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Solve the system of equations
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$\dfrac { 2 x + 4 } { 5 } = \dfrac { 2 x - y } { 2 }$
$\dfrac { 2 x - y } { 2 } = \dfrac { 4 x + y } { 3 }$
$x$Intercept
$\left ( \dfrac { 4 } { 3 } , 0 \right )$
$y$Intercept
$\left ( 0 , - \dfrac { 8 } { 5 } \right )$
$x$Intercept
$\left ( 0 , 0 \right )$
$y$Intercept
$\left ( 0 , 0 \right )$
$x = 1 , y = - \dfrac { 2 } { 5 }$
Solve the system of equations
$\begin{cases} \dfrac { 2 x + 4 } { 5 } = \dfrac { 2 x - y } { 2 } \\ \dfrac { 2 x - y } { 2 } = \dfrac { 4 x + y } { 3 } \end{cases}$
$ $ Solve a solution to $ x$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 5 } } { \color{#FF6800}{ 6 } } } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 4 } } { \color{#FF6800}{ 3 } } } \\ \dfrac { 2 x - y } { 2 } = \dfrac { 4 x + y } { 3 } \end{cases}$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 5 } } { \color{#FF6800}{ 6 } } } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 4 } } { \color{#FF6800}{ 3 } } } \\ \color{#FF6800}{ \dfrac { \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ y } } { \color{#FF6800}{ 2 } } } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 4 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ y } } { \color{#FF6800}{ 3 } } } \end{cases}$
$ $ Substitute the given $ x $ value into the equation $ \dfrac { 2 x - y } { 2 } = \dfrac { 4 x + y } { 3 }$
$\color{#FF6800}{ \dfrac { \color{#FF6800}{ 2 } \left ( \color{#FF6800}{ \dfrac { \color{#FF6800}{ 5 } } { \color{#FF6800}{ 6 } } } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 4 } } { \color{#FF6800}{ 3 } } } \right ) \color{#FF6800}{ - } \color{#FF6800}{ y } } { \color{#FF6800}{ 2 } } } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 4 } \left ( \color{#FF6800}{ \dfrac { \color{#FF6800}{ 5 } } { \color{#FF6800}{ 6 } } } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 4 } } { \color{#FF6800}{ 3 } } } \right ) \color{#FF6800}{ + } \color{#FF6800}{ y } } { \color{#FF6800}{ 3 } } }$
$\color{#FF6800}{ \dfrac { \color{#FF6800}{ 2 } \left ( \color{#FF6800}{ \dfrac { \color{#FF6800}{ 5 } } { \color{#FF6800}{ 6 } } } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 4 } } { \color{#FF6800}{ 3 } } } \right ) \color{#FF6800}{ - } \color{#FF6800}{ y } } { \color{#FF6800}{ 2 } } } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 4 } \left ( \color{#FF6800}{ \dfrac { \color{#FF6800}{ 5 } } { \color{#FF6800}{ 6 } } } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 4 } } { \color{#FF6800}{ 3 } } } \right ) \color{#FF6800}{ + } \color{#FF6800}{ y } } { \color{#FF6800}{ 3 } } }$
$ $ Solve a solution to $ y$
$\color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 2 } } { \color{#FF6800}{ 5 } } }$
$\color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 2 } } { \color{#FF6800}{ 5 } } }$
$ $ Substitute the given $ y $ value into the equation $ x = \dfrac { 5 } { 6 } y + \dfrac { 4 } { 3 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 5 } } { \color{#FF6800}{ 6 } } } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 2 } } { \color{#FF6800}{ 5 } } } \right ) \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 4 } } { \color{#FF6800}{ 3 } } }$
$x = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 5 } } { \color{#FF6800}{ 6 } } } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 2 } } { \color{#FF6800}{ 5 } } } \right ) + \dfrac { 4 } { 3 }$
$ $ Calculate the product of rational numbers $ $
$x = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 3 } } } + \dfrac { 4 } { 3 }$
$x = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 3 } } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 4 } } { \color{#FF6800}{ 3 } } }$
$ $ Find the sum or difference of the fractions $ $
$x = \color{#FF6800}{ 1 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ 1 }$
$ $ The possible solutions are as follows $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ 1 } , \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 2 } } { \color{#FF6800}{ 5 } } }$
$\color{#FF6800}{ x } = \color{#FF6800}{ 1 } , \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 2 } } { \color{#FF6800}{ 5 } } }$
$ $ Check if it is the solution to the system of equations $ $
$\begin{cases} \color{#FF6800}{ \dfrac { \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ + } \color{#FF6800}{ 4 } } { \color{#FF6800}{ 5 } } } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ - } \left ( \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 2 } } { \color{#FF6800}{ 5 } } } \right ) } { \color{#FF6800}{ 2 } } } \\ \color{#FF6800}{ \dfrac { \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ - } \left ( \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 2 } } { \color{#FF6800}{ 5 } } } \right ) } { \color{#FF6800}{ 2 } } } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 2 } } { \color{#FF6800}{ 5 } } } } { \color{#FF6800}{ 3 } } } \end{cases}$
$\begin{cases} \color{#FF6800}{ \dfrac { \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ + } \color{#FF6800}{ 4 } } { \color{#FF6800}{ 5 } } } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ - } \left ( \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 2 } } { \color{#FF6800}{ 5 } } } \right ) } { \color{#FF6800}{ 2 } } } \\ \color{#FF6800}{ \dfrac { \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ - } \left ( \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 2 } } { \color{#FF6800}{ 5 } } } \right ) } { \color{#FF6800}{ 2 } } } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 2 } } { \color{#FF6800}{ 5 } } } } { \color{#FF6800}{ 3 } } } \end{cases}$
$ $ Simplify the equality $ $
$\begin{cases} \color{#FF6800}{ \dfrac { \color{#FF6800}{ 6 } } { \color{#FF6800}{ 5 } } } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 6 } } { \color{#FF6800}{ 5 } } } \\ \color{#FF6800}{ \dfrac { \color{#FF6800}{ 6 } } { \color{#FF6800}{ 5 } } } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 6 } } { \color{#FF6800}{ 5 } } } \end{cases}$
$\begin{cases} \color{#FF6800}{ \dfrac { \color{#FF6800}{ 6 } } { \color{#FF6800}{ 5 } } } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 6 } } { \color{#FF6800}{ 5 } } } \\ \color{#FF6800}{ \dfrac { \color{#FF6800}{ 6 } } { \color{#FF6800}{ 5 } } } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 6 } } { \color{#FF6800}{ 5 } } } \end{cases}$
$ $ Since it is true in both equations, it is the solution of the system of equations $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ 1 } , \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 2 } } { \color{#FF6800}{ 5 } } }$
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