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$1- \dfrac{ 2-3 }{ 5 } = \dfrac{ 2+a }{ 2 }$
$a = \dfrac { 2 } { 5 }$
$ $ Solve a solution to $ a$
$1 - \dfrac { \color{#FF6800}{ 2 } \color{#FF6800}{ - } \color{#FF6800}{ 3 } } { 5 } = \dfrac { 2 + a } { 2 }$
$ $ Subtract $ 3 $ from $ 2$
$1 - \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 1 } } { 5 } = \dfrac { 2 + a } { 2 }$
$\color{#FF6800}{ 1 } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { - 1 } { 5 } } = \dfrac { 2 + a } { 2 }$
$ $ Subtract $ \dfrac { - 1 } { 5 } $ from $ 1$
$\color{#FF6800}{ \dfrac { 6 } { 5 } } = \dfrac { 2 + a } { 2 }$
$\dfrac { 6 } { 5 } = \dfrac { \color{#FF6800}{ 2 } \color{#FF6800}{ + } \color{#FF6800}{ a } } { 2 }$
$ $ Organize the expression $ $
$\dfrac { 6 } { 5 } = \dfrac { \color{#FF6800}{ a } \color{#FF6800}{ + } \color{#FF6800}{ 2 } } { 2 }$
$\color{#FF6800}{ \dfrac { 6 } { 5 } } = \color{#FF6800}{ \dfrac { a + 2 } { 2 } }$
$ $ Multiply both sides by the least common multiple for the denominators to eliminate the fraction $ $
$\color{#FF6800}{ \dfrac { 12 } { 5 } } = \color{#FF6800}{ a } \color{#FF6800}{ + } \color{#FF6800}{ 2 }$
$\dfrac { 12 } { 5 } = \color{#FF6800}{ a } + 2$
$ $ Move the variable to the left-hand side and change the symbol $ $
$\dfrac { 12 } { 5 } - a = 2$
$\color{#FF6800}{ \dfrac { 12 } { 5 } } - a = 2$
$ $ Move the constant to the right side and change the sign $ $
$- a = 2 \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 12 } { 5 } }$
$- a = \color{#FF6800}{ 2 } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 12 } { 5 } }$
$ $ Subtract $ \dfrac { 12 } { 5 } $ from $ 2$
$- a = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 } { 5 } }$
$\color{#FF6800}{ - } \color{#FF6800}{ a } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 } { 5 } }$
$ $ Change the sign of both sides of the equation $ $
$\color{#FF6800}{ a } = \color{#FF6800}{ - } \left ( \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 } { 5 } } \right )$
$a = \color{#FF6800}{ - } \left ( \color{#FF6800}{ - } \dfrac { 2 } { 5 } \right )$
$ $ Simplify Minus $ $
$a = \dfrac { 2 } { 5 }$
$a = \dfrac { 2 } { 5 }$
Solve the fractional equation
$\color{#FF6800}{ 1 } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 - 3 } { 5 } } = \color{#FF6800}{ \dfrac { 2 + a } { 2 } }$
$ $ Reverse the left and right terms of the equation (or inequality) $ $
$\color{#FF6800}{ \dfrac { 2 + a } { 2 } } = \color{#FF6800}{ 1 } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 - 3 } { 5 } }$
$\color{#FF6800}{ \dfrac { 2 + a } { 2 } } = \color{#FF6800}{ 1 } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 - 3 } { 5 } }$
$ $ 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}$
$\begin{cases} \color{#FF6800}{ 2 } \color{#FF6800}{ + } \color{#FF6800}{ a } = \color{#FF6800}{ 2 } \left ( \color{#FF6800}{ 1 } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 - 3 } { 5 } } \right ) \\ \color{#FF6800}{ 2 } \neq \color{#FF6800}{ 0 } \end{cases}$
$\begin{cases} \color{#FF6800}{ 2 } \color{#FF6800}{ + } \color{#FF6800}{ a } = \color{#FF6800}{ 2 } \left ( \color{#FF6800}{ 1 } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 - 3 } { 5 } } \right ) \\ \color{#FF6800}{ 2 } \neq \color{#FF6800}{ 0 } \end{cases}$
$ $ Simplify the expression $ $
$\begin{cases} \color{#FF6800}{ a } \color{#FF6800}{ + } \color{#FF6800}{ 2 } = \color{#FF6800}{ \dfrac { 12 } { 5 } } \\ \color{#FF6800}{ 2 } \neq \color{#FF6800}{ 0 } \end{cases}$
$\begin{cases} \color{#FF6800}{ a } \color{#FF6800}{ + } \color{#FF6800}{ 2 } = \color{#FF6800}{ \dfrac { 12 } { 5 } } \\ 2 \neq 0 \end{cases}$
$ $ Solve a solution to $ a$
$\begin{cases} \color{#FF6800}{ a } = \color{#FF6800}{ \dfrac { 2 } { 5 } } \\ 2 \neq 0 \end{cases}$
$\begin{cases} a = \dfrac { 2 } { 5 } \\ \color{#FF6800}{ 2 } \neq \color{#FF6800}{ 0 } \end{cases}$
$ $ There are infinitely many solutions if both sides of $ \ne $ are different. $ $
$\begin{cases} a = \dfrac { 2 } { 5 } \\ \text{해가 무수히 많습니다} \end{cases}$
$\begin{cases} \color{#FF6800}{ a } = \color{#FF6800}{ \dfrac { 2 } { 5 } } \\ \text{해가 무수히 많습니다} \end{cases}$
$ $ Ignore the cases where the system of equations where there are infinitely many solutions. $ $
$\color{#FF6800}{ a } = \color{#FF6800}{ \dfrac { 2 } { 5 } }$
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