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Formula
Solve the system of equations
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$\dfrac { x - 3 y } { 4 } = - 3 - y$
$- 2 \left ( x - 1 \right ) = 3 \left ( x - y \right ) + 2$
$x$-intercept
$\left ( - 12 , 0 \right )$
$y$-intercept
$\left ( 0 , - 12 \right )$
$x$-intercept
$\left ( 0 , 0 \right )$
$y$-intercept
$\left ( 0 , 0 \right )$
$\begin{cases} \dfrac{ x-3y }{ 4 } = -3-y \\-2 \left( x-1 \right) = 3 \left( x-y \right) +2 \end{cases}$
$x = - \dfrac { 9 } { 2 } , y = - \dfrac { 15 } { 2 }$
Solve quadratic equations using the square root
$\begin{cases} \color{#FF6800}{ \dfrac { x - 3 y } { 4 } } = \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ - } \color{#FF6800}{ y } \\ \color{#FF6800}{ - } \color{#FF6800}{ 2 } \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) = \color{#FF6800}{ 3 } \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ y } \right ) \color{#FF6800}{ + } \color{#FF6800}{ 2 } \end{cases}$
 Organize the expression 
$\begin{cases} \color{#FF6800}{ \dfrac { x - 3 y } { 4 } } = \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ - } \color{#FF6800}{ y } \\ \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } = \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \end{cases}$
$\begin{cases} \dfrac { x - 3 y } { 4 } = - 3 - y \\ - 2 x + 2 = 3 x - 3 y + 2 \end{cases}$
 Solve a solution to $y$
$\begin{cases} \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 12 } \\ - 2 x + 2 = 3 x - 3 y + 2 \end{cases}$
$\begin{cases} \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 12 } \\ \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } = \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \end{cases}$
 Substitute the given $y$ value into the equation $- 2 x + 2 = 3 x - 3 y + 2$
$\color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } = \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \left ( \color{#FF6800}{ - } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 12 } \right ) \color{#FF6800}{ + } \color{#FF6800}{ 2 }$
$\color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } = \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \left ( \color{#FF6800}{ - } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 12 } \right ) \color{#FF6800}{ + } \color{#FF6800}{ 2 }$
 Solve a solution to $x$
$\color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 9 } { 2 } }$
$\color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 9 } { 2 } }$
 Substitute the given $x$ value into the equation $y = - x - 12$
$\color{#FF6800}{ y } = \color{#FF6800}{ - } \left ( \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 9 } { 2 } } \right ) \color{#FF6800}{ - } \color{#FF6800}{ 12 }$
$\color{#FF6800}{ y } = \color{#FF6800}{ - } \left ( \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 9 } { 2 } } \right ) \color{#FF6800}{ - } \color{#FF6800}{ 12 }$
 Organize the expression 
$\color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 15 } { 2 } }$
$\color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 15 } { 2 } }$
 The possible solutions are as follows 
$\color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 9 } { 2 } } , \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 15 } { 2 } }$
$\color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 9 } { 2 } } , \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 15 } { 2 } }$
 Check if it is the solution to the system of equations 
$\begin{cases} \color{#FF6800}{ \dfrac { - \dfrac { 9 } { 2 } - 3 \times \left ( - \dfrac { 15 } { 2 } \right ) } { 4 } } = \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ - } \left ( \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 15 } { 2 } } \right ) \\ \color{#FF6800}{ - } \color{#FF6800}{ 2 } \left ( \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 9 } { 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) = \color{#FF6800}{ 3 } \left ( \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 9 } { 2 } } \color{#FF6800}{ - } \left ( \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 15 } { 2 } } \right ) \right ) \color{#FF6800}{ + } \color{#FF6800}{ 2 } \end{cases}$
$\begin{cases} \color{#FF6800}{ \dfrac { - \dfrac { 9 } { 2 } - 3 \times \left ( - \dfrac { 15 } { 2 } \right ) } { 4 } } = \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ - } \left ( \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 15 } { 2 } } \right ) \\ \color{#FF6800}{ - } \color{#FF6800}{ 2 } \left ( \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 9 } { 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) = \color{#FF6800}{ 3 } \left ( \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 9 } { 2 } } \color{#FF6800}{ - } \left ( \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 15 } { 2 } } \right ) \right ) \color{#FF6800}{ + } \color{#FF6800}{ 2 } \end{cases}$
 Simplify the equality 
$\begin{cases} \color{#FF6800}{ \dfrac { 9 } { 2 } } = \color{#FF6800}{ \dfrac { 9 } { 2 } } \\ \color{#FF6800}{ 11 } = \color{#FF6800}{ 11 } \end{cases}$
$\begin{cases} \color{#FF6800}{ \dfrac { 9 } { 2 } } = \color{#FF6800}{ \dfrac { 9 } { 2 } } \\ \color{#FF6800}{ 11 } = \color{#FF6800}{ 11 } \end{cases}$
 Since it is true in both equations, it is the solution of the system of equations 
$\color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 9 } { 2 } } , \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 15 } { 2 } }$
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