (+ 2)
% Row 1
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The distance, {\emph{d}}, between any two Cartesian coordinates (x1, y1) and (x2, y2) is: & The midpoint M(x,y) between 2 points (x1, y1) and (x2, y2) is given by: \tn 
% Row Count 7 (+ 5)
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d = & M(x,y) = \tn 
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\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Gradient}}  \tn
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\mymulticolumn{1}{x{5.377cm}}{The gradient of a line can be found by the ratio of the vertical {\bf{rise}} to the horizontal {\bf{run}} between any two points on the line. \newline % Row Count 3 (+ 3)
Gradient AB = (vertical rise)/(horizontal run) \newline % Row Count 4 (+ 1)
The letter {\emph{m}} is used to represent the gradient. {\bf{y=mx+b}} \newline % Row Count 6 (+ 2)
          m = (rise)/(run) \newline % Row Count 7 (+ 1)
                = (change in {\emph{y}}-value)/(change in {\emph{x}}-value) \newline % Row Count 9 (+ 2)
              = (y2-y1)/(x2-x1)% Row Count 10 (+ 1)
} \tn 
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\begin{tabularx}{5.377cm}{X}
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\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Linear Graphs: Gradient and Y-Intercept}}  \tn
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\mymulticolumn{1}{x{5.377cm}}{The {\bf{gradient-intercept form:}} \newline % Row Count 1 (+ 1)
     {\emph{y=mx+b}}, where {\emph{m}} is the gradient and {\emph{b}} is the y-intercept. \newline % Row Count 3 (+ 2)
The {\bf{general form:}} \newline % Row Count 4 (+ 1)
      {\emph{ax+by+c=0}}, where {\emph{a}}, {\emph{b}} and {\emph{c}} are constants and {\emph{a}} ≠ 0, b ≠ 0.  \newline % Row Count 6 (+ 2)
In this equation, the gradient {\emph{m}} = (-a)/(b) and the y-intercept = (-c)/(b)can be found by rearranging the equation to make {\emph{y}} the subject. \newline % Row Count 9 (+ 3)
{\bf{Worked Example 15}} \newline % Row Count 10 (+ 1)
{\emph{(a)}} y = -3x+2 \newline % Row Count 11 (+ 1)
gradient = -3 \newline % Row Count 12 (+ 1)
y-intercept: (0,2) \newline % Row Count 13 (+ 1)
{\emph{(b)}} y = x-1 \newline % Row Count 14 (+ 1)
= 1x+(-1) \newline % Row Count 15 (+ 1)
gradient = 1 \newline % Row Count 16 (+ 1)
y-intercept: (0,-1) \newline % Row Count 17 (+ 1)
{\emph{(c)}} 2y = 5x - 4 \newline % Row Count 18 (+ 1)
y = (5)/(2)x - 2 \newline % Row Count 19 (+ 1)
gradient = (5)/(2) \newline % Row Count 20 (+ 1)
y-intercept: (0,-2) \newline % Row Count 21 (+ 1)
{\emph{(d)}} y+2x =6 \newline % Row Count 22 (+ 1)
y = 6-2x \newline % Row Count 23 (+ 1)
y = -2x+6 \newline % Row Count 24 (+ 1)
gradient: -2 \newline % Row Count 25 (+ 1)
y-intercept: (0,6)% Row Count 26 (+ 1)
} \tn 
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\begin{tabularx}{5.377cm}{X}
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\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Sketching Linear Graphs}}  \tn
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\mymulticolumn{1}{x{5.377cm}}{{\bf{Graphing using the y-intercept and gradient}} \newline % Row Count 1 (+ 1)
There is no need to plot a number of points to graph linear relationships. Because only 2 points are needed to define a line, use the y-intercept as one point on the graph and the gradient, {\emph{m}} to locate the second point. Once 2 points are known, a straight line can be drawn through them. \newline % Row Count 7 (+ 6)
{\bf{Sketching other Linear Graphs, y=mx+b, b ≠ 0}} \newline % Row Count 9 (+ 2)
If the y-intercept is not the origin, then first identify the y-intercept from the equation before using the gradient to find the second point.  \newline % Row Count 12 (+ 3)
Using the equation {\emph{y=mx+b}} and substituting x = 0, you get y = b. \newline % Row Count 14 (+ 2)
x = 0 is the x=coordinate of a point on the y-axis, so {\emph{b}} is the y-value at the point at the point at which the graph crosses the y-axis. This is the y-intercept \newline % Row Count 18 (+ 4)
{\bf{Worked Example 17}} \newline % Row Count 19 (+ 1)
{\emph{Use the y-intercept and the gradient to sketch the graph of y = 2x+1}} \newline % Row Count 21 (+ 2)
y-intercept: (0,1) \newline % Row Count 22 (+ 1)
Gradient = 2 \newline % Row Count 23 (+ 1)
= (2)/(1) = (rise)/(run)% Row Count 24 (+ 1)
} \tn 
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\begin{tabularx}{5.377cm}{X}
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\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Parallel and Perpendicular Lines}}  \tn
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\mymulticolumn{1}{x{5.377cm}}{{\bf{Worked Example 19}} \newline % Row Count 1 (+ 1)
{\emph{Find the gradient of the line parallel to each of the following}} \newline % Row Count 3 (+ 2)
{\emph{(a)}} y = 2x-5 \newline % Row Count 4 (+ 1)
m = 2 \newline % Row Count 5 (+ 1)
Any parallel line will have a gradient of 2. \newline % Row Count 6 (+ 1)
{\emph{(b)}} 2x + 3y = 6 \newline % Row Count 7 (+ 1)
2x-2x+3y = 6 - 2x \newline % Row Count 8 (+ 1)
(3y)/(3) = - (2)/(3)x + (6)/(3) \newline % Row Count 9 (+ 1)
y = - (2)/(3)x + 2 \newline % Row Count 10 (+ 1)
m1 =  - (2)/(3) \newline % Row Count 11 (+ 1)
Any parallel line will have a gradient of - (2)/(3).  \newline % Row Count 13 (+ 2)
{\bf{Worked Example 20}} \newline % Row Count 14 (+ 1)
{\emph{Find the gradient of a line perpendicular to each of the following}} \newline % Row Count 16 (+ 2)
{\emph{(a)}} y = 4x + 3 \newline % Row Count 17 (+ 1)
m1 = 4 \newline % Row Count 18 (+ 1)
m2 = - (1)/(m1) \newline % Row Count 19 (+ 1)
= - (1)/(4) \newline % Row Count 20 (+ 1)
{\emph{(b)}} 3x-5y = 10 \newline % Row Count 21 (+ 1)
3x-3x-5y = 10-3x \newline % Row Count 22 (+ 1)
-5y = -3x + 10 \newline % Row Count 23 (+ 1)
(-5y)/(-5) =  (-3x)/(-5) + (10)/(-5) \newline % Row Count 24 (+ 1)
y = (3)/(5)x - 2     m1= (3)/(5) \newline % Row Count 25 (+ 1)
m2 = - (1)/(m1)    = -(5)/(3)% Row Count 26 (+ 1)
} \tn 
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% That's all folks
\end{multicols*}

\end{document}