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Solve the system of equations
Answer
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Graph
$\dfrac { 1 } { 2 } x - \dfrac { 1 } { 3 } y = \dfrac { 2 } { 3 }$
$\dfrac { 1 } { 3 } x + \dfrac { 1 } { 6 } y = \dfrac { 5 } { 6 }$
$x$Intercept
$\left ( \dfrac { 4 } { 3 } , 0 \right )$
$y$Intercept
$\left ( 0 , - 2 \right )$
$x$Intercept
$\left ( \dfrac { 5 } { 2 } , 0 \right )$
$y$Intercept
$\left ( 0 , 5 \right )$
$\begin{cases} \dfrac{ 1 }{ 2 } x- \dfrac{ 1 }{ 3 } y = \dfrac{ 2 }{ 3 } \\ \dfrac{ 1 }{ 3 } x+ \dfrac{ 1 }{ 6 } y = \dfrac{ 5 }{ 6 } \end{cases}$
$x = 2 , y = 1$
Solve quadratic equations using the square root
$\begin{cases} \color{#FF6800}{ \dfrac { 1 } { 2 } } \color{#FF6800}{ x } - \dfrac { 1 } { 3 } y = \dfrac { 2 } { 3 } \\ \dfrac { 1 } { 3 } x + \dfrac { 1 } { 6 } y = \dfrac { 5 } { 6 } \end{cases}$
$ $ Calculate the multiplication expression $ $
$\begin{cases} \color{#FF6800}{ \dfrac { x } { 2 } } - \dfrac { 1 } { 3 } y = \dfrac { 2 } { 3 } \\ \dfrac { 1 } { 3 } x + \dfrac { 1 } { 6 } y = \dfrac { 5 } { 6 } \end{cases}$
$\begin{cases} \dfrac { x } { 2 } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 3 } } \color{#FF6800}{ y } = \dfrac { 2 } { 3 } \\ \dfrac { 1 } { 3 } x + \dfrac { 1 } { 6 } y = \dfrac { 5 } { 6 } \end{cases}$
$ $ Calculate the multiplication expression $ $
$\begin{cases} \dfrac { x } { 2 } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { y } { 3 } } = \dfrac { 2 } { 3 } \\ \dfrac { 1 } { 3 } x + \dfrac { 1 } { 6 } y = \dfrac { 5 } { 6 } \end{cases}$
$\begin{cases} \dfrac { x } { 2 } - \dfrac { y } { 3 } = \dfrac { 2 } { 3 } \\ \color{#FF6800}{ \dfrac { 1 } { 3 } } \color{#FF6800}{ x } + \dfrac { 1 } { 6 } y = \dfrac { 5 } { 6 } \end{cases}$
$ $ Calculate the multiplication expression $ $
$\begin{cases} \dfrac { x } { 2 } - \dfrac { y } { 3 } = \dfrac { 2 } { 3 } \\ \color{#FF6800}{ \dfrac { x } { 3 } } + \dfrac { 1 } { 6 } y = \dfrac { 5 } { 6 } \end{cases}$
$\begin{cases} \dfrac { x } { 2 } - \dfrac { y } { 3 } = \dfrac { 2 } { 3 } \\ \dfrac { x } { 3 } + \color{#FF6800}{ \dfrac { 1 } { 6 } } \color{#FF6800}{ y } = \dfrac { 5 } { 6 } \end{cases}$
$ $ Calculate the multiplication expression $ $
$\begin{cases} \dfrac { x } { 2 } - \dfrac { y } { 3 } = \dfrac { 2 } { 3 } \\ \dfrac { x } { 3 } + \color{#FF6800}{ \dfrac { y } { 6 } } = \dfrac { 5 } { 6 } \end{cases}$
$\begin{cases} \dfrac { x } { 2 } - \dfrac { y } { 3 } = \dfrac { 2 } { 3 } \\ \dfrac { x } { 3 } + \dfrac { y } { 6 } = \dfrac { 5 } { 6 } \end{cases}$
$ $ Solve a solution to $ x$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 2 } { 3 } } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 4 } { 3 } } \\ \dfrac { x } { 3 } + \dfrac { y } { 6 } = \dfrac { 5 } { 6 } \end{cases}$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 2 } { 3 } } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 4 } { 3 } } \\ \color{#FF6800}{ \dfrac { x } { 3 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { y } { 6 } } = \color{#FF6800}{ \dfrac { 5 } { 6 } } \end{cases}$
$ $ Substitute the given $ x $ value into the equation $ \dfrac { x } { 3 } + \dfrac { y } { 6 } = \dfrac { 5 } { 6 }$
$\color{#FF6800}{ \dfrac { \dfrac { 2 } { 3 } y + \dfrac { 4 } { 3 } } { 3 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { y } { 6 } } = \color{#FF6800}{ \dfrac { 5 } { 6 } }$
$\color{#FF6800}{ \dfrac { \dfrac { 2 } { 3 } y + \dfrac { 4 } { 3 } } { 3 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { y } { 6 } } = \color{#FF6800}{ \dfrac { 5 } { 6 } }$
$ $ 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 { 2 } { 3 } y + \dfrac { 4 } { 3 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 2 } { 3 } } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 4 } { 3 } }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 2 } { 3 } } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 4 } { 3 } }$
$ $ Organize the expression $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ 2 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ 2 }$
$ $ The possible solutions are as follows $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ 2 } , \color{#FF6800}{ y } = \color{#FF6800}{ 1 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ 2 } , \color{#FF6800}{ y } = \color{#FF6800}{ 1 }$
$ $ Check if it is the solution to the system of equations $ $
$\begin{cases} \color{#FF6800}{ \dfrac { 1 } { 2 } } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 3 } } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } = \color{#FF6800}{ \dfrac { 2 } { 3 } } \\ \color{#FF6800}{ \dfrac { 1 } { 3 } } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 1 } { 6 } } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } = \color{#FF6800}{ \dfrac { 5 } { 6 } } \end{cases}$
$\begin{cases} \color{#FF6800}{ \dfrac { 1 } { 2 } } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 3 } } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } = \color{#FF6800}{ \dfrac { 2 } { 3 } } \\ \color{#FF6800}{ \dfrac { 1 } { 3 } } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 1 } { 6 } } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } = \color{#FF6800}{ \dfrac { 5 } { 6 } } \end{cases}$
$ $ Simplify the equality $ $
$\begin{cases} \color{#FF6800}{ \dfrac { 2 } { 3 } } = \color{#FF6800}{ \dfrac { 2 } { 3 } } \\ \color{#FF6800}{ \dfrac { 5 } { 6 } } = \color{#FF6800}{ \dfrac { 5 } { 6 } } \end{cases}$
$\begin{cases} \color{#FF6800}{ \dfrac { 2 } { 3 } } = \color{#FF6800}{ \dfrac { 2 } { 3 } } \\ \color{#FF6800}{ \dfrac { 5 } { 6 } } = \color{#FF6800}{ \dfrac { 5 } { 6 } } \end{cases}$
$ $ Since it is true in both equations, it is the solution of the system of equations $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ 2 } , \color{#FF6800}{ y } = \color{#FF6800}{ 1 }$
$ $ 그래프 보기 $ $
Graph
Solution search results
search-thumbnail-$\left(2\right)$ $ \begin{cases} \dfrac {1} {2}x+\dfrac {3} {4}y=1 \\ \dfrac {2} {3}x-\dfrac {1} {6}y=\dfrac {1} {6} \end{cases} $
10th-13th grade
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