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Line breaks, Chapters SLE, V

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src/section-LC.xml

 
 <p>Step 3.  For each dependent variable, use the augmented matrix to formulate an equation expressing the dependent variable as a constant plus multiples of the free variables.  Convert this equation into entries of the vectors that ensure equality for each dependent variable, one at a time.
 <alignmath>
-<![CDATA[x_1&=15-2x_3+3x_4-9x_7]]>
-<![CDATA[&]]>
-<![CDATA[\Rightarrow&]]>
-<![CDATA[&]]>
-<![CDATA[\vect{x}&=]]>
+<![CDATA[x_1&=15-2x_3+3x_4-9x_7\ \Rightarrow\\&\vect{x}=]]>
 \colvector{x_1\\x_2\\x_3\\x_4\\x_5\\x_6\\x_7}=
 \colvector{15\\ \\ 0\\ 0\\ \\ \\ 0}+
 x_3\colvector{-2\\ \\ 1\\ 0\\ \\ \\ 0}+
 x_4\colvector{3\\ \\ 0\\ 1\\ \\ \\ 0}+
 x_7\colvector{-9\\ \\ 0\\ 0\\ \\ \\ 1}\\
-<![CDATA[x_2&=-10+5x_3-4x_4+8x_7]]>
-<![CDATA[&]]>
-<![CDATA[\Rightarrow&]]>
-<![CDATA[&]]>
-<![CDATA[\vect{x}&=]]>
+<![CDATA[x_2&=-10+5x_3-4x_4+8x_7\ \Rightarrow\\&\vect{x}=]]>
 \colvector{x_1\\x_2\\x_3\\x_4\\x_5\\x_6\\x_7}=
 \colvector{15\\ -10\\ 0\\ 0\\ \\ \\ 0}+
 x_3\colvector{-2\\ 5\\ 1\\ 0\\ \\ \\ 0}+
 x_4\colvector{3\\ -4\\ 0\\ 1\\ \\ \\ 0}+
 x_7\colvector{-9\\ 8\\ 0\\ 0\\ \\ \\ 1}\\
-<![CDATA[x_5&=11+6x_7]]>
-<![CDATA[&]]>
-<![CDATA[\Rightarrow&]]>
-<![CDATA[&]]>
-<![CDATA[\vect{x}&=]]>
+<![CDATA[x_5&=11+6x_7\ \Rightarrow\\&\vect{x}=]]>
 \colvector{x_1\\x_2\\x_3\\x_4\\x_5\\x_6\\x_7}=
 \colvector{15\\ -10\\ 0\\ 0\\ 11\\ \\ 0}+
 x_3\colvector{-2\\ 5\\ 1\\ 0\\ 0\\ \\ 0}+
 x_4\colvector{3\\ -4\\ 0\\ 1\\ 0\\ \\ 0}+
 x_7\colvector{-9\\ 8\\ 0\\ 0\\ 6\\ \\ 1}\\
-<![CDATA[x_6&=-21-7x_7]]>
-<![CDATA[&]]>
-<![CDATA[\Rightarrow&]]>
-<![CDATA[&]]>
-<![CDATA[\vect{x}&=]]>
+<![CDATA[x_6&=-21-7x_7\ \Rightarrow\\&\vect{x}=]]>
 \colvector{x_1\\x_2\\x_3\\x_4\\x_5\\x_6\\x_7}=
 \colvector{15\\ -10\\ 0\\ 0\\ 11\\ -21\\ 0}+
 x_3\colvector{-2\\ 5\\ 1\\ 0\\ 0\\ 0\\ 0}+
 \vectorentry{\vect{c}}{f_i}+
 \alpha_1\vectorentry{\vect{u}_1}{f_i}+
 \alpha_2\vectorentry{\vect{u}_2}{f_i}+
-\alpha_3\vectorentry{\vect{u}_3}{f_i}+
 \cdots+
 \alpha_i\vectorentry{\vect{u}_i}{f_i}+
 \cdots+
 0+
 \alpha_1(0)+
 \alpha_2(0)+
-\alpha_3(0)+
 \cdots+
 \alpha_i(1)+
 \cdots+
 \cdots+
 \matrixentry{B}{\ell f_{n-r}}\alpha_{n-r}\\
 <![CDATA[&=]]>
-\matrixentry{B}{\ell,{n+1}}+
-\alpha_1(-\matrixentry{B}{\ell,f_1})+
-\alpha_2(-\matrixentry{B}{\ell,f_2})+
-\alpha_3(-\matrixentry{B}{\ell,f_3})+
+\matrixentry{B}{\ell,{n+1}}+\\
+<![CDATA[&\quad\quad]]>
+\alpha_1(-\matrixentry{B}{\ell f_1})+
+\alpha_2(-\matrixentry{B}{\ell f_2})+
+\alpha_3(-\matrixentry{B}{\ell f_3})+
 \cdots+
-\alpha_{n-r}(-\matrixentry{B}{\ell,f_{n-r}})
+\alpha_{n-r}(-\matrixentry{B}{\ell f_{n-r}})
 +\\
 <![CDATA[&\quad\quad]]>
 \matrixentry{B}{\ell f_1}\alpha_1+
 \cdots+
 \vectorentry{\vect{w}}{n}\vect{A}_n
 <![CDATA[&&]]>\text{<acroref type="theorem" acro="SLSLC" />}\\
-<![CDATA[&\quad\quad\quad\quad]]>
+<![CDATA[&\quad\quad]]>
 +
 \vectorentry{\vect{z}}{1}\vect{A}_1+
 \vectorentry{\vect{z}}{2}\vect{A}_2+
 \vectorentry{\vect{z}}{n}\vect{A}_n\\
 <![CDATA[&=]]>
 \left(\vectorentry{\vect{w}}{1}+\vectorentry{\vect{z}}{1}\right)\vect{A}_1+
-\left(\vectorentry{\vect{w}}{2}+\vectorentry{\vect{z}}{2}\right)\vect{A}_2+
+\left(\vectorentry{\vect{w}}{2}+\vectorentry{\vect{z}}{2}\right)\vect{A}_2
+<![CDATA[&&]]>\text{<acroref type="theorem" acro="VSPCV" />}\\
+<![CDATA[&\quad\quad]]>
+\left(\vectorentry{\vect{w}}{3}+\vectorentry{\vect{z}}{3}\right)\vect{A}_3+
 \cdots+
-\left(\vectorentry{\vect{w}}{n}+\vectorentry{\vect{z}}{n}\right)\vect{A}_n
-<![CDATA[&&]]>\text{<acroref type="theorem" acro="VSPCV" />}\\
+\left(\vectorentry{\vect{w}}{n}+\vectorentry{\vect{z}}{n}\right)\vect{A}_n\\
 <![CDATA[&=]]>
-\vectorentry{\vect{w}+\vect{z}}{1}\vect{A}_1+
++\vectorentry{\vect{w}+\vect{z}}{1}\vect{A}_1+
 \vectorentry{\vect{w}+\vect{z}}{2}\vect{A}_2+
-\vectorentry{\vect{w}+\vect{z}}{3}\vect{A}_3+
 \cdots+
 \vectorentry{\vect{w}+\vect{z}}{n}\vect{A}_n
 <![CDATA[&&]]>\text{<acroref type="definition" acro="CVA" />}\\
 \cdots+
 \vectorentry{\vect{y}}{n}\vect{A}_n
 <![CDATA[&&]]>\text{<acroref type="theorem" acro="SLSLC" />}\\
-<![CDATA[&\quad\quad\quad\quad]]>
+<![CDATA[&\quad\quad]]>
 -
 \left(
 \vectorentry{\vect{w}}{1}\vect{A}_1+
 \right)\\
 <![CDATA[&=]]>
 \left(\vectorentry{\vect{y}}{1}-\vectorentry{\vect{w}}{1}\right)\vect{A}_1+
-\left(\vectorentry{\vect{y}}{2}-\vectorentry{\vect{w}}{2}\right)\vect{A}_2+
+\left(\vectorentry{\vect{y}}{2}-\vectorentry{\vect{w}}{2}\right)\vect{A}_2
+<![CDATA[&&]]>\text{<acroref type="theorem" acro="VSPCV" />}\\
+<![CDATA[&\quad\quad]]>
++
+\left(\vectorentry{\vect{y}}{3}-\vectorentry{\vect{w}}{3}\right)\vect{A}_3+
 \cdots+
-\left(\vectorentry{\vect{y}}{n}-\vectorentry{\vect{w}}{n}\right)\vect{A}_n
-<![CDATA[&&]]>\text{<acroref type="theorem" acro="VSPCV" />}\\
+\left(\vectorentry{\vect{y}}{n}-\vectorentry{\vect{w}}{n}\right)\vect{A}_n\\
 <![CDATA[&=]]>
 \vectorentry{\vect{y}-\vect{w}}{1}\vect{A}_1+
-\vectorentry{\vect{y}-\vect{w}}{2}\vect{A}_2+
-\vectorentry{\vect{y}-\vect{w}}{3}\vect{A}_3+
+\vectorentry{\vect{y}-\vect{w}}{2}\vect{A}_2
+<![CDATA[&&]]>\text{<acroref type="definition" acro="CVA" />}\\
+<![CDATA[&\quad\quad]]>
++\vectorentry{\vect{y}-\vect{w}}{3}\vect{A}_3+
 \cdots+
 \vectorentry{\vect{y}-\vect{w}}{n}\vect{A}_n
-<![CDATA[&&]]>\text{<acroref type="definition" acro="CVA" />}
 </alignmath>
 By <acroref type="theorem" acro="SLSLC" /> we see that the vector $\vect{y}-\vect{w}$ is a solution to the homogeneous system $\homosystem{A}$ and by <acroref type="definition" acro="NSM" />, $\vect{y}-\vect{w}\in\nsp{A}$.  In other words, $\vect{y}-\vect{w}=\vect{z}$ for some vector $\vect{z}\in\nsp{A}$.  Rewritten, this is
 $\vect{y}=\vect{w}+\vect{z}$, as desired.</p>

src/section-LDS.xml

 <![CDATA[\vect{w}&=]]>
 \alpha_1\vect{v}_{d_1}+
 \alpha_2\vect{v}_{d_2}+
-\alpha_3\vect{v}_{d_3}+
 \ldots+
 \alpha_r\vect{v}_{d_r}+
 \beta_1\vect{v}_{f_1}+
 \beta_2\vect{v}_{f_2}+
-\beta_3\vect{v}_{f_3}+
 \ldots+
 \beta_{n-r}\vect{v}_{f_{n-r}}
 </alignmath>
 
 <p>From the above, we can replace each $\vect{v}_{f_j}$ by a linear combination of the $\vect{v}_{d_i}$,
 <alignmath>
-<![CDATA[\vect{w}=&]]>
-\ \alpha_1\vect{v}_{d_1}+
+<![CDATA[\vect{w}&=]]>
+\alpha_1\vect{v}_{d_1}+
 \alpha_2\vect{v}_{d_2}+
-\alpha_3\vect{v}_{d_3}+
 \ldots+
 \alpha_r\vect{v}_{d_r}+\\
 <![CDATA[&\beta_1\left(]]>
 \ldots+
 \matrixentry{B}{r,f_2}\vect{v}_{d_r}
 \right)+\\
-<![CDATA[&\beta_3\left(]]>
-\matrixentry{B}{1,f_3}\vect{v}_{d_1}+
-\matrixentry{B}{2,f_3}\vect{v}_{d_2}+
-\matrixentry{B}{3,f_3}\vect{v}_{d_3}+
-\ldots+
-\matrixentry{B}{r,f_3}\vect{v}_{d_r}
-\right)+\\
 <![CDATA[&\quad\quad\vdots\\]]>
 <![CDATA[&\beta_{n-r}\left(]]>
 \matrixentry{B}{1,f_{n-r}}\vect{v}_{d_1}+
 \ldots+
 \beta_{n-r}\matrixentry{B}{2,f_{n-r}}
 \right)\vect{v}_{d_2}+\\
-<![CDATA[&\left(\alpha_3+]]>
-\beta_1\matrixentry{B}{3,f_1}+
-\beta_2\matrixentry{B}{3,f_2}+
-\beta_3\matrixentry{B}{3,f_3}+
-\ldots+
-\beta_{n-r}\matrixentry{B}{3,f_{n-r}}
-\right)\vect{v}_{d_3}+\\
 <![CDATA[&\quad\quad\vdots\\]]>
 <![CDATA[&\left(\alpha_r+]]>
 \beta_1\matrixentry{B}{r,f_1}+

src/section-NM.xml

 
 
 </sageadvice>
-<p>The next theorem combines with our main computational technique (row-reducing a matrix) to make it easy to recognize a nonsingular matrix.  But first a definition.</p>
+<p>The next theorem combines with our main computational technique (row reducing a matrix) to make it easy to recognize a nonsingular matrix.  But first a definition.</p>
 
 <definition acro="IM" index="matrix!identity">
 <title>Identity Matrix</title>
 <p>Given the coefficient matrix from <acroref type="archetype" acro="B" />,
 <equation>
 A=<archetypepart acro="B" part="purematrix" /></equation>
-the homogeneous system $\homosystem{A}$ has a solution set constructed in <acroref type="example" acro="HUSAB" /> that contains only the trivial solution, so the null space has only a single element,
+the solution set to the homogeneous system $\homosystem{A}$ is constructed in <acroref type="example" acro="HUSAB" /> and contains only the trivial solution, so the null space has only a single element,
 <equation>
 \nsp{A}=\set{\colvector{0\\0\\0}}
 </equation></p>

src/section-O.xml

 <example acro="CSIP" index="inner product">
 <title>Computing some inner products</title>
 
-<p>The scalar product of
+<p>The inner product of
 <alignmath>
 <![CDATA[\vect{u}=\colvector{2+3i\\5+2i\\-3+i}&&\text{and}&&]]>
 \vect{v}=\colvector{1+2i\\-4+5i\\0+5i}
 </alignmath>
 </p>
 
-<p>The scalar product of
+<p>The inner product of
 <alignmath>
 <![CDATA[\vect{w}=\colvector{2\\4\\-3\\2\\8}&&\text{and}&&]]>
 \vect{x}=\colvector{3\\1\\0\\-1\\-2}
 </alignmath>
 is
-<equation>
-\innerproduct{\vect{w}}{\vect{x}}=
-(\conjugate{2})3+(\conjugate{4})1+(\conjugate{-3})0+(\conjugate{2})(-1)+(\conjugate{8})(-2)
-=2(3)+4(1)+(-3)0+2(-1)+8(-2)=-8.
-</equation>
+<alignmath>
+<![CDATA[\innerproduct{\vect{w}}{\vect{x}}&=]]>
+(\conjugate{2})3+(\conjugate{4})1+(\conjugate{-3})0+(\conjugate{2})(-1)+(\conjugate{8})(-2)\\
+<![CDATA[&=2(3)+4(1)+(-3)0+2(-1)+8(-2)=-8.]]>
+</alignmath>
 </p>
 
 </example>
 \vect{u}=\colvector{3+2i\\1-6i\\2+4i\\2+i}
 </equation>
 is
-<equation>
-\norm{\vect{u}}=
-\sqrt{\modulus{3+2i}^2+\modulus{1-6i}^2+\modulus{2+4i}^2+\modulus{2+i}^2}
-=\sqrt{13+37+20+5}=\sqrt{75}=5\sqrt{3}.
-</equation>
+<alignmath>
+<![CDATA[\norm{\vect{u}}&=]]>
+\sqrt{\modulus{3+2i}^2+\modulus{1-6i}^2+\modulus{2+4i}^2+\modulus{2+i}^2}\\
+<![CDATA[&=\sqrt{13+37+20+5}=\sqrt{75}=5\sqrt{3}]]>
+</alignmath>
 </p>
 
 <p>The norm of
 
 <p>Then, for every $1\leq i\leq n$, we have
 <alignmath>
-\alpha_i
-<![CDATA[&=]]>
-\frac{1}{\innerproduct{\vect{u}_i}{\vect{u}_i}}\left(
-\alpha_i\innerproduct{\vect{u}_i}{\vect{u}_i}
-\right)
-<![CDATA[&&]]>\text{<acroref type="theorem" acro="PIP" />}\\
-<![CDATA[&=]]>
-\frac{1}{\innerproduct{\vect{u}_i}{\vect{u}_i}}\left(
-\alpha_1(0)+\alpha_2(0)+\cdots+\alpha_i\innerproduct{\vect{u}_i}{\vect{u}_i}+\cdots+\alpha_n(0)
-\right)
+<![CDATA[&\alpha_i\innerproduct{\vect{u}_i}{\vect{u}_i}\\]]>
+<![CDATA[&\quad\quad=\alpha_1(0)+\alpha_2(0)+\cdots+\alpha_i\innerproduct{\vect{u}_i}{\vect{u}_i}+\cdots+\alpha_n(0)]]>
 <![CDATA[&&]]>\text{<acroref type="property" acro="ZCN" />}\\
-<![CDATA[&=]]>
-\frac{1}{\innerproduct{\vect{u}_i}{\vect{u}_i}}\left(
+<![CDATA[&\quad\quad=]]>
 \alpha_1\innerproduct{\vect{u}_i}{\vect{u}_1}+
 \cdots+
 \alpha_i\innerproduct{\vect{u}_i}{\vect{u}_i}+
 \cdots+
 \alpha_n\innerproduct{\vect{u}_i}{\vect{u}_n}
-\right)
 <![CDATA[&&]]>\text{<acroref type="definition" acro="OSV" />}\\
-<![CDATA[&=]]>
-\frac{1}{\innerproduct{\vect{u}_i}{\vect{u}_i}}\left(
+<![CDATA[&\quad\quad=]]>
 \innerproduct{\vect{u}_i}{\alpha_1\vect{u}_1}+
 \innerproduct{\vect{u}_i}{\alpha_2\vect{u}_2}+
 \cdots+
 \innerproduct{\vect{u}_i}{\alpha_n\vect{u}_n}
-\right)
 <![CDATA[&&]]>\text{<acroref type="theorem" acro="IPSM" />}\\
-<![CDATA[&=]]>
-\frac{1}{\innerproduct{\vect{u}_i}{\vect{u}_i}}
+<![CDATA[&\quad\quad=]]>
 \innerproduct{\vect{u}_i}{\lincombo{\alpha}{u}{n}}
 <![CDATA[&&]]>\text{<acroref type="theorem" acro="IPVA" />}\\
-<![CDATA[&=]]>
-\frac{1}{\innerproduct{\vect{u}_i}{\vect{u}_i}}
+<![CDATA[&\quad\quad=]]>
 \innerproduct{\vect{u}_i}{\zerovector}
 <![CDATA[&&]]>\text{<acroref type="definition" acro="RLDCV" />}\\
-<![CDATA[&=]]>
-\frac{1}{\innerproduct{\vect{u}_i}{\vect{u}_i}}\,0
-<![CDATA[&&]]>\text{<acroref type="definition" acro="IP" />}\\
-<![CDATA[&=0&&]]>\text{<acroref type="property" acro="ZCN" />}
+<![CDATA[&\quad\quad=0]]>
+<![CDATA[&&]]>\text{<acroref type="definition" acro="IP" />}
 </alignmath></p>
 
-<p>So we conclude that $\alpha_i=0$ for all $1\leq i\leq n$ in any relation of linear dependence on $S$.  But this says that $S$ is a linearly independent set since the only way to form a relation of linear dependence is the trivial way (<acroref type="definition" acro="LICV" />).  Boom!</p>
+<p>Because $\vect{u}_i$ was assumed to be nozero, <acroref type="theorem" acro="PIP" /> says $\innerproduct{\vect{u}_i}{\vect{u}_i}$ is nonzero and thus $\alpha_i$ must be zero.  So we conclude that $\alpha_i=0$ for all $1\leq i\leq n$ in any relation of linear dependence on $S$.  But this says that $S$ is a linearly independent set since the only way to form a relation of linear dependence is the trivial way (<acroref type="definition" acro="LICV" />).  Boom!</p>
 
 </proof>
 </theorem>

src/section-RREF.xml

 r := 0
 for j := 1 to n
    i := r+1
-   while i <![CDATA[<= m]]> and A[i,j] = 0
+   while i <![CDATA[<= m]]> and A[i,j] == 0
        i := i+1
    if i <![CDATA[<]]> m+1
        r := r+1
        swap rows i and r of A (row op 1)
-       scale entry in row r, column j of A to a leading 1 (row op 2)
-           for k := 1 to m, k <![CDATA[<>]]> r
-          zero out entry in row k, column j of A (row op 3 using row r)
+       scale A[r,j] to a leading 1 (row op 2)
+       for k := 1 to m, k <![CDATA[<>]]> r
+           make A[k,j] zero (row op 3, employing row r)
 output r and A
 </programlisting>
 

src/section-SS.xml

 
 <p>This equation says that whenever we encounter the vector $\vect{w}_4$, we can replace it with a specific linear combination of the vectors $\vect{w}_1$ and $\vect{w}_2$.  So using $\vect{w}_4$ in the set $T$, along with $\vect{w}_1$ and $\vect{w}_2$, is excessive.  An example of what we mean here can be illustrated by the computation,
 <alignmath>
-<![CDATA[5\vect{w}_1+(-4)\vect{w}_2+6\vect{w}_3+(-3)\vect{w}_4&=]]>
-5\vect{w}_1+(-4)\vect{w}_2+6\vect{w}_3+(-3)\left((-2)\vect{w}_1+(-3)\vect{w}_2\right)\\
+<![CDATA[5\vect{w}_1&+(-4)\vect{w}_2+6\vect{w}_3+(-3)\vect{w}_4\\]]>
+<![CDATA[&=5\vect{w}_1+(-4)\vect{w}_2+6\vect{w}_3+(-3)\left((-2)\vect{w}_1+(-3)\vect{w}_2\right)\\]]>
 <![CDATA[&=5\vect{w}_1+(-4)\vect{w}_2+6\vect{w}_3+\left(6\vect{w}_1+9\vect{w}_2\right)\\]]>
 <![CDATA[&=11\vect{w}_1+5\vect{w}_2+6\vect{w}_3]]>
 </alignmath></p>

src/section-SSLE.xml

 <ol><li> Show $S\subseteq T$.  Suppose $(x_1,\,x_2,\,\,x_3,\,\ldots,x_n)=(\beta_1,\,\beta_2,\,\,\beta_3,\,\ldots,\beta_n)\in S$ is a solution to the original system.  Ignoring the $j$-th equation for a moment, we know this solution makes all the other equations of the transformed system true.  Using the fact that the solution makes the $i$-th and $j$-th equations of the original system true, we find
 <alignmath>
 <![CDATA[&(\alpha a_{i1}+a_{j1})\beta_1+(\alpha a_{i2}+a_{j2})\beta_2+\dots+(\alpha a_{in}+a_{jn})\beta_n\\]]>
-<![CDATA[&\quad\quad=(\alpha a_{i1}\beta_1+\alpha a_{i2}\beta_2+\dots+\alpha a_{in}\beta_n)+]]>
+<![CDATA[&\quad=(\alpha a_{i1}\beta_1+\alpha a_{i2}\beta_2+\dots+\alpha a_{in}\beta_n)+]]>
 (a_{j1}\beta_1+a_{j2}\beta_2+\dots+a_{jn}\beta_n)\\
-<![CDATA[&\quad\quad=\alpha(a_{i1}\beta_1+a_{i2}\beta_2+\dots+a_{in}\beta_n)+]]>
+<![CDATA[&\quad=\alpha(a_{i1}\beta_1+a_{i2}\beta_2+\dots+a_{in}\beta_n)+]]>
 (a_{j1}\beta_1+a_{j2}\beta_2+\dots+a_{jn}\beta_n)\\
-<![CDATA[&\quad\quad=\alpha b_i+b_j.]]>
+<![CDATA[&\quad=\alpha b_i+b_j.]]>
 </alignmath>
 This says that the $j$-th equation of the transformed system is also true, so we have established that $(\beta_1,\,\beta_2,\,\,\beta_3,\,\ldots,\beta_n)\in T$, and therefore $S\subseteq T$.
 </li><li> Now show $T\subseteq S$.  Suppose $(x_1,\,x_2,\,\,x_3,\,\ldots,x_n)=(\beta_1,\,\beta_2,\,\,\beta_3,\,\ldots,\beta_n)\in T$ is a solution to the transformed system.  Ignoring the $j$-th equation for a moment, we know it makes all the other equations of the original system true.  We then find
 <alignmath>
-<![CDATA[&a_{j1}\beta_1+a_{j2}\beta_2+\dots+a_{jn}\beta_n\\]]>
-<![CDATA[&\quad\quad=a_{j1}\beta_1+a_{j2}\beta_2+\dots+a_{jn}\beta_n +\alpha b_i -\alpha b_i\\]]>
-<![CDATA[&\quad\quad=a_{j1}\beta_1+a_{j2}\beta_2+\dots+a_{jn}\beta_n +(\alpha a_{i1}\beta_1+\alpha a_{i2}\beta_2+\dots+\alpha a_{in}\beta_n) -\alpha b_i\\]]>
-<![CDATA[&\quad\quad=a_{j1}\beta_1+\alpha a_{i1}\beta_1+ a_{j2}\beta_2+\alpha a_{i2}\beta_2+ \dots+ a_{jn}\beta_n+\alpha a_{in}\beta_n-\alpha b_i\\]]>
-<![CDATA[&\quad\quad=(\alpha a_{i1}+a_{j1})\beta_1+(\alpha a_{i2}+a_{j2})\beta_2+\dots+(\alpha a_{in}+a_{jn})\beta_n -\alpha b_i\\]]>
+<![CDATA[&a_{j1}\beta_1+\dots+a_{jn}\beta_n\\]]>
+<![CDATA[&\quad\quad=a_{j1}\beta_1+\dots+a_{jn}\beta_n +\alpha b_i -\alpha b_i\\]]>
+<![CDATA[&\quad\quad=a_{j1}\beta_1+\dots+a_{jn}\beta_n +(\alpha a_{i1}\beta_1+\dots+\alpha a_{in}\beta_n) -\alpha b_i\\]]>
+<![CDATA[&\quad\quad=a_{j1}\beta_1+\alpha a_{i1}\beta_1+\dots+a_{jn}\beta_n+\alpha a_{in}\beta_n-\alpha b_i\\]]>
+<![CDATA[&\quad\quad=(\alpha a_{i1}+a_{j1})\beta_1+\dots+(\alpha a_{in}+a_{jn})\beta_n -\alpha b_i\\]]>
 <![CDATA[&\quad\quad=\alpha b_i + b_j -\alpha b_i\\]]>
 <![CDATA[&\quad\quad=b_j]]>
 </alignmath>
 <![CDATA[0&=0]]>
 </alignmath></p>
 
-<p>What does the equation $0=0$ mean?  We can choose <em>any</em> values for $x_1,\,x_2,\,x_3,\,x_4$ and this equation will be true, so we only need to consider further the first two equations, since the third is true no matter what.  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's choose $x_3$ and $x_4$ to be <em>anything</em> we please, say $x_3=a$ and $x_4=b$.</p>
+<p>What does the equation $0=0$ mean?  We can choose <em>any</em> values for $x_1$, $x_2$, $x_3$, $x_4$ and this equation will be true, so we only need to consider further the first two equations, since the third is true no matter what.  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's choose $x_3$ and $x_4$ to be <em>anything</em> we please, say $x_3=a$ and $x_4=b$.</p>
 
 <p>Now we can take these arbitrary values for $x_3$ and $x_4$, substitute them in equation 1,
 to obtain

src/section-WILA.xml

 
 <p>Recalling your weekly meeting with the CEO suggests that you might want to choose a production schedule that yields the biggest possible profit for the company.  So you compute an expression for the profit based on your as yet undetermined decision for the value of $f$,
 <alignmath>
-<![CDATA[(4f-3300)(4.99-3.70)+(-5f+4800)(5.50-3.85)+(f)(6.50-4.45)&=-1.04f + 3663]]>
+<![CDATA[&(4f-3300)(4.99-3.70)+(-5f+4800)(5.50-3.85)+(f)(6.50-4.45)\\]]>
+<![CDATA[&\quad\quad=-1.04f + 3663]]>
 </alignmath>
 </p>
 
 
 <p>In the food industry, things do not stay the same for long, and now the sales department says that increased competition has led to the decision to stay competitive and charge just \$5.25 for a kilogram of the standard mix, rather than the previous \$5.50 per kilogram.  This decision has no effect on the possibilities for the production schedule, but will affect the decision based on profit considerations.  So you revisit just the profit computation, suitably adjusted for the new selling price of standard mix,
 <alignmath>
-<![CDATA[(4f-3300)(4.99-3.70)+(-5f+4800)(5.25-3.85)+(f)(6.50-4.45)&=0.21f+2463]]>
+<![CDATA[&(4f-3300)(4.99-3.70)+(-5f+4800)(5.25-3.85)+(f)(6.50-4.45)\\]]>
+<![CDATA[&\quad\quad=0.21f+2463]]>
 </alignmath>
 </p>
 
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