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Formula
Solve the equation
Graph
$y = \dfrac { x - 1 } { 2 } - 1$
$y = \dfrac { x + 1 } { 3 }$
$x$Intercept
$\left ( 3 , 0 \right )$
$y$Intercept
$\left ( 0 , - \dfrac { 3 } { 2 } \right )$
$x$Intercept
$\left ( - 1 , 0 \right )$
$y$Intercept
$\left ( 0 , \dfrac { 1 } { 3 } \right )$
$\dfrac{ x-1 }{ 2 } -1 = \dfrac{ x+1 }{ 3 }$
$x = 11$
 Solve a solution to $x$
$\dfrac { x - 1 } { 2 } \color{#FF6800}{ - } \color{#FF6800}{ 1 } = \dfrac { x + 1 } { 3 }$
 Convert an equation to a fraction using $a=\dfrac{a}{1}$
$\dfrac { x - 1 } { 2 } + \color{#FF6800}{ \dfrac { - 1 } { 1 } } = \dfrac { x + 1 } { 3 }$
$\color{#FF6800}{ \dfrac { x - 1 } { 2 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { - 1 } { 1 } } = \dfrac { x + 1 } { 3 }$
 Write all numerators above the least common denominator 
$\color{#FF6800}{ \dfrac { x - 1 - 2 } { 2 } } = \dfrac { x + 1 } { 3 }$
$\dfrac { x \color{#FF6800}{ - } \color{#FF6800}{ 1 } \color{#FF6800}{ - } \color{#FF6800}{ 2 } } { 2 } = \dfrac { x + 1 } { 3 }$
 Find the sum of the negative numbers 
$\dfrac { x \color{#FF6800}{ - } \color{#FF6800}{ 3 } } { 2 } = \dfrac { x + 1 } { 3 }$
$\color{#FF6800}{ \dfrac { x - 3 } { 2 } } = \color{#FF6800}{ \dfrac { x + 1 } { 3 } }$
 Multiply both sides by the least common multiple for the denominators to eliminate the fraction 
$\color{#FF6800}{ 3 } \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right ) = \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 }$
$\color{#FF6800}{ 3 } \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right ) = \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 }$
 Organize the expression 
$\color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } = \color{#FF6800}{ 2 } \color{#FF6800}{ + } \color{#FF6800}{ 9 }$
$\color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } = 2 + 9$
 Organize the expression 
$\color{#FF6800}{ x } = 2 + 9$
$x = \color{#FF6800}{ 2 } \color{#FF6800}{ + } \color{#FF6800}{ 9 }$
 Add $2$ and $9$
$x = \color{#FF6800}{ 11 }$
$x = 11$
Solve the fractional equation
$\color{#FF6800}{ \dfrac { x - 1 } { 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 1 } = \color{#FF6800}{ \dfrac { x + 1 } { 3 } }$
 Reverse the left and right terms of the equation (or inequality) 
$\color{#FF6800}{ \dfrac { x + 1 } { 3 } } = \color{#FF6800}{ \dfrac { x - 1 } { 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 1 }$
$\color{#FF6800}{ \dfrac { x + 1 } { 3 } } = \color{#FF6800}{ \dfrac { x - 1 } { 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 1 }$
 If $\frac{a(x)}{b(x)} = c(x)$ is valid, it is $\begin{cases} a(x) = b(x) c(x) \\ b(x) \ne 0 \end{cases}$
$\begin{cases} \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } = \color{#FF6800}{ 3 } \left ( \color{#FF6800}{ \dfrac { x - 1 } { 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) \\ \color{#FF6800}{ 3 } \neq \color{#FF6800}{ 0 } \end{cases}$
$\begin{cases} \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } = \color{#FF6800}{ 3 } \left ( \color{#FF6800}{ \dfrac { x - 1 } { 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) \\ \color{#FF6800}{ 3 } \neq \color{#FF6800}{ 0 } \end{cases}$
 Simplify the expression 
$\begin{cases} \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } = \color{#FF6800}{ \dfrac { 3 x - 9 } { 2 } } \\ \color{#FF6800}{ 3 } \neq \color{#FF6800}{ 0 } \end{cases}$
$\begin{cases} \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } = \color{#FF6800}{ \dfrac { 3 x - 9 } { 2 } } \\ 3 \neq 0 \end{cases}$
 Reverse the left and right terms of the equation (or inequality) 
$\begin{cases} \color{#FF6800}{ \dfrac { 3 x - 9 } { 2 } } = \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \\ 3 \neq 0 \end{cases}$
$\begin{cases} \color{#FF6800}{ \dfrac { 3 x - 9 } { 2 } } = \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \\ 3 \neq 0 \end{cases}$
 If $\frac{a(x)}{b(x)} = c(x)$ is valid, it is $\begin{cases} a(x) = b(x) c(x) \\ b(x) \ne 0 \end{cases}$
$\begin{cases} \begin{cases} \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 9 } = \color{#FF6800}{ 2 } \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right ) \\ \color{#FF6800}{ 2 } \neq \color{#FF6800}{ 0 } \end{cases} \\ 3 \neq 0 \end{cases}$
$\begin{cases} \begin{cases} \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 9 } = \color{#FF6800}{ 2 } \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right ) \\ \color{#FF6800}{ 2 } \neq \color{#FF6800}{ 0 } \end{cases} \\ \color{#FF6800}{ 3 } \neq \color{#FF6800}{ 0 } \end{cases}$
 If there is a system of equations (inequality) in the system of equations (inequality), take it out. 
$\begin{cases} \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 9 } = \color{#FF6800}{ 2 } \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right ) \\ \color{#FF6800}{ 2 } \neq \color{#FF6800}{ 0 } \\ \color{#FF6800}{ 3 } \neq \color{#FF6800}{ 0 } \end{cases}$
$\begin{cases} \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 9 } = \color{#FF6800}{ 2 } \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right ) \\ \color{#FF6800}{ 2 } \neq \color{#FF6800}{ 0 } \\ \color{#FF6800}{ 3 } \neq \color{#FF6800}{ 0 } \end{cases}$
 Simplify the expression 
$\begin{cases} \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 9 } = \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \\ \color{#FF6800}{ 2 } \neq \color{#FF6800}{ 0 } \\ \color{#FF6800}{ 3 } \neq \color{#FF6800}{ 0 } \end{cases}$
$\begin{cases} \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 9 } = \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \\ 2 \neq 0 \\ 3 \neq 0 \end{cases}$
 Solve a solution to $x$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ 11 } \\ 2 \neq 0 \\ 3 \neq 0 \end{cases}$
$\begin{cases} x = 11 \\ \color{#FF6800}{ 2 } \neq \color{#FF6800}{ 0 } \\ 3 \neq 0 \end{cases}$
 There are infinitely many solutions if both sides of $\ne$ are different. 
$\begin{cases} x = 11 \\ \text{해가 무수히 많습니다} \\ 3 \neq 0 \end{cases}$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ 11 } \\ \text{해가 무수히 많습니다} \\ \color{#FF6800}{ 3 } \neq \color{#FF6800}{ 0 } \end{cases}$
 Ignore the cases where the system of equations where there are infinitely many solutions. 
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ 11 } \\ \color{#FF6800}{ 3 } \neq \color{#FF6800}{ 0 } \end{cases}$
$\begin{cases} x = 11 \\ \color{#FF6800}{ 3 } \neq \color{#FF6800}{ 0 } \end{cases}$
 There are infinitely many solutions if both sides of $\ne$ are different. 
$\begin{cases} x = 11 \\ \text{해가 무수히 많습니다} \end{cases}$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ 11 } \\ \text{해가 무수히 많습니다} \end{cases}$
 Ignore the cases where the system of equations where there are infinitely many solutions. 
$\color{#FF6800}{ x } = \color{#FF6800}{ 11 }$
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