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<document xmlns="http://www.math.cuhk.edu.hk/~pschan/cranach">
<course environment="" chapter_type="" type="Course" wbtag="course" title="MATH 1030">
<keywords slide="all" environ="course" chapter="" course="MATH 1030"/>
<title environment="title" chapter_type="" wbtag="course" scope="course" course="MATH 1030" chapter="" section="" subsection="" subsubsection="" serial="" slide="">MATH 1030</title>
<slides environment="" chapter_type="">
<slide environment="" chapter_type="" canon_num="1" scope="course" course="MATH 1030" title="MATH 1030" id="s1" slide="1" chapter="" section="" subsection="" subsubsection="" chapter_title="" section_title="">
<title environment="title" chapter_type="" wbtag="course" scope="course" course="" chapter="" section="" subsection="" subsubsection="" serial="" slide="1">MATH 1030</title>
<keywords slide="1" environ="slide" chapter="" course="MATH 1030"/>
<setchapter course="MATH 1030" chapter="" chapter_type="" section="" subsection="" subsubsection="" environment="" wbtag="setchapter" argument="1"/>
</slide>
</slides>
<chapter environment="" chapter_type="Chapter" type="Chapter" wbtag="chapter" num="1" title="Technique of solving system of linear equations" course="MATH 1030" serial="1">
<keywords slide="all" environ="chapter" chapter="1" course="MATH 1030"/>
<title environment="title" chapter_type="Chapter" wbtag="chapter" scope="chapter" course="MATH 1030" chapter="1" section="" subsection="" subsubsection="" serial="" slide="">Technique of solving system of linear equations</title>
<slides environment="" chapter_type="">
<slide environment="" chapter_type="Chapter" canon_num="2" scope="chapter" id="s2" slide="2" course="MATH 1030" chapter="1" section="" subsection="" subsubsection="" chapter_title="Technique of solving system of linear equations" section_title="">
<title environment="title" chapter_type="Chapter" wbtag="chapter" scope="chapter" course="" chapter="" section="" subsection="" subsubsection="" serial="" slide="2">Technique of solving system of linear equations</title>
<keywords slide="2" environ="slide" chapter="1" course="MATH 1030"/>
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="" subsection="" subsubsection="" environment="" wbtag="paragraphs"> The lecture is based on Beezer, A first course in Linear algebra. Ver 3.5 Downloadable at </paragraphs>
<href course="MATH 1030" chapter="1" chapter_type="" section="" subsection="" subsubsection="" environment="" src="http://linear.ups.edu/download.html" name="http://linear.ups.edu/download.html" wbtag="href" argument="http://linear.ups.edu/download.html"/>
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="" subsection="" subsubsection="" environment="" wbtag="paragraphs"> .</paragraphs>
<newline course="MATH 1030" chapter="1" chapter_type="" section="" subsection="" subsubsection="" environment=""/>
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="" subsection="" subsubsection="" environment="" wbtag="paragraphs">The print version can be downloaded at </paragraphs>
<href course="MATH 1030" chapter="1" chapter_type="" section="" subsection="" subsubsection="" environment="" src="http://linear.ups.edu/download/fcla-3.50-print.pdf" name="http://linear.ups.edu/download/fcla-3.50-print.pdf" wbtag="href" argument="http://linear.ups.edu/download/fcla-3.50-print.pdf"/>
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="" subsection="" subsubsection="" environment="" wbtag="paragraphs"> .</paragraphs>
<newline course="MATH 1030" chapter="1" chapter_type="" section="" subsection="" subsubsection="" environment=""/>
<h5 xmlns="http://www.w3.org/1999/xhtml" class="notkw" environment="">
<paragraphs xmlns="http://www.math.cuhk.edu.hk/~pschan/cranach" course="MATH 1030" chapter="1" chapter_type="" section="" subsection="" subsubsection="" environment="" wbtag="paragraphs">Reference:</paragraphs>
</h5>
<enumerate course="MATH 1030" chapter="1" chapter_type="" section="" subsection="" subsubsection="" environment="enumerate" wbtag="enumerate">
<item course="MATH 1030" chapter="1" chapter_type="" section="" subsection="" subsubsection="" environment="enumerate" wbtag="item">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="" subsection="" subsubsection="" environment="enumerate" wbtag="paragraphs"> Gilbert Strang, </paragraphs>
<i xmlns="http://www.w3.org/1999/xhtml" environment="enumerate">
<paragraphs xmlns="http://www.math.cuhk.edu.hk/~pschan/cranach" course="MATH 1030" chapter="1" chapter_type="" section="" subsection="" subsubsection="" environment="enumerate" wbtag="paragraphs">Linear Algebra and Its Applications</paragraphs>
</i>
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="" subsection="" subsubsection="" environment="enumerate" wbtag="paragraphs">. Section 1.3. </paragraphs>
</item>
<item course="MATH 1030" chapter="1" chapter_type="" section="" subsection="" subsubsection="" environment="enumerate" wbtag="item">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="" subsection="" subsubsection="" environment="enumerate" wbtag="paragraphs"> Robert A. Beezer, </paragraphs>
<i xmlns="http://www.w3.org/1999/xhtml" environment="enumerate">
<paragraphs xmlns="http://www.math.cuhk.edu.hk/~pschan/cranach" course="MATH 1030" chapter="1" chapter_type="" section="" subsection="" subsubsection="" environment="enumerate" wbtag="paragraphs">A First Course in Linear Algebra</paragraphs>
</i>
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="" subsection="" subsubsection="" environment="enumerate" wbtag="paragraphs">. Section </paragraphs>
<a xmlns="http://www.w3.org/1999/xhtml" target="_blank" href="http://linear.ups.edu/html/section-SSLE.html" environment="enumerate">
<paragraphs xmlns="http://www.math.cuhk.edu.hk/~pschan/cranach" course="MATH 1030" chapter="1" chapter_type="" section="" subsection="" subsubsection="" environment="enumerate" wbtag="paragraphs">Solving Systems of Linear Equations</paragraphs>
</a>
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="" subsection="" subsubsection="" environment="enumerate" wbtag="paragraphs">. </paragraphs>
</item>
</enumerate>
<newline course="MATH 1030" chapter="1" chapter_type="" section="" subsection="" subsubsection="" environment=""/>
</slide>
</slides>
<section environment="" chapter_type="" type="Section" wbtag="section" title="Introduction" course="MATH 1030" chapter="1" num="1" serial="1.1">
<title environment="title" chapter_type="Chapter" wbtag="title" scope="section" course="MATH 1030" chapter="1" section="1" subsection="" subsubsection="" serial="" slide="">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="1" subsection="" subsubsection="" environment="title" wbtag="paragraphs">Introduction</paragraphs>
</title>
<slides environment="" chapter_type="">
<slide environment="" chapter_type="Chapter" canon_num="3" scope="section" id="s3" slide="3" course="MATH 1030" chapter="1" section="1" subsection="" subsubsection="" chapter_title="Technique of solving system of linear equations" section_title="Introduction">
<keywords slide="3" environ="slide" chapter="1" course="MATH 1030"/>
<title environment="title" chapter_type="Chapter" wbtag="title" scope="section" course="MATH 1030" chapter="1" section="1" subsection="" subsubsection="" serial="" slide="3">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="1" subsection="" subsubsection="" environment="title" wbtag="paragraphs">Introduction</paragraphs>
</title>
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="1" subsection="" subsubsection="" environment="" wbtag="paragraphs"> In this section we give examples of systems of linear equations and solve one example. </paragraphs>
<statement environment="statement" chapter_type="Chapter" type="Example" wbtag="eg" num="1" md5="e05f03a153c037dda34d0d9271255645" course="MATH 1030" chapter="1" slide="3" item="1.1">
<label wbtag="label" name="ex2" type="Example"/>
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="1" subsection="" subsubsection="" environment="statement" wbtag="paragraphs"> Solve the system of equations:
\begin{eqnarray}
\label{eq:eg2-1}
2x_{1}+3x_{2}&=3\\
\label{eq:eg2-2}
x_{1}-x_{2}&=4\\
\end{eqnarray} </paragraphs>
</statement>
<substatement environment="substatement" chapter_type="Chapter" type="Solution" wbtag="sol" md5="1f481128acf016648786a4c5c7107bbe" course="MATH 1030" chapter="1" slide="3">
<newcol course="MATH 1030" chapter="1" chapter_type="" section="1" subsection="" subsubsection="" environment="newcol" wbtag="newcol">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="1" subsection="" subsubsection="" environment="newcol" wbtag="paragraphs"> Adding Equation $\eqref{eq:eg2-1}$ to $-2$ $\times$
Equation $\eqref{eq:eg2-2}$ gives:
\[
5x_{2}=-5.
\]
So $x_{2}=-1$. Substituting $x_{2}=-1$ into Equation $\eqref{eq:eg2-1}$ gives
\[
2x_{1}+3(-1)=3
\]
Hence, $x_{1}=3$ and $x_{2}=-1$ is a solution.</paragraphs>
<newline course="MATH 1030" chapter="1" chapter_type="" section="1" subsection="" subsubsection="" environment="newcol"/>
<collapse course="MATH 1030" chapter="1" chapter_type="" section="1" subsection="" subsubsection="" environment="newcol" wbtag="collapse">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="1" subsection="" subsubsection="" environment="newcol" wbtag="paragraphs"> Indeed:
\[2(3)+3(-1)=3\]
\[(3)-(-1)=4.\]
In fact, it is the unique solution. </paragraphs>
</collapse>
</newcol>
</substatement>
</slide>
<slide environment="" chapter_type="Chapter" wbtag="slide" canon_num="4" id="s4" slide="4" course="MATH 1030" chapter="1" section="1" subsection="" subsubsection="" chapter_title="Technique of solving system of linear equations" section_title="Introduction">
<keywords slide="4" environ="slide" chapter="1" course="MATH 1030"/>
<strong xmlns="http://www.w3.org/1999/xhtml" environment="">
<paragraphs xmlns="http://www.math.cuhk.edu.hk/~pschan/cranach" course="MATH 1030" chapter="1" chapter_type="" section="1" subsection="" subsubsection="" environment="" wbtag="paragraphs">Main goal</paragraphs>
</strong>
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="1" subsection="" subsubsection="" environment="" wbtag="paragraphs">: One of the main goals of this course is to solve systems of linear equations with more variables and more equations. </paragraphs>
<statement environment="statement" chapter_type="Chapter" type="Example" wbtag="eg" num="2" md5="ab6efe948a924f1ab4876b4bb8278108" course="MATH 1030" chapter="1" slide="4" item="1.2">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="1" subsection="" subsubsection="" environment="statement" wbtag="paragraphs"> \begin{align*}
\displaystyle x_{1}+2x_{2}+2x_{3}&=4\\
\displaystyle x_{1}+3x_{2}+3x_{3}&=5\\
\displaystyle 2x_{1}+6x_{2}+5x_{3}&=6
\end{align*} </paragraphs>
</statement>
<statement environment="statement" chapter_type="Chapter" type="Example" wbtag="eg" num="3" md5="902490b20fc8496037015e9634ea1a23" course="MATH 1030" chapter="1" slide="4" item="1.3">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="1" subsection="" subsubsection="" environment="statement" wbtag="paragraphs"> \begin{align*}
\displaystyle x_{1}+2x_{2}+x_{4}&=7\\
\displaystyle x_{1}+x_{2}+x_{3}-x_{4}&=3\\
\displaystyle 3x_{1}+x_{2}+5x_{3}-7x_{4}&=1
\end{align*} </paragraphs>
</statement>
</slide>
</slides>
</section>
<section environment="" chapter_type="" type="Section" wbtag="section" title="System of linear equations of two unknowns" course="MATH 1030" chapter="1" num="2" serial="1.2">
<title environment="title" chapter_type="Chapter" wbtag="title" scope="section" course="MATH 1030" chapter="1" section="2" subsection="" subsubsection="" serial="" slide="">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="" subsubsection="" environment="title" wbtag="paragraphs">System of linear equations of two unknowns</paragraphs>
</title>
<slides environment="" chapter_type="">
<slide environment="" chapter_type="Chapter" canon_num="5" scope="section" id="s5" slide="5" course="MATH 1030" chapter="1" section="2" subsection="" subsubsection="" chapter_title="Technique of solving system of linear equations" section_title="System of linear equations of two unknowns">
<keywords slide="5" environ="slide" chapter="1" course="MATH 1030"/>
<title environment="title" chapter_type="Chapter" wbtag="title" scope="section" course="MATH 1030" chapter="1" section="2" subsection="" subsubsection="" serial="" slide="5">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="" subsubsection="" environment="title" wbtag="paragraphs">System of linear equations of two unknowns</paragraphs>
</title>
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="" subsubsection="" environment="" wbtag="paragraphs"> You should all be very familiar with the procedure of solving a system equations of two unknowns. </paragraphs>
</slide>
</slides>
<subsection environment="" chapter_type="" type="Section" wbtag="subsection" title="Substitution" course="MATH 1030" chapter="1" section="2" num="1" serial="1.2.1">
<title environment="title" chapter_type="Chapter" wbtag="title" scope="subsection" course="MATH 1030" chapter="1" section="2" subsection="" subsubsection="" serial="1.2.1" slide="">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="" subsubsection="" environment="title" wbtag="paragraphs">Substitution</paragraphs>
</title>
<slides environment="" chapter_type="">
<slide environment="" chapter_type="Chapter" canon_num="6" scope="subsection" id="s6" slide="6" course="MATH 1030" chapter="1" section="2" subsection="1" subsubsection="" chapter_title="Technique of solving system of linear equations" section_title="System of linear equations of two unknowns">
<keywords slide="6" environ="slide" chapter="1" course="MATH 1030"/>
<title environment="title" chapter_type="Chapter" wbtag="title" scope="subsection" course="MATH 1030" chapter="1" section="2" subsection="1" subsubsection="" serial="" slide="6">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="1" subsubsection="" environment="title" wbtag="paragraphs">Substitution</paragraphs>
</title>
<statement environment="statement" chapter_type="Chapter" type="Example" wbtag="eg" num="4" md5="c0978a5b47c33086e0c55378e9e6d9bd" course="MATH 1030" chapter="1" slide="6" item="1.4">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="1" subsubsection="" environment="statement" wbtag="paragraphs"> \begin{equation}\label{v2sube1}3x+4y=2\end{equation}
\begin{equation}\label{v2sube2}4x+5y=3\end{equation}
Use $\eqref{v2sube2}$, we can solve for $y$ in terms of $x$:
\begin{equation}\label{v2sube3}y=\frac{3}{5}-\frac{4}{5}x.\end{equation}
Substitution this into $\eqref{v2sube1}$:</paragraphs>
<newline course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="1" subsubsection="" environment="statement"/>
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="1" subsubsection="" environment="statement" wbtag="paragraphs">\[3x+4(\frac{3}{5}-\frac{4}{5}x)=2\]
\[\frac{12}{5}-\frac{x}{5}=2\]
\[x=2.\]
Substituting $x=2$ into $\eqref{v2sube3}$, we can solve for $y$:
\[y=\frac{3}{5}-\frac{4}{5}\times 2=-1.\]
So the solution is $x=2$, $y=-1$. </paragraphs>
</statement>
<newline course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="1" subsubsection="" environment=""/>
<substatement environment="substatement" chapter_type="Chapter" type="Remark" wbtag="remark" md5="96de9389dca667727e5ee220f94d456b" course="MATH 1030" chapter="1" slide="6">
<newcol course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="1" subsubsection="" environment="newcol" wbtag="newcol">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="1" subsubsection="" environment="newcol" wbtag="paragraphs"> There are other ways to use substitution, for example </paragraphs>
<enumerate course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="1" subsubsection="" environment="enumerate" wbtag="enumerate">
<item course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="1" subsubsection="" environment="enumerate" wbtag="item">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="1" subsubsection="" environment="enumerate" wbtag="paragraphs"> Solve $x$ by $\eqref{v2sube2}$ in terms of $y$ and substitute it into $\eqref{v2sube1}$ </paragraphs>
</item>
<item course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="1" subsubsection="" environment="enumerate" wbtag="item">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="1" subsubsection="" environment="enumerate" wbtag="paragraphs"> Solve $y$ by $\eqref{v2sube1}$ in terms of $x$ and substitute it into $\eqref{v2sube2}$ </paragraphs>
</item>
<item course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="1" subsubsection="" environment="enumerate" wbtag="item">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="1" subsubsection="" environment="enumerate" wbtag="paragraphs"> Solve $x$ by $\eqref{v2sube1}$ in terms of $y$ and substitute it into $\eqref{v2sube2}$ </paragraphs>
</item>
</enumerate>
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="1" subsubsection="" environment="newcol" wbtag="paragraphs"> But you </paragraphs>
<strong xmlns="http://www.w3.org/1999/xhtml" environment="newcol">
<paragraphs xmlns="http://www.math.cuhk.edu.hk/~pschan/cranach" course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="1" subsubsection="" environment="newcol" wbtag="paragraphs">cannot</paragraphs>
</strong>
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="1" subsubsection="" environment="newcol" wbtag="paragraphs"> get the solution by </paragraphs>
<enumerate course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="1" subsubsection="" environment="enumerate" wbtag="enumerate">
<item course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="1" subsubsection="" environment="enumerate" wbtag="item">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="1" subsubsection="" environment="enumerate" wbtag="paragraphs"> Solve $y$ by $\eqref{v2sube2}$ in terms of $x$ and substitute it into $\eqref{v2sube2}$ (No substitution back to the original equation) </paragraphs>
</item>
<item course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="1" subsubsection="" environment="enumerate" wbtag="item">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="1" subsubsection="" environment="enumerate" wbtag="paragraphs"> Solve $y$ by $\eqref{v2sube1}$ in terms of $x$ and substitute it into $\eqref{v2sube1}$</paragraphs>
<newline course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="1" subsubsection="" environment="enumerate"/>
</item>
<item course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="1" subsubsection="" environment="enumerate" wbtag="item">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="1" subsubsection="" environment="enumerate" wbtag="paragraphs"> Solve $x$ by $\eqref{v2sube2}$ in terms of $y$ and substitute it into $\eqref{v2sube2}$ </paragraphs>
</item>
<item course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="1" subsubsection="" environment="enumerate" wbtag="item">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="1" subsubsection="" environment="enumerate" wbtag="paragraphs"> Solve $x$ by $\eqref{v2sube1}$ in terms of $y$ and substitute it into $\eqref{v2sube1}$ </paragraphs>
</item>
</enumerate>
</newcol>
</substatement>
<newline course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="1" subsubsection="" environment=""/>
</slide>
</slides>
</subsection>
<subsection environment="" chapter_type="" type="Section" wbtag="subsection" title="Elimination" course="MATH 1030" chapter="1" section="2" num="2" serial="1.2.2">
<title environment="title" chapter_type="Chapter" wbtag="title" scope="subsection" course="MATH 1030" chapter="1" section="2" subsection="" subsubsection="" serial="1.2.2" slide="">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="" subsubsection="" environment="title" wbtag="paragraphs">Elimination</paragraphs>
</title>
<slides environment="" chapter_type="">
<slide environment="" chapter_type="Chapter" canon_num="7" scope="subsection" id="s7" slide="7" course="MATH 1030" chapter="1" section="2" subsection="2" subsubsection="" chapter_title="Technique of solving system of linear equations" section_title="System of linear equations of two unknowns">
<keywords slide="7" environ="slide" chapter="1" course="MATH 1030"/>
<title environment="title" chapter_type="Chapter" wbtag="title" scope="subsection" course="MATH 1030" chapter="1" section="2" subsection="2" subsubsection="" serial="" slide="7">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="2" subsubsection="" environment="title" wbtag="paragraphs">Elimination</paragraphs>
</title>
<statement environment="statement" chapter_type="Chapter" type="Example" wbtag="eg" num="5" md5="de814b8113f364aaf2f1850642fc0c9f" course="MATH 1030" chapter="1" slide="7" item="1.5">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="2" subsubsection="" environment="statement" wbtag="paragraphs"> Again, let’s solve the system of linear equations from the previous example.
\begin{equation}\label{v2elime1}3x+4y=2\end{equation}
\begin{equation}\label{v2elime2}4x+5y=3\end{equation}
Consider $\eqref{v2elime1}-\frac{3}{4}\eqref{v2elime2}$.</paragraphs>
<newline course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="2" subsubsection="" environment="statement"/>
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="2" subsubsection="" environment="statement" wbtag="paragraphs">\begin{align*}
& 3x & + & 4y & = & 2 \\
-)\quad & 3x & + & \frac{15}4 y & = & \frac{9}4 \\
\hline
& & & \frac{1}{4} y &= &-\frac{1}{4}
\end{align*}
Thus $y=-1$. Substituting it into $\eqref{v2elime1}$:
\[3x+4(-1)=2\]
\[x=2.\]
So we obtain the solution $x=2$, $y=-1$. </paragraphs>
</statement>
<newline course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="2" subsubsection="" environment=""/>
<substatement environment="substatement" chapter_type="Chapter" type="Remark" wbtag="remark" md5="340adadda21a12c6d241f5149471dcf9" course="MATH 1030" chapter="1" slide="7">
<enumerate course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="2" subsubsection="" environment="enumerate" wbtag="enumerate">
<item course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="2" subsubsection="" environment="enumerate" wbtag="item">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="2" subsubsection="" environment="enumerate" wbtag="paragraphs"> The number $\frac{3}{4}$ is so chosen such that the coefficient of $x$ is eliminated. </paragraphs>
</item>
<item course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="2" subsubsection="" environment="enumerate" wbtag="item">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="2" subsubsection="" environment="enumerate" wbtag="paragraphs"> At some point, we still need to use substitution to get the solution. </paragraphs>
</item>
</enumerate>
</substatement>
<newline course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="2" subsubsection="" environment=""/>
</slide>
</slides>
</subsection>
<subsection environment="" chapter_type="" type="Section" wbtag="subsection" title="Substitution" course="MATH 1030" chapter="1" section="2" num="3" serial="1.2.3">
<title environment="title" chapter_type="Chapter" wbtag="title" scope="subsection" course="MATH 1030" chapter="1" section="2" subsection="" subsubsection="" serial="1.2.3" slide="">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="" subsubsection="" environment="title" wbtag="paragraphs">Substitution</paragraphs>
</title>
<slides environment="" chapter_type="">
<slide environment="" chapter_type="Chapter" canon_num="8" scope="subsection" data-lecture-skip="true" id="s8" slide="8" course="MATH 1030" chapter="1" section="2" subsection="3" subsubsection="" chapter_title="Technique of solving system of linear equations" section_title="System of linear equations of two unknowns">
<keywords slide="8" environ="slide" chapter="1" course="MATH 1030"/>
<title environment="title" chapter_type="Chapter" wbtag="title" scope="subsection" course="MATH 1030" chapter="1" section="2" subsection="3" subsubsection="" serial="" slide="8">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="3" subsubsection="" environment="title" wbtag="paragraphs">Substitution</paragraphs>
</title>
<statement environment="statement" chapter_type="Chapter" type="Example" wbtag="eg" num="6" md5="1b371e06004d09ae1d723d41f7f76612" course="MATH 1030" chapter="1" slide="8" item="1.6">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="3" subsubsection="" environment="statement" wbtag="paragraphs"> Solve the following system of linear equations
\begin{equation}\label{v3sube1}x+2y+2z=4\end{equation}
\begin{equation}\label{v3sube2}x+3y+3z=5\end{equation}
\begin{equation}\label{v3sube3}2x+6y+5z=6\end{equation}</paragraphs>
<newline course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="3" subsubsection="" environment="statement"/>
<newcol course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="3" subsubsection="" environment="newcol" wbtag="newcol">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="3" subsubsection="" environment="newcol" wbtag="paragraphs"> Find $x$ in terms of $y,z$ by $\eqref{v3sube1}$ :
\begin{equation}\label{v3sube4}x=4-2y-2z.\end{equation}
Substituting $\eqref{v3sube4}$ into $\eqref{v3sube2}$, we obtain
\[(4-2y-2z)+3y+3z=5\]
i.e.,
\begin{equation}
\label{v3sube5}y+z=1.
\end{equation} </paragraphs>
<collapse course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="3" subsubsection="" environment="newcol" wbtag="collapse">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="3" subsubsection="" environment="newcol" wbtag="paragraphs"> Substituting $\eqref{v3sube4}$ into $\eqref{v3sube3}$, we obtain
\[2(4-2y-2z)+6y+5z=6,\]
i.e.,
\begin{equation}\label{v3sube6}2y+z=-2\end{equation}
The equations are reduced to solving a linear system of equations with two unknowns:
\[
\begin{cases}
y+z=1\\
2y+z=-2
\end{cases}
\] </paragraphs>
<collapse course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="3" subsubsection="" environment="newcol" wbtag="collapse">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="3" subsubsection="" environment="newcol" wbtag="paragraphs"> Solve $y$ in terms of $z$ by $\eqref{v3sube5}$:
\begin{equation}\label{v3sube7}y=1-z\end{equation}
Then substitute $y=1-z$ into $\eqref{v3sube6}$:
\[2(1-z)+z=-2\]
i.e.
\[
z=4.
\] </paragraphs>
<collapse course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="3" subsubsection="" environment="newcol" wbtag="collapse">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="3" subsubsection="" environment="newcol" wbtag="paragraphs"> By $\eqref{v3sube7}$
\[y=1-z=-3.\]
Substitute $y=-3$, $z=4$ into $\eqref{v3sube4}$,
\[x=4-2y-2z=4-2\times(-3)-2\times 4=2.\]
Hence $x=2,y=-3,z=4$ is a solution. </paragraphs>
</collapse>
</collapse>
</collapse>
</newcol>
</statement>
</slide>
</slides>
</subsection>
<subsection environment="" chapter_type="" type="Section" wbtag="subsection" title="Elimination" course="MATH 1030" chapter="1" section="2" num="4" serial="1.2.4">
<title environment="title" chapter_type="Chapter" wbtag="title" scope="subsection" course="MATH 1030" chapter="1" section="2" subsection="" subsubsection="" serial="1.2.4" slide="">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="" subsubsection="" environment="title" wbtag="paragraphs">Elimination</paragraphs>
</title>
<slides environment="" chapter_type="">
<slide environment="" chapter_type="Chapter" canon_num="9" scope="subsection" id="s9" slide="9" course="MATH 1030" chapter="1" section="2" subsection="4" subsubsection="" chapter_title="Technique of solving system of linear equations" section_title="System of linear equations of two unknowns">
<keywords slide="9" environ="slide" chapter="1" course="MATH 1030"/>
<title environment="title" chapter_type="Chapter" wbtag="title" scope="subsection" course="MATH 1030" chapter="1" section="2" subsection="4" subsubsection="" serial="" slide="9">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="4" subsubsection="" environment="title" wbtag="paragraphs">Elimination</paragraphs>
</title>
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="4" subsubsection="" environment="" wbtag="paragraphs"> Using substitution all the way to solve linear equations is not the best way. Instead, we can use elimination to simplify the system of linear equations first. </paragraphs>
<statement environment="statement" chapter_type="Chapter" type="Example" wbtag="eg" num="7" md5="60f7320d5ebfe0d66a893c04ac2e3bc0" course="MATH 1030" chapter="1" slide="9" item="1.7">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="4" subsubsection="" environment="statement" wbtag="paragraphs"> We solve the following system by a sequence of equation operations.
\[\displaystyle x+2y+2z=4 \tag{1}\]
\[\displaystyle x+3y+3z=5 \tag{2}\]
\[\displaystyle 2x+6y+5z=6 \tag{3}\] </paragraphs>
<newcol course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="4" subsubsection="" environment="newcol" wbtag="newcol">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="4" subsubsection="" environment="newcol" wbtag="paragraphs"> $-1$ times equation 1, add to equation 2:
\[\displaystyle x+2y+2z=4\]
\[\displaystyle 0x+1y+1z=1\]
\[\displaystyle 2x+6y+5z=6\] </paragraphs>
<collapse course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="4" subsubsection="" environment="newcol" wbtag="collapse">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="4" subsubsection="" environment="newcol" wbtag="paragraphs"> $-2$ times equation 1, add to equation 3:
\[\displaystyle x+2y+2z=4\]
\[\displaystyle 0x+1y+1z=1\]
\[\displaystyle 0x+2y+1z=-2\] </paragraphs>
<collapse course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="4" subsubsection="" environment="newcol" wbtag="collapse">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="4" subsubsection="" environment="newcol" wbtag="paragraphs"> $-2$ times equation 2, add to equation 3:
\[\displaystyle x+2y+2z=4\]
\[\displaystyle 0x+1y+1z=1\]
\[\displaystyle 0x+0y-1z=-4\] </paragraphs>
<collapse course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="4" subsubsection="" environment="newcol" wbtag="collapse">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="4" subsubsection="" environment="newcol" wbtag="paragraphs"> $-1$ times equation 3:
\[\displaystyle x+2y+2z=4\]
\[\displaystyle 0x+1y+1z=1\]
\[\displaystyle 0x+0y+1z=4\]
which can be written more clearly as:</paragraphs>
<newline course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="4" subsubsection="" environment="newcol"/>
<collapse course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="4" subsubsection="" environment="newcol" wbtag="collapse">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="4" subsubsection="" environment="newcol" wbtag="paragraphs"> \[\displaystyle x+2y+2z=4\]
\[\displaystyle y+z=1\]
\[\displaystyle z=4\] </paragraphs>
<collapse course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="4" subsubsection="" environment="newcol" wbtag="collapse">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="4" subsubsection="" environment="newcol" wbtag="paragraphs"> The third equation requires that $z=4$ to be true. Making this substitution into equation 2 we arrive at $y=-3$, and finally, substituting these values of $y$ and $z$ into the first equation, we find that $x=2$. </paragraphs>
</collapse>
</collapse>
</collapse>
</collapse>
</collapse>
</newcol>
</statement>
<substatement environment="substatement" chapter_type="Chapter" type="Remark" wbtag="remark" md5="bab56abc382edc1829bb73afed2cc561" course="MATH 1030" chapter="1" slide="9">
<newcol course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="4" subsubsection="" environment="newcol" wbtag="newcol">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="2" subsection="4" subsubsection="" environment="newcol" wbtag="paragraphs"> We can add several more eliminations to solve $x,y,z$ without substitution:
\[\displaystyle x+2y+2z=4\]
\[\displaystyle 0x+1y+1z=1\]
\[\displaystyle 0x+0y+1z=4\]
$-1$ times equation 3, add to equation 2 and $-2$ times equation 3, add to equation 1
\[\displaystyle x+2y+0z=-4\]
\[\displaystyle 0x+1y+0z=-3\]
\[\displaystyle 0x+0y+1z=4\]
$-2$ times equation 2, add to equation 1
\[\displaystyle x+0y+0z=2\]
\[\displaystyle 0x+1y+0z=-3\]
\[\displaystyle 0x+0y+1z=4\]
So $x=2,y=-3,z=4$ is a solution. </paragraphs>
</newcol>
</substatement>
</slide>
</slides>
</subsection>
</section>
<section environment="" chapter_type="" type="Section" wbtag="section" title="More Examples" course="MATH 1030" chapter="1" num="3" serial="1.3">
<title environment="title" chapter_type="Chapter" wbtag="title" scope="section" course="MATH 1030" chapter="1" section="3" subsection="" subsubsection="" serial="" slide="">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="title" wbtag="paragraphs">More Examples</paragraphs>
</title>
<slides environment="" chapter_type="">
<slide environment="" chapter_type="Chapter" canon_num="10" scope="section" id="s10" slide="10" course="MATH 1030" chapter="1" section="3" subsection="" subsubsection="" chapter_title="Technique of solving system of linear equations" section_title="More Examples">
<keywords slide="10" environ="slide" chapter="1" course="MATH 1030"/>
<title environment="title" chapter_type="Chapter" wbtag="title" scope="section" course="MATH 1030" chapter="1" section="3" subsection="" subsubsection="" serial="" slide="10">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="title" wbtag="paragraphs">More Examples</paragraphs>
</title>
<statement environment="statement" chapter_type="Chapter" type="Example" wbtag="eg" num="8" md5="6d0972bf488da29acaf91983069fbe67" course="MATH 1030" chapter="1" slide="10" item="1.8">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="statement" wbtag="paragraphs"> Solve:
\begin{align}
\displaystyle x_{1}-5x_{2}+3x_{3}&=1 \tag{1}\\
\displaystyle 2x_{1}-4x_{2}+x_{3}&=0 \tag{2}\\
\displaystyle x_{1}+x_{2}-2x_{3}&=-1 \tag{3}
\end{align} </paragraphs>
<newcol course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="newcol">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="paragraphs"> $-2$ times equation 1, add to equation 2:
\begin{align*}
x_1 - 5x_2 + 3x_3 &= 1 \\
0x_1 + 6x_2 -5 x_3 &= -2 \\
x_1 + x_2 -2x_3 &= -1
\end{align*} </paragraphs>
<collapse course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="collapse">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="paragraphs"> $-1$ times equation 1, add to equation 3:
\begin{align*}
x_1 - 5x_2 + 3x_3 &= 1 \\
0x_1+ 6x_2 -5 x_3 &= -2 \\
0x_1+ 6x_2 -5x_3 &= -2
\end{align*} </paragraphs>
<collapse course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="collapse">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="paragraphs"> $-1$ times equation 2, add to equation 3:
\begin{align*}
x_1 - 5x_2 + 3x_3 &= 1 \\
0x_1+ 6x_2 -5 x_3 &= -2 \\
0x_1+0x_2+0x_3 &= 0
\end{align*}
$\frac{5}{6}$ times equation 2, add to equation 1:
\begin{align*}
x_1 +0x_2 -\frac{7}{6}x_3 &= -\frac{2}{3} \\
0x_1+ 6x_2 -5 x_3 &= -2 \\
0x_1+0x_2+0x_3 &= 0
\end{align*} </paragraphs>
<collapse course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="collapse">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="paragraphs"> We can express $x_{1},x_{2}$ in terms of $x_{3}$:
\[\displaystyle x_{1}=-\frac{2}{3}+\frac{7}{6}x_{3}\]
\[\displaystyle x_{2}=-\frac{1}{3}+\frac{5}{6}x_{3}\] </paragraphs>
<collapse course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="collapse">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="paragraphs"> The solution set is:
\[\left\{(-\frac{2}{3}+\frac{7}{6}a,-\frac{1}{3}+\frac{5}{6}a,a)\,|\,a\text{ real numbers}.\right\}\] </paragraphs>
</collapse>
</collapse>
</collapse>
</collapse>
</newcol>
</statement>
</slide>
<slide environment="" chapter_type="Chapter" wbtag="slide" canon_num="11" id="s11" slide="11" course="MATH 1030" chapter="1" section="3" subsection="" subsubsection="" chapter_title="Technique of solving system of linear equations" section_title="More Examples">
<keywords slide="11" environ="slide" chapter="1" course="MATH 1030"/>
<statement environment="statement" chapter_type="Chapter" type="Example" wbtag="eg" num="9" md5="c80d4b93749104c199f19d39e0ba900a" course="MATH 1030" chapter="1" slide="11" item="1.9">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="statement" wbtag="paragraphs"> Solve:
\begin{align*}
x_1 - 5x_2 + 3x_3 &= 1 \\
2x_1 - 4x_2 + x_3 &= 0 \\
x_1 + x_2 -2x_3 &= -2
\end{align*} </paragraphs>
<newcol course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="newcol">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="paragraphs"> $-2$ times equation 1, add to equation 2:
\begin{align*}
x_1 - 5x_2 + 3x_3 &= 1 \\
0x_1+6x_2 -5 x_3 &= -2 \\
x_1 + x_2 -2x_3 &= -2
\end{align*} </paragraphs>
<collapse course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="collapse">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="paragraphs"> $-1$ times equation 1, add to equation 3:
\begin{align*}
x_1 - 5x_2 + 3x_3 &= 1 \\
0x_1+6x_2 -5 x_3 &= -2 \\
0x_1+6x_2 -5x_3 &= -3
\end{align*} </paragraphs>
<collapse course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="collapse">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="paragraphs"> $-1$ times equation 2, add to equation 3:
\begin{align*}
x_1 - 5x_2 + 3x_3 &= 1 \\
0x_1+6x_2 -5 x_3 &= -2 \\
0x_1+0x_2+0x_3 &= -1
\end{align*} </paragraphs>
<collapse course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="collapse">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="paragraphs"> The last equation, $0=-1$ has no solution. So the system of linear equations has no solution. </paragraphs>
</collapse>
</collapse>
</collapse>
</newcol>
</statement>
</slide>
<slide environment="" chapter_type="Chapter" wbtag="slide" canon_num="12" id="s12" slide="12" course="MATH 1030" chapter="1" section="3" subsection="" subsubsection="" chapter_title="Technique of solving system of linear equations" section_title="More Examples">
<keywords slide="12" environ="slide" chapter="1" course="MATH 1030"/>
<statement environment="statement" chapter_type="Chapter" type="Example" wbtag="eg" num="10" md5="622ecea1cde9099bd77067aa715ff5cc" course="MATH 1030" chapter="1" slide="12" item="1.10">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="statement" wbtag="paragraphs"> Solve:
\begin{align*}
x_1+2x_2 +0x_3+ x_4&= 7\\
x_1+x_2+x_3-x_4&=3\\
3x_1+x_2+5x_3-7x_4&=1
\end{align*} </paragraphs>
<newcol course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="newcol">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="paragraphs"> $-1$ times equation 1, add to equation 2:
\begin{align*}
x_1+2x_2 +0x_3+ x_4&= 7\\
0x_1-x_2+x_3-2x_4&=-4\\
3x_1+x_2+5x_3-7x_4&=1
\end{align*} </paragraphs>
<collapse course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="collapse">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="paragraphs"> $-3$ times equation 1, add to equation 3:
\begin{align*}
x_1+2x_2 +0x_3+ x_4&= 7\\
0x_1-x_2+x_3-2x_4&=-4\\
0x_1-5x_2+5x_3-10x_4&=-20
\end{align*} </paragraphs>
<collapse course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="collapse">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="paragraphs"> $-5$ times equation 2, add to equation 3:
\begin{align*}
x_1+2x_2 +0x_3+ x_4&= 7\\
0x_1-x_2+x_3-2x_4&=-4\\
0x_1+0x_2+0x_3+0x_4&=0
\end{align*} </paragraphs>
<collapse course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="collapse">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="paragraphs"> $-1$ times equation 2:
\begin{align*}
x_1+2x_2 +0x_3+ x_4&= 7\\
0x_1+x_2-x_3+2x_4&=4\\
0x_1+0x_2+0x_3+0x_4&=0
\end{align*} </paragraphs>
<collapse course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="collapse">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="paragraphs"> $-2$ times equation 2, add to equation 1:
\begin{align*}
x_1+0x_2 +2x_3-3x_4&= -1\\
0x_1+x_2-x_3+2x_4&=4\\
0x_1+0x_2+0x_3+0x_4&=0
\end{align*}
which can be written more clearly as:</paragraphs>
<newline course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol"/>
<collapse course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="collapse">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="paragraphs"> \begin{align*}
x_1+2x_3 - 3x_4&= -1\\
x_2-x_3+2x_4&=4\\
0&=0
\end{align*} </paragraphs>
<collapse course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="collapse">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="paragraphs"> The last equation $0=0$ is always true, so we can ignore it and only consider the first two equations. We can analyze the second equation without consideration of the variable $x_{1}$. It would appear that there is considerable latitude in how we can choose $x_{2}$, $x_{3}$, $x_{4}$ and make this equation true. Let us choose $x_{3}$ and $x_{4}$ to be </paragraphs>
<strong xmlns="http://www.w3.org/1999/xhtml" environment="newcol">
<paragraphs xmlns="http://www.math.cuhk.edu.hk/~pschan/cranach" course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="paragraphs">anything</paragraphs>
</strong>
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="paragraphs"> we please, say $x_{3}=a$ and $x_{4}=b$.</paragraphs>
<newline course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol"/>
<collapse course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="collapse">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="paragraphs"> Now we can take these arbitrary values for $x_{3}$ and $x_{4}$, substitute them in equation 1, to obtain
\[x_{1}+2a-3b=-1\]
\[x_{1}=-1-2a+3b\] </paragraphs>
<collapse course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="collapse">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="paragraphs"> Similarly, equation 2 becomes
\[x_{2}-a+2b=4\]
\[x_{2}=4+a-2b\] </paragraphs>
<collapse course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="collapse">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="paragraphs"> So our arbitrary choices of values for $x_{3}$ and $x_{4}$ ($a$ and $b$) translate into specific values of $x_{1}$ and $x_{2}$.</paragraphs>
<newline course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol"/>
<collapse course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="collapse">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="paragraphs"> Now we can easily and quickly find many more (infinitely more). Suppose we choose $a=5$ and $b=-2$, then we compute
\[\displaystyle x_{1}=-1-2(5)+3(-2)=-17\]
\[\displaystyle x_{2}=4+5-2(-2)=13\] </paragraphs>
<collapse course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="collapse">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="paragraphs"> and you can verify that $(x_{1},\,x_{2},\,x_{3},\,x_{4})=(-17,\,13,\,5,\,-2)$ makes all three equations true. The entire solution set is written as
\[\{(-1-2a+3b,\,4+a-2b,\,a,\,b)\,|\,a,b\text{ real numbers}\}.\] </paragraphs>
</collapse>
</collapse>
</collapse>
</collapse>
</collapse>
</collapse>
</collapse>
</collapse>
</collapse>
</collapse>
</collapse>
</newcol>
</statement>
</slide>
<slide environment="" chapter_type="Chapter" wbtag="slide" canon_num="13" data-lecture-skip="true" id="s13" slide="13" course="MATH 1030" chapter="1" section="3" subsection="" subsubsection="" chapter_title="Technique of solving system of linear equations" section_title="More Examples">
<keywords slide="13" environ="slide" chapter="1" course="MATH 1030"/>
<statement environment="statement" chapter_type="Chapter" type="Example" wbtag="eg" num="11" md5="540ab3f62927ff3ef5f81892e26b2bf5" course="MATH 1030" chapter="1" slide="13" item="1.11">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="statement" wbtag="paragraphs"> Solve the following system of linear equations:
\[\begin{alignedat}{6}
& {} {} & x_2 & {}+{} & x_3 & {}+{} & 2x_4 & {}+{} & 2x_5 &{}={} & 2 \\
x_1 & {}+{} & 2x_2 & {}+{} & 3x_3 & {}+{} & 2x_4 & {}+{}& 3x_5&{}={} & 4 \\
-2x_1 & {}-{} & x_2 & {}-{} & 3x_3 & {}+{} & 3x_4 & {}+{} & x_5 &{}={} & 3
\end{alignedat}\] </paragraphs>
<newcol course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="newcol">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="paragraphs"> Swap equation 1 and equation 2:
\[\begin{alignedat}{6}
x_1 & {}+{} & 2x_2 & {}+{} & 3x_3 & {}+{} & 2x_4 & {}+{}& 3x_5&{}={} & 4 \\
& {} {} & x_2 & {}+{} & x_3 & {}+{} & 2x_4 & {}+{} & 2x_5 &{}={} & 2 \\
-2x_1 & {}-{} & x_2 & {}-{} & 3x_3 & {}+{} & 3x_4 & {}+{} & x_5 &{}={} & 3
\end{alignedat}\] </paragraphs>
<collapse course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="collapse">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="paragraphs"> 2 times equation 1, and add it to equation 3:
\[\begin{alignedat}{6}
x_1 & {}+{} & 2x_2 & {}+{} & 3x_3 & {}+{} & 2x_4 & {}+{}& 3x_5&{}={} & 4 \\
& {} {} & x_2 & {}+{} & x_3 & {}+{} & 2x_4 & {}+{} & 2x_5 &{}={} & 2 \\
& {} {} & 3x_2 & {}+{} & 3x_3 & {}+{} & 7x_4 & {}+{} &7x_5 &{}={} & 11
\end{alignedat}\] </paragraphs>
<collapse course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="collapse">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="paragraphs"> -3 times equation 2 and add it to equation 3:
\[\begin{alignedat}{6}
x_1 & {}+{} & 2x_2 & {}+{} & 3x_3 & {}+{} & 2x_4 & {}+{}& 3x_5&{}={} & 4 \\
& {} {} & x_2 & {}+{} & x_3 & {}+{} & 2x_4 & {}+{} & 2x_5 &{}={} & 2 \\
& {} {} & & {} {} & & {} {} & x_4 & {}+{} & x_5 &{}={} & 5
\end{alignedat}\] </paragraphs>
<collapse course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="collapse">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="paragraphs"> Now the system of linear equations looks like an ”inverted stair”.
We can then solve the system of linear equations by substitution.
(A better method well be given later.)</paragraphs>
<newline course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol"/>
<collapse course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="collapse">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="paragraphs"> By the last equation:
\[x_{4}=5-x_{5}.\]
Solve $x_{2}$ in terms of other variables by equation 2:
\[\displaystyle x_{2}=2-x_{3}-2x_{4}-2x_{5}\]
\[\displaystyle=2-x_{3}-2(5-x_{5})-2x_{5}\]
\[\displaystyle=-8-x_{3}\] </paragraphs>
<collapse course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="collapse">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="paragraphs"> Solve $x_{1}$ in terms of other variables by equation 1:
\[\displaystyle x_{1}=4-2x_{2}-3x_{3}-2x_{4}-3x_{5}\]
\[\displaystyle=4-2(-8-x_{3})-3x_{3}-2(5-x_{5})-3x_{5}\]
\[\displaystyle=10-x_{3}-x_{5}.\]
$x_{3},x_{5}$ can be taken as any values.</paragraphs>
<newline course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol"/>
<collapse course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="collapse">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="paragraphs"> Set $x_{3}=a$, $x_{5}=b$, the solution set can be given by
\[\{(10-a-b,-8-a,a,5-b,b)\,|\,a,b\text{ real numbers}\}.\] </paragraphs>
<collapse course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="collapse">
<strong xmlns="http://www.w3.org/1999/xhtml" environment="newcol">
<paragraphs xmlns="http://www.math.cuhk.edu.hk/~pschan/cranach" course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="paragraphs">A better method:</paragraphs>
</strong>
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="paragraphs"> Instead of subsitution, we could use more elimination:
\[\begin{alignedat}{6}
x_1 & {}+{} & 2x_2 & {}+{} & 3x_3 & {}+{} & 2x_4 & {}+{}& 3x_5&{}={} & 4 \\
& {} {} & x_2 & {}+{} & x_3 & {}+{} & 2x_4 & {}+{} & 2x_5 &{}={} & 2 \\
& {} {} & & {} {} & & {} {} & x_4 & {}+{} & x_5 &{}={} & 5
\end{alignedat}\] </paragraphs>
<collapse course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="collapse">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="paragraphs"> -2 times equation 3 and add it to equation 2:
\[\begin{alignedat}{6}
x_1 & {}+{} & 2x_2 & {}+{} & 3x_3 & {}+{} & 2x_4 & {}+{}& 3x_5&{}={} & 4 \\
& {} {} & x_2 & {}+{} & x_3 & {} {} & & {} {} & &{}={} & -8 \\
& {} {} & & {} {} & & {} {} & x_4 & {}+{} & x_5 &{}={} & 5
\end{alignedat}\] </paragraphs>
<collapse course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="collapse">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="paragraphs"> -2 times equation 3 and add it to equation 1:
\[\begin{alignedat}{6}
x_1 & {}+{} & 2x_2 & {}+{} & 3x_3 & {}+{} & & {}+{}& x_5&{}={} & -6 \\
& {} {} & x_2 & {}+{} & x_3 & {} {} & & {} {} & &{}={} & -8 \\
& {} {} & & {} {} & & {} {} & x_4 & {}+{} & x_5 &{}={} & 5
\end{alignedat}\] </paragraphs>
<collapse course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="collapse">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="paragraphs"> -2 times equation 2 and add it to equation 1:
\[\begin{alignedat}{6}
x_1 & {}+{} & & {}+{} & x_3 & {} {} & & {}+{}& x_5&{}={} &10 \\
& {} {} & x_2 & {}+{} & x_3 & {} {} & & {} {} & &{}={} & -8 \\
& {} {} & & {} {} & & {} {} & x_4 & {}+{} & x_5 &{}={} & 5
\end{alignedat}\] </paragraphs>
<collapse course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="collapse">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="paragraphs"> Notice the following: </paragraphs>
<enumerate course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="enumerate" wbtag="enumerate">
<item course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="enumerate" wbtag="item">
<newcol course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="newcol">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="paragraphs"> The system of equations looks like an ”inverted” stairs. </paragraphs>
</newcol>
</item>
<item course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="enumerate" wbtag="item">
<newcol course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="newcol">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="paragraphs"> The leftmost variables in the equations are $x_{1}$, $x_{2}$ and $x_{4}$. </paragraphs>
</newcol>
</item>
<item course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="enumerate" wbtag="item">
<newcol course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="newcol">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="paragraphs"> Only the first equation has variable $x_{1}$. </paragraphs>
</newcol>
</item>
<item course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="enumerate" wbtag="item">
<newcol course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="newcol">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="paragraphs"> Only the second equation has variable $x_{2}$. </paragraphs>
</newcol>
</item>
<item course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="enumerate" wbtag="item">
<newcol course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="newcol">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="paragraphs"> Only the third equation has variable $x_{4}$. </paragraphs>
</newcol>
</item>
</enumerate>
<collapse course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="collapse">
<paragraphs course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="newcol" wbtag="paragraphs"> Move $x_{3},x_{5}$ to another side.
\begin{align*}
x_1 &= 10-x_3-x_5\\
x_2 &= -8-x_3\\
x_4 &= 5-x_5
\end{align*}
The right hand sides have $x_{3},x_{5}$ as variables only and $x_{3},x_{5}$ can be taken as any values.
Set $x_{3}=a$, $x_{5}=b$, the solution set can be given by
\[\{(10-a-b,-8-a,a,5-b,b)\,|\,a,b\text{ real numbers}\}.\] </paragraphs>
</collapse>
</collapse>
</collapse>
</collapse>
</collapse>
</collapse>
</collapse>
</collapse>
</collapse>
</collapse>
</collapse>
</collapse>
</newcol>
</statement>
<newline course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment=""/>
<comment course="MATH 1030" chapter="1" chapter_type="" section="3" subsection="" subsubsection="" environment="">DELIMITER</comment>
</slide>
</slides>
</section>
</chapter>
</course>
</document>