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Solve the equation
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$\dfrac { x - y - 2 } { x + y } = \dfrac { 1 } { 3 }$
$x$-intercept
$\left ( 3 , 0 \right )$
$y$-intercept
$\left ( 0 , - \dfrac { 3 } { 2 } \right )$
$\dfrac{ x-y-2 }{ x+y } = \dfrac{ 1 }{ 3 }$
$\begin{cases} 3 x - 3 y - 6 = x + y \\ x + y \neq 0 \end{cases}$
Solve the fractional equation
$\color{#FF6800}{ \dfrac { x - y - 2 } { x + y } } = \color{#FF6800}{ \dfrac { 1 } { 3 } }$
 If $\frac{a(x)}{b(x)} = \frac{c(x)}{d(x)}$ is valid, it is $\begin{cases} a(x) d(x) = b(x) c(x) \\ b(x) \ne 0 \\ d(x) \ne 0 \end{cases}$
$\begin{cases} \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ y } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \right ) \color{#FF6800}{ \times } \color{#FF6800}{ 3 } = \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ y } \right ) \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \\ \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ y } \neq \color{#FF6800}{ 0 } \\ \color{#FF6800}{ 3 } \neq \color{#FF6800}{ 0 } \end{cases}$
$\begin{cases} \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ y } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \right ) \color{#FF6800}{ \times } \color{#FF6800}{ 3 } = \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ y } \right ) \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \\ \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ y } \neq \color{#FF6800}{ 0 } \\ \color{#FF6800}{ 3 } \neq \color{#FF6800}{ 0 } \end{cases}$
 Simplify the expression 
$\begin{cases} \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ y } \color{#FF6800}{ - } \color{#FF6800}{ 6 } = \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ y } \\ \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ y } \neq \color{#FF6800}{ 0 } \\ \color{#FF6800}{ 3 } \neq \color{#FF6800}{ 0 } \end{cases}$
$\begin{cases} 3 x - 3 y - 6 = x + y \\ x + y \neq 0 \\ \color{#FF6800}{ 3 } \neq \color{#FF6800}{ 0 } \end{cases}$
 There are infinitely many solutions if both sides of $\ne$ are different. 
$\begin{cases} 3 x - 3 y - 6 = x + y \\ x + y \neq 0 \\ \text{There are countless solutions} \end{cases}$
$\begin{cases} \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ y } \color{#FF6800}{ - } \color{#FF6800}{ 6 } = \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ y } \\ \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ y } \neq \color{#FF6800}{ 0 } \\ \text{There are countless solutions} \end{cases}$
 Ignore the cases where the system of equations where there are infinitely many solutions. 
$\begin{cases} \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ y } \color{#FF6800}{ - } \color{#FF6800}{ 6 } = \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ y } \\ \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ y } \neq \color{#FF6800}{ 0 } \end{cases}$
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