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Solve the system of equations
Answer
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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 $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 9 } { 2 } } , \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 15 } { 2 } }$
$ $ 그래프 보기 $ $
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Solution search results
search-thumbnail-$.$ $\left(2,4\right)$ $7.$ $\left(0.25$ $0.5\right)$ 
$ \begin{cases} 2y<-x+10 \\ y>-x+2 \end{cases} $ $ \begin{cases} y\leq 2x \\ 2y\geq x \end{cases} $ 
$\left(-\dfrac {1} {3},\dfrac {1} {2}\right)$ $8$ $\left(1,1\right)$ 
$ \begin{cases} y+4<-3x \\ y>-x+2 \end{cases} $ $ \begin{cases} y+10<11x \\ y>-x+2 \end{cases} $
7th-9th grade
Other
search-thumbnail-$∩$ $∩1$ 
$ \begin{cases} 3\left(x-1\right)-2\left(1+x\right)<1 \\ 3x-11>0 \end{cases} $ $ \begin{cases} 3x-3-2+2x47 \\ 3x-4>0 \end{cases} $ 
$ \begin{cases} x<7+5 \\ 222l+4 \end{cases} $ $ \begin{cases} x<6 \\ 3x>4 \end{cases} $
1st-6th grade
Other
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