From 723b6ded0066651ac8521ce1320371c8430b1434 Mon Sep 17 00:00:00 2001 From: Christopher Baines Date: Tue, 1 Apr 2014 19:10:41 +0100 Subject: Initial commit --- linear.tex | 360 +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ 1 file changed, 360 insertions(+) create mode 100644 linear.tex (limited to 'linear.tex') diff --git a/linear.tex b/linear.tex new file mode 100644 index 0000000..7710933 --- /dev/null +++ b/linear.tex @@ -0,0 +1,360 @@ +\documentclass[12pt,a4paper,twocolumn,landscape]{exam} + +\usepackage{xparse} +\usepackage{tikz} + \usetikzlibrary{arrows,calc,fit} +\usepackage{mathtools} +\usepackage{fullpage} +\usepackage{todonotes} +\usepackage{float} +\usepackage[compact,explicit]{titlesec} % http://ctan.org/pkg/titlesec +\usepackage[utf8]{inputenc} +\usepackage{titling} +\usepackage{wrapfig} +\usepackage{boxedminipage} +\usepackage{parskip} % Don't indent paragraphs + +\usepackage{fib} +\usepackage{common} + +\pagestyle{headandfoot} + +\firstpageheadrule +\firstpageheader{Solving Linear Equations} + {Level 7-8} + {\today} + +\runningheadrule +\runningheader{Solving Linear Equations} + {Solving Linear Equations, Page \thepage\ of \numpages} + {\today} + +\firstpagefooter{}{}{} +\runningfooter{}{}{} + +\begin{document} + +\section*{Solving Linear Equations - Level 7-8} + +\begin{minipage}{\linewidth} + +\begin{wrapfigure}{R}{0.3\linewidth} + \centering + \begin{boxedminipage}{\linewidth} + Keywords: + \begin{itemize} + \item Linear + \item Solve + \item Unknown + \end{itemize} + \end{boxedminipage} +\end{wrapfigure} + +Linear equations are written with one unknown variable which is shown by a +letter. + +It is necessary to \textbf{rearrange} the equation in order to find the value +of x. + +\end{minipage} + +\begin{questions} + +\question Solve 2x+3 = 7 for the unknown. + +\begin{minipage}{\linewidth} + + \begin{wrapfigure}{R}{0.3\linewidth} + \vspace{-25pt} + \centering + \begin{tikzpicture}[->,>=stealth',shorten >=1pt,auto,node distance=1.5cm, + thick,main node/.style={draw=none,fill=none}] + + \node[main node] (1) {$2x + 3 = 7$}; + \node[main node] (2) [below of=1] {$2x = 4$}; + + \path[every node/.style={font=\sffamily\small}] + (1.west) edge [bend right] node[left]{$-3$} (2.west) + (1.east) edge [bend left] node[right] {$-3$} (2.east); + \end{tikzpicture} + \end{wrapfigure} + + Start by looking at the section of the equation that contains the unknown. In + this case the left hand side. + +\end{minipage} + +The next step is to collect all constants (numbers) on the side of the equation +that does not contain the unknown. We want to move the 3 to the right hand +side. Because it is +3, to move it to the other side we must -3. + +\begin{minipage}{\linewidth} + + \begin{wrapfigure}{R}{0.3\linewidth} + \centering + \begin{tikzpicture}[->,>=stealth',shorten >=1pt,auto,node distance=1.5cm, + thick,main node/.style={draw=none,fill=none}] + + \node[main node] (1) {$2x = 4$}; + \node[main node] (2) [below of=1] {$x = 2$}; + + \path[every node/.style={font=\sffamily\small}] + (1.west) edge [bend right] node[left]{$\div 2$} (2.west) + (1.east) edge [bend left] node[right] {$\div 2$} (2.east); + \end{tikzpicture} + \end{wrapfigure} + + We can see that there are still numbers on the right hand side with the + unknown. The coefficient (multiplier) of x is 2. Since the 2 is times x we + need to divide by 2 to get the x by itself. + +\end{minipage} + +We can see that there are no more numbers on the side of x so the solution is: +$x = 2$ + +To check the solution, put the unknown back into the original equation. $2 +\times 2 + 3 = 7$ + +\begin{keypoints}[t] + Key things to remember: + \begin{itemize} + \item Collect all \textbf{terms} involving the \textbf{unknown} to one + side of the equation. + \item The opposite of addition is subtraction. + \item The opposite of multiplication is division. + \item If you do something to one side of the equation you MUST do the + same thing to the other side. + \end{itemize} +\end{keypoints} + +\subsection*{Higher Level Questions} + +A harder question is shown below, the steps taken are shown on the arrows. The +first step is to bring all \textbf{unknowns} to the same side of the equation. + +\question Find the value of x: $6x - 12 = x + 8$ + +\begin{figure}[H] + \centering + \begin{tikzpicture}[->,>=stealth',shorten >=1pt,auto,node distance=1.5cm, + thick,main node/.style={draw=none,fill=none}] + + \node[main node] (1) {$6x - 12 = x + 8$}; + \node[main node] (2) [below of=1] {$5x - 12 = 8$}; + \node[main node] (3) [below of=2] {$5x = 20$}; + \node[main node] (4) [below of=3] {$x = 4$}; + + \path[every node/.style={font=\sffamily\small}] + (1.west) edge [bend right] node[left]{$-x$} (2.west) + (1.east) edge [bend left] node[right] {$-x$} (2.east) + (2.west) edge [bend right] node[left]{$+12$} (3.west) + (2.east) edge [bend left] node[right] {$+12$} (3.east) + (3.west) edge [bend right] node[left]{$\div 5$} (4.west) + (3.east) edge [bend left] node[right] {$\div 5$} (4.east); + \end{tikzpicture} +\end{figure} + +Check: + +On the left hand side of the \textbf{original} equation:\\ +$6(4) - 12 = 24 - 12 = 12$ + +On the right hand side of the \textbf{original} equation:\\ +$4 + 8 = 12$ + +The left hand side and right hand side of the equation are equal so you know +that your solution is correct: $x = 4$ + +\end{questions} + +\newpage + +\subsection*{Practice Questions} + +\begin{questions} + +\question Find the value of x: $6+2x = x-6$ + +\begin{figure}[H] + \centering + \begin{tikzpicture}[->,>=stealth',shorten >=1pt,auto,node distance=1.5cm, + thick,main node/.style={draw=none,fill=none}] + + \node[main node] (1) {$6 + 2x = x - 6$}; + \node[main node] (2) [below of=1] {$\fib{6 + x} = -6$}; + \node[main node] (3) [below of=2] {$\fib{x} = \fib{-12}$}; + + \path[every node/.style={font=\sffamily\small}] + (1.west) edge [bend right] node[left]{$-x$} (2.west) + (1.east) edge [bend left] node[right] {$-x$} (2.east) + (2.west) edge [bend right] node[left]{$-6$} (3.west) + (2.east) edge [bend left] node[right] {$\fib{-6}$} (3.east); + \end{tikzpicture} +\end{figure} + +Check: + +$6 + \fib{2}x = 6 + 2 \times -12 = \fib{-18}$\\ +$\fib{x} - 6 = \fib{-12} - 6 = \fib{-18}$ + +\end{questions} + +\begin{questions} + +\newpage + +\subsection*{Solving Linear Equations} + +\question[3] Find the value of x: $6x + 13 = 4 + 9x$ \droppoints + +\begin{minipage}{0.69\linewidth} +\begin{figure}[H] + \centering + \begin{tikzpicture}[->,>=stealth',shorten >=1pt,auto,node distance=1.5cm, + thick,main node/.style={draw=none,fill=none}] + + \node[main node] (1) {$6x + 13 = 4 + 9x$}; + \node[main node] (2) [below of=1] {$\fib{13} = 4 + 3x$}; + \node[main node] (3) [below of=2] {$\fib{9} = \fib{3x}$}; + \node[main node] (4) [below of=3] {$\fib{x} = \fib{3}$}; + + \path[every node/.style={font=\sffamily\small}] + (1.west) edge [bend right] node[left]{$-6x$} (2.west) + (1.east) edge [bend left] node[right] {$-6x$} (2.east) + (2.west) edge [bend right] node[left]{$-4$} (3.west) + (2.east) edge [bend left] node[right] {$-4$} (3.east) + (3.west) edge [bend right] node[left]{$\fib{\div 3}$} (4.west) + (3.east) edge [bend left] node[right] {$\fib{\div 3}$} (4.east); + \end{tikzpicture} +\end{figure} +\end{minipage} +\begin{boxedminipage}{0.3\linewidth} + TIP: + + When you have unknowns on both sides of the equations you can choose which + one to move. + + Try and move the lower number of unknowns to avoid negatives. + +\end{boxedminipage} + +Check: + +$6\fib{x} + 13 = 6\fib{3} + 13 = 31$\\ +$4 + 9\fib{x} = 4 + \fib{27} = 31$ + +\newpage + +\question[3] Find the value of x: $7x + 8 = 2x - 2$ \droppoints + +\begin{figure}[H] + \centering + \begin{tikzpicture}[->,>=stealth',shorten >=1pt,auto,node distance=1.5cm, + thick,main node/.style={draw=none,fill=none}] + + \node[main node] (1) {$7x + 8 = 2x - 2$}; + \node[main node] (2) [below of=1] {$5x + 8 = \fib{-2}$}; + \node[main node] (3) [below of=2] {$\fib{5x} = \fib{-10}$}; + \node[main node] (4) [below of=3] {$\fib{x} = \fib{-2}$}; + + \path[every node/.style={font=\sffamily\small}] + (1.west) edge [bend right] node[left]{$-2x$} (2.west) + (1.east) edge [bend left] node[right] {$-2x$} (2.east) + (2.west) edge [bend right] node[left]{$\fib{-8}$} (3.west) + (2.east) edge [bend left] node[right] {$\fib{-8}$} (3.east) + (3.west) edge [bend right] node[left]{$\div \fib{5}$} (4.west) + (3.east) edge [bend left] node[right] {$\div \fib{5}$} (4.east); + \end{tikzpicture} +\end{figure} + +Check: + +$7\fib{x} + 8 = \fib{-14} + 8 = \fib{-4}$\\ +$2\fib{x} - 2 = \fib{-6} - 2 = \fib{-4}$ + +\newpage + +\question[3] + +Susie clears out boxes on the weekend. She gets paid £10 for the weekend and an +extra £2 for each box she clears. If Susie is paid £40, how many boxes did she +clear? \droppoints + +\begin{minipage}{0.69\linewidth} +\begin{figure}[H] + \centering + \begin{tikzpicture}[->,>=stealth',shorten >=1pt,auto,node distance=1.5cm, + thick,main node/.style={draw=none,fill=none}] + + \node[main node] (1) {$\fib{10 + 2x} = 40$}; + \node[main node] (2) [below of=1] {$\fib{2x} = \fib{30}$}; + \node[main node] (3) [below of=2] {$\fib{x} = \fib{15}$}; + + \path[every node/.style={font=\sffamily\small}] + (1.west) edge [bend right] node[left]{$-\fib{10}$} (2.west) + (1.east) edge [bend left] node[right] {$-\fib{10}$} (2.east) + (2.west) edge [bend right] node[left]{$\div\fib{2}$} (3.west) + (2.east) edge [bend left] node[right] {$\div\fib{2}$} (3.east); + \end{tikzpicture} +\end{figure} +\end{minipage} +\begin{boxedminipage}{0.3\linewidth} + LOOK!: + + This question is asking you to make the linear equation yourself. + + HINT: + + What is the unknown? +\end{boxedminipage} + +Check: + +$\fib{2}\fib{x} + \fib{10} = \fib{30} + \fib{10} = \fib{40}$ + +\newpage + +\question[3] + +Rearrange to find the unknown:\\ +$3z - 18 = 7z - 10$ \droppoints + +\begin{minipage}{0.69\linewidth} +\begin{figure}[H] + \centering + \begin{tikzpicture}[->,>=stealth',shorten >=1pt,auto,node distance=1.5cm, + thick,main node/.style={draw=none,fill=none}] + + \node[main node] (1) {$3z - 18 = 7z - 10$}; + \node[main node] (2) [below of=1] {$\fib{3z} = \fib{7z + 8}$}; + \node[main node] (3) [below of=2] {$\fib{-4z} = \fib{8}$}; + \node[main node] (4) [below of=3] {$\fib{z} = \fib{-2}$}; + + + \path[every node/.style={font=\sffamily\small}] + (1.west) edge [bend right] node[left]{$\fib{+ 18}$} (2.west) + (1.east) edge [bend left] node[right] {$\fib{+ 18}$} (2.east) + (2.west) edge [bend right] node[left]{$\fib{- 7z}$} (3.west) + (2.east) edge [bend left] node[right] {$\fib{- 7z}$} (3.east) + (3.west) edge [bend right] node[left]{$\fib{\div -2}$} (4.west) + (3.east) edge [bend left] node[right] {$\fib{\div -2}$} (4.east); + \end{tikzpicture} +\end{figure} +\end{minipage} +\begin{boxedminipage}{0.3\linewidth} + NOTE: + + This question uses z as the unknown. Any letter may be used as the + unknown. + +\end{boxedminipage} + +Check: + +$3\fib{z} - 18 = \fib{-6} - 18 = \fib{-24}$\\ +$7\fib{z} - 10 = \fib{-14} - 10 = \fib{-24}$ + +\end{questions} + +\end{document} -- cgit v1.2.3