%html %head %title Linear Equations 2 %script{:type => "text/x-mathjax-config"} MathJax.Hub.Config({tex2jax: {inlineMath: [['$','$'], ['\\(','\\)']]}}); %script{:type => "text/javascript", :src => "http://cdn.mathjax.org/mathjax/latest/MathJax.js?config=TeX-AMS-MML_HTMLorMML"} %link{:rel => 'stylesheet', :type => 'text/css', :href => './style.css'} %body %h2 Solve for $x$: #solve-for-x-01.solution :escaped $$ \begin{eqnarray} 6x + 8 & = & 5 \\ -8 + 6x + 8 & = & 5 - 8 \\ 6x & = & 5 - 8 \\ 6x & = & -3 \\ 6x \times \frac{1}{6} & = & -3 \times \frac{1}{6} \\ x & = & -3 \times \frac{1}{6} \\ x & = & -3 \over 6 \\ x & = & \frac{-1}{2} \\ \end{eqnarray} $$ %h2 Solve for $x$ (again): #solve-for-x-02.solution :escaped $$ \begin{eqnarray} 10x - 2 & = & 8 \\ 10x - 2 + 2 & = & 8 + 2 \\ 10x & = & 10 \\ 10x \over 10 & = & 10 \over 10 \\ x & = & 10 \over 10 \\ x & = & 1 \\ \end{eqnarray} $$ %h2 Solve for $x$ (again): #solve-for-x-03.solution :escaped $$ \begin{eqnarray} 6x - 4 & = & 3 \\ 6x - 4 + 4 & = & 3 + 4 \\ 6x & = & 7 \\ 6x \over 6 & = & 7 \over 6 \\ x & = & 7 \over 6 \\ \end{eqnarray} $$ %h2 Solve for $x$ (again): #solve-for-x-04.solution :escaped $$ \begin{eqnarray} 10x + 7 & = & 8 \\ 10x + 7 - 7 & = & 8 - 7 \\ 10x & = & 1 \\ 10x \over 10 & = & 1 \over 10 \\ x & = & 1 \over 10 \\ \end{eqnarray} $$ %h2 Solve for $x$ (again): #solve-for-x-05.solution :escaped $$ \begin{eqnarray} 10x + 3 & = & 5 \\ 10x + 3 - 3 & = & 5 - 3 \\ 10x & = & 2 \\ 10x \over 10 & = & 2 \over 10 \\ x & = & 1 \over 5 \\ \end{eqnarray} $$ %h2 Solve for $x$ (again): #solve-for-x-06.solution :escaped $$ \begin{eqnarray} 4x + 9 & = & 10 \\ 4x + 9 - 9 & = & 10 - 9 \\ 4x & = & 1 \\ 4x \over 4 & = & 1 \over 4 \\ x & = & 1 \over 4 \\ \end{eqnarray} $$ %h2 Solve for $x$ (again): #solve-for-x-07.solution :escaped $$ \begin{eqnarray} 7x - 5 & = & 9 \\ 7x - 5 + 5 & = & 9 + 5 \\ 7x & = & 14 \\ 7x \over 7 & = & 14 \over 7 \\ x & = & 2 \\ \end{eqnarray} $$ %h2 Solve for $x$ (again): #solve-for-x-08.solution :escaped $$ \begin{eqnarray} 3x + 9 & = & 9 \\ 3x + 9 - 9 & = & 9 - 9 \\ 3x & = & 0 \\ 3x \over 3 & = & 0 \over 3 \\ x & = & 0 \\ \end{eqnarray} $$ %h2 Solve for $x$ (again): #solve-for-x-09.solution :escaped $$ \begin{eqnarray} 3x + 10 & = & 7 \\ 3x + 10 - 10 & = & 7 - 10 \\ 3x & = & -3 \\ 3x \over 3 & = & -3 \over 3 \\ x & = & -1 \\ \end{eqnarray} $$