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Formula
Solve the system of equations
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$x - \dfrac { y - 5 } { 2 } = 8$
$\dfrac { 5 } { 6 } x - \dfrac { y } { 4 } = \dfrac { 19 } { 4 }$
$x$-intercept
$\left ( \dfrac { 11 } { 2 } , 0 \right )$
$y$-intercept
$\left ( 0 , - 11 \right )$
$x$-intercept
$\left ( \dfrac { 57 } { 10 } , 0 \right )$
$y$-intercept
$\left ( 0 , - 19 \right )$
$\begin{cases} x- \dfrac{ y-5 }{ 2 } = 8 \\ \dfrac{ 5 }{ 6 } x- \dfrac{ y }{ 4 } = \dfrac{ 19 }{ 4 } \end{cases}$
$x = 6 , y = 1$
Solve quadratic equations using the square root
$\begin{cases} x - \dfrac { y - 5 } { 2 } = 8 \\ \color{#FF6800}{ \dfrac { 5 } { 6 } } \color{#FF6800}{ x } - \dfrac { y } { 4 } = \dfrac { 19 } { 4 } \end{cases}$
 Calculate the multiplication expression 
$\begin{cases} x - \dfrac { y - 5 } { 2 } = 8 \\ \color{#FF6800}{ \dfrac { 5 x } { 6 } } - \dfrac { y } { 4 } = \dfrac { 19 } { 4 } \end{cases}$
$\begin{cases} x - \dfrac { y - 5 } { 2 } = 8 \\ \dfrac { 5 x } { 6 } - \dfrac { y } { 4 } = \dfrac { 19 } { 4 } \end{cases}$
 Solve a solution to $x$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 1 } { 2 } } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 11 } { 2 } } \\ \dfrac { 5 x } { 6 } - \dfrac { y } { 4 } = \dfrac { 19 } { 4 } \end{cases}$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 1 } { 2 } } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 11 } { 2 } } \\ \color{#FF6800}{ \dfrac { 5 x } { 6 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { y } { 4 } } = \color{#FF6800}{ \dfrac { 19 } { 4 } } \end{cases}$
 Substitute the given $x$ value into the equation $\dfrac { 5 x } { 6 } - \dfrac { y } { 4 } = \dfrac { 19 } { 4 }$
$\color{#FF6800}{ \dfrac { 5 \left ( \dfrac { 1 } { 2 } y + \dfrac { 11 } { 2 } \right ) } { 6 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { y } { 4 } } = \color{#FF6800}{ \dfrac { 19 } { 4 } }$
$\color{#FF6800}{ \dfrac { 5 \left ( \dfrac { 1 } { 2 } y + \dfrac { 11 } { 2 } \right ) } { 6 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { y } { 4 } } = \color{#FF6800}{ \dfrac { 19 } { 4 } }$
 Solve a solution to $y$
$\color{#FF6800}{ y } = \color{#FF6800}{ 1 }$
$\color{#FF6800}{ y } = \color{#FF6800}{ 1 }$
 Substitute the given $y$ value into the equation $x = \dfrac { 1 } { 2 } y + \dfrac { 11 } { 2 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 1 } { 2 } } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 11 } { 2 } }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 1 } { 2 } } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 11 } { 2 } }$
 Organize the expression 
$\color{#FF6800}{ x } = \color{#FF6800}{ 6 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ 6 }$
 The possible solutions are as follows 
$\color{#FF6800}{ x } = \color{#FF6800}{ 6 } , \color{#FF6800}{ y } = \color{#FF6800}{ 1 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ 6 } , \color{#FF6800}{ y } = \color{#FF6800}{ 1 }$
 Check if it is the solution to the system of equations 
$\begin{cases} \color{#FF6800}{ 6 } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 - 5 } { 2 } } = \color{#FF6800}{ 8 } \\ \color{#FF6800}{ \dfrac { 5 } { 6 } } \color{#FF6800}{ \times } \color{#FF6800}{ 6 } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 4 } } = \color{#FF6800}{ \dfrac { 19 } { 4 } } \end{cases}$
$\begin{cases} \color{#FF6800}{ 6 } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 - 5 } { 2 } } = \color{#FF6800}{ 8 } \\ \color{#FF6800}{ \dfrac { 5 } { 6 } } \color{#FF6800}{ \times } \color{#FF6800}{ 6 } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 4 } } = \color{#FF6800}{ \dfrac { 19 } { 4 } } \end{cases}$
 Simplify the equality 
$\begin{cases} \color{#FF6800}{ 8 } = \color{#FF6800}{ 8 } \\ \color{#FF6800}{ \dfrac { 19 } { 4 } } = \color{#FF6800}{ \dfrac { 19 } { 4 } } \end{cases}$
$\begin{cases} \color{#FF6800}{ 8 } = \color{#FF6800}{ 8 } \\ \color{#FF6800}{ \dfrac { 19 } { 4 } } = \color{#FF6800}{ \dfrac { 19 } { 4 } } \end{cases}$
 Since it is true in both equations, it is the solution of the system of equations 
$\color{#FF6800}{ x } = \color{#FF6800}{ 6 } , \color{#FF6800}{ y } = \color{#FF6800}{ 1 }$
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