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Formula
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
$\left ( 0 , 0 \right )$
$\begin{cases} \dfrac{ 2x+4 }{ 5 } = \dfrac{ 2x-y }{ 2 } \\ \dfrac{ 2x-y }{ 2 } = \dfrac{ 4x+y }{ 3 } \end{cases}$
$x = 1 , y = - \dfrac { 2 } { 5 }$
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 $\dfrac { 2 x + 4 } { 5 } , \dfrac { 2 x - y } { 2 } , \dfrac { 4 x + y } { 3 }$
$\begin{cases} \color{#FF6800}{ \dfrac { - 14 x + 12 - 5 y } { 15 } } = \color{#FF6800}{ 0 } \\ \color{#FF6800}{ \dfrac { 2 x - y } { 2 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 4 x + y } { 3 } } = \color{#FF6800}{ 0 } \end{cases}$
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