$\begin{cases} \dfrac { 2 x + 4 } { 5 } = \dfrac { 2 x - y } { 2 } \\ \dfrac { 2 x - y } { 2 } = \dfrac { 4 x + y } { 3 } \end{cases}$
$ $ Solve a solution to $ x$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 5 } { 6 } } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 4 } { 3 } } \\ \dfrac { 2 x - y } { 2 } = \dfrac { 4 x + y } { 3 } \end{cases}$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 5 } { 6 } } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 4 } { 3 } } \\ \color{#FF6800}{ \dfrac { 2 x - y } { 2 } } = \color{#FF6800}{ \dfrac { 4 x + y } { 3 } } \end{cases}$
$ $ Substitute the given $ x $ value into the equation $ \dfrac { 2 x - y } { 2 } = \dfrac { 4 x + y } { 3 }$
$\color{#FF6800}{ \dfrac { 2 \left ( \dfrac { 5 } { 6 } y + \dfrac { 4 } { 3 } \right ) - y } { 2 } } = \color{#FF6800}{ \dfrac { 4 \left ( \dfrac { 5 } { 6 } y + \dfrac { 4 } { 3 } \right ) + y } { 3 } }$
$\color{#FF6800}{ \dfrac { 2 \left ( \dfrac { 5 } { 6 } y + \dfrac { 4 } { 3 } \right ) - y } { 2 } } = \color{#FF6800}{ \dfrac { 4 \left ( \dfrac { 5 } { 6 } y + \dfrac { 4 } { 3 } \right ) + y } { 3 } }$
$ $ Solve a solution to $ y$
$\color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 } { 5 } }$
$\color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 } { 5 } }$
$ $ Substitute the given $ y $ value into the equation $ x = \dfrac { 5 } { 6 } y + \dfrac { 4 } { 3 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 5 } { 6 } } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 } { 5 } } \right ) \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 4 } { 3 } }$
$x = \color{#FF6800}{ \dfrac { 5 } { 6 } } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 } { 5 } } \right ) + \dfrac { 4 } { 3 }$
$ $ Calculate the product of rational numbers $ $
$x = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 3 } } + \dfrac { 4 } { 3 }$
$x = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 3 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 4 } { 3 } }$
$ $ Find the sum or difference of the fractions $ $
$x = \color{#FF6800}{ 1 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ 1 }$
$ $ The possible solutions are as follows $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ 1 } , \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 } { 5 } }$
$\color{#FF6800}{ x } = \color{#FF6800}{ 1 } , \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 } { 5 } }$
$ $ Check if it is the solution to the system of equations $ $
$\begin{cases} \color{#FF6800}{ \dfrac { 2 \times 1 + 4 } { 5 } } = \color{#FF6800}{ \dfrac { 2 \times 1 - \left ( - \dfrac { 2 } { 5 } \right ) } { 2 } } \\ \color{#FF6800}{ \dfrac { 2 \times 1 - \left ( - \dfrac { 2 } { 5 } \right ) } { 2 } } = \color{#FF6800}{ \dfrac { 4 \times 1 - \dfrac { 2 } { 5 } } { 3 } } \end{cases}$
$\begin{cases} \color{#FF6800}{ \dfrac { 2 \times 1 + 4 } { 5 } } = \color{#FF6800}{ \dfrac { 2 \times 1 - \left ( - \dfrac { 2 } { 5 } \right ) } { 2 } } \\ \color{#FF6800}{ \dfrac { 2 \times 1 - \left ( - \dfrac { 2 } { 5 } \right ) } { 2 } } = \color{#FF6800}{ \dfrac { 4 \times 1 - \dfrac { 2 } { 5 } } { 3 } } \end{cases}$
$ $ Simplify the equality $ $
$\begin{cases} \color{#FF6800}{ \dfrac { 6 } { 5 } } = \color{#FF6800}{ \dfrac { 6 } { 5 } } \\ \color{#FF6800}{ \dfrac { 6 } { 5 } } = \color{#FF6800}{ \dfrac { 6 } { 5 } } \end{cases}$
$\begin{cases} \color{#FF6800}{ \dfrac { 6 } { 5 } } = \color{#FF6800}{ \dfrac { 6 } { 5 } } \\ \color{#FF6800}{ \dfrac { 6 } { 5 } } = \color{#FF6800}{ \dfrac { 6 } { 5 } } \end{cases}$
$ $ Since it is true in both equations, it is the solution of the system of equations $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ 1 } , \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 } { 5 } }$