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Solve the system of equations
Answer
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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
$\left ( 0 , 0 \right )$
$\begin{cases} \dfrac{ 2x+4 }{ 5 } = \dfrac{ 2x-y }{ 2 } \\ \dfrac{ 2x-y }{ 2 } = \dfrac{ 4x+y }{ 3 } \end{cases}$
$x = 1 , y = - \dfrac { 2 } { 5 }$
Solve quadratic equations using the square root
$\begin{cases} \dfrac { 2 x + 4 } { 5 } = \dfrac { 2 x - y } { 2 } \\ \dfrac { 2 x - y } { 2 } = \dfrac { 4 x + y } { 3 } \end{cases}$
$ $ 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 $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ 1 } , \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 } { 5 } }$
$\begin{cases} \dfrac { - 14 x + 12 - 5 y } { 15 } = 0 \\ \dfrac { 2 x - y } { 2 } - \dfrac { 4 x + y } { 3 } = 0 \end{cases}$
Solve quadratic equations using the square root
$\begin{cases} \color{#FF6800}{ \dfrac { 2 x + 4 } { 5 } } = \color{#FF6800}{ \dfrac { 2 x - y } { 2 } } \\ \dfrac { 2 x - y } { 2 } = \dfrac { 4 x + y } { 3 } \end{cases}$
$ $ Do transposition $ $
$\begin{cases} \color{#FF6800}{ \dfrac { 2 x + 4 } { 5 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 x - y } { 2 } } = \color{#FF6800}{ 0 } \\ \dfrac { 2 x - y } { 2 } = \dfrac { 4 x + y } { 3 } \end{cases}$
$\begin{cases} \dfrac { 2 x + 4 } { 5 } - \dfrac { 2 x - y } { 2 } = 0 \\ \color{#FF6800}{ \dfrac { 2 x - y } { 2 } } = \color{#FF6800}{ \dfrac { 4 x + y } { 3 } } \end{cases}$
$ $ Do transposition $ $
$\begin{cases} \dfrac { 2 x + 4 } { 5 } - \dfrac { 2 x - y } { 2 } = 0 \\ \color{#FF6800}{ \dfrac { 2 x - y } { 2 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 4 x + y } { 3 } } = \color{#FF6800}{ 0 } \end{cases}$
$\begin{cases} \color{#FF6800}{ \dfrac { 2 x + 4 } { 5 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 x - y } { 2 } } = \color{#FF6800}{ 0 } \\ \color{#FF6800}{ \dfrac { 2 x - y } { 2 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 4 x + y } { 3 } } = \color{#FF6800}{ 0 } \end{cases}$
$ $ Solve the system of linear equations for $ \dfrac { 2 x + 4 } { 5 } , \dfrac { 2 x - y } { 2 } , \dfrac { 4 x + y } { 3 } $
$\begin{cases} \color{#FF6800}{ \dfrac { - 14 x + 12 - 5 y } { 15 } } = \color{#FF6800}{ 0 } \\ \color{#FF6800}{ \dfrac { 2 x - y } { 2 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 4 x + y } { 3 } } = \color{#FF6800}{ 0 } \end{cases}$
$ $ 그래프 보기 $ $
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Solution search results
search-thumbnail-$ \begin{cases} \dfrac {x+y} {2}-\dfrac {x+y} {3}=\dfrac {1} {6} \\ \dfrac {x+y} {3}-\dfrac {x-y} {4}=\dfrac {1} {12} \end{cases} $
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