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Solve the system of equations
Answer
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Graph
$0.6 x + 0.5 y = 2.8$
$\dfrac { 1 } { 3 } x + \dfrac { 1 } { 2 } y = 2$
$x$Intercept
$\left ( \dfrac { 14 } { 3 } , 0 \right )$
$y$Intercept
$\left ( 0 , \dfrac { 28 } { 5 } \right )$
$x$Intercept
$\left ( 6 , 0 \right )$
$y$Intercept
$\left ( 0 , 4 \right )$
$\begin{cases} 0.6x+0.5y = 2.8 \\ \dfrac{ 1 }{ 3 } x+ \dfrac{ 1 }{ 2 } y = 2 \end{cases}$
$x = 3 , y = 2$
Solve quadratic equations using the square root
$\begin{cases} \color{#FF6800}{ 0.6 } \color{#FF6800}{ x } + 0.5 y = 2.8 \\ \dfrac { 1 } { 3 } x + \dfrac { 1 } { 2 } y = 2 \end{cases}$
$ $ Calculate the multiplication expression $ $
$\begin{cases} \color{#FF6800}{ \dfrac { 3 x } { 5 } } + 0.5 y = 2.8 \\ \dfrac { 1 } { 3 } x + \dfrac { 1 } { 2 } y = 2 \end{cases}$
$\begin{cases} \dfrac { 3 x } { 5 } + \color{#FF6800}{ 0.5 } \color{#FF6800}{ y } = 2.8 \\ \dfrac { 1 } { 3 } x + \dfrac { 1 } { 2 } y = 2 \end{cases}$
$ $ Calculate the multiplication expression $ $
$\begin{cases} \dfrac { 3 x } { 5 } + \color{#FF6800}{ \dfrac { y } { 2 } } = 2.8 \\ \dfrac { 1 } { 3 } x + \dfrac { 1 } { 2 } y = 2 \end{cases}$
$\begin{cases} \dfrac { 3 x } { 5 } + \dfrac { y } { 2 } = \color{#FF6800}{ 2.8 } \\ \dfrac { 1 } { 3 } x + \dfrac { 1 } { 2 } y = 2 \end{cases}$
$ $ Convert decimals to fractions $ $
$\begin{cases} \dfrac { 3 x } { 5 } + \dfrac { y } { 2 } = \color{#FF6800}{ \dfrac { 14 } { 5 } } \\ \dfrac { 1 } { 3 } x + \dfrac { 1 } { 2 } y = 2 \end{cases}$
$\begin{cases} \dfrac { 3 x } { 5 } + \dfrac { y } { 2 } = \dfrac { 14 } { 5 } \\ \color{#FF6800}{ \dfrac { 1 } { 3 } } \color{#FF6800}{ x } + \dfrac { 1 } { 2 } y = 2 \end{cases}$
$ $ Calculate the multiplication expression $ $
$\begin{cases} \dfrac { 3 x } { 5 } + \dfrac { y } { 2 } = \dfrac { 14 } { 5 } \\ \color{#FF6800}{ \dfrac { x } { 3 } } + \dfrac { 1 } { 2 } y = 2 \end{cases}$
$\begin{cases} \dfrac { 3 x } { 5 } + \dfrac { y } { 2 } = \dfrac { 14 } { 5 } \\ \dfrac { x } { 3 } + \color{#FF6800}{ \dfrac { 1 } { 2 } } \color{#FF6800}{ y } = 2 \end{cases}$
$ $ Calculate the multiplication expression $ $
$\begin{cases} \dfrac { 3 x } { 5 } + \dfrac { y } { 2 } = \dfrac { 14 } { 5 } \\ \dfrac { x } { 3 } + \color{#FF6800}{ \dfrac { y } { 2 } } = 2 \end{cases}$
$\begin{cases} \dfrac { 3 x } { 5 } + \dfrac { y } { 2 } = \dfrac { 14 } { 5 } \\ \dfrac { x } { 3 } + \dfrac { y } { 2 } = 2 \end{cases}$
$ $ Solve a solution to $ x$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 6 } } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 14 } { 3 } } \\ \dfrac { x } { 3 } + \dfrac { y } { 2 } = 2 \end{cases}$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 6 } } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 14 } { 3 } } \\ \color{#FF6800}{ \dfrac { x } { 3 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { y } { 2 } } = \color{#FF6800}{ 2 } \end{cases}$
$ $ Substitute the given $ x $ value into the equation $ \dfrac { x } { 3 } + \dfrac { y } { 2 } = 2$
$\color{#FF6800}{ \dfrac { - \dfrac { 5 } { 6 } y + \dfrac { 14 } { 3 } } { 3 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { y } { 2 } } = \color{#FF6800}{ 2 }$
$\color{#FF6800}{ \dfrac { - \dfrac { 5 } { 6 } y + \dfrac { 14 } { 3 } } { 3 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { y } { 2 } } = \color{#FF6800}{ 2 }$
$ $ Solve a solution to $ y$
$\color{#FF6800}{ y } = \color{#FF6800}{ 2 }$
$\color{#FF6800}{ y } = \color{#FF6800}{ 2 }$
$ $ Substitute the given $ y $ value into the equation $ x = - \dfrac { 5 } { 6 } y + \dfrac { 14 } { 3 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 6 } } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 14 } { 3 } }$
$x = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 6 } } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } + \dfrac { 14 } { 3 }$
$ $ Calculate the product of rational numbers $ $
$x = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 3 } } + \dfrac { 14 } { 3 }$
$x = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 3 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 14 } { 3 } }$
$ $ Find the sum or difference of the fractions $ $
$x = \color{#FF6800}{ 3 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ 3 }$
$ $ The possible solutions are as follows $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ 3 } , \color{#FF6800}{ y } = \color{#FF6800}{ 2 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ 3 } , \color{#FF6800}{ y } = \color{#FF6800}{ 2 }$
$ $ Check if it is the solution to the system of equations $ $
$\begin{cases} \color{#FF6800}{ 0.6 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } \color{#FF6800}{ + } \color{#FF6800}{ 0.5 } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } = \color{#FF6800}{ 2.8 } \\ \color{#FF6800}{ \dfrac { 1 } { 3 } } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 1 } { 2 } } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } = \color{#FF6800}{ 2 } \end{cases}$
$\begin{cases} \color{#FF6800}{ 0.6 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } \color{#FF6800}{ + } \color{#FF6800}{ 0.5 } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } = \color{#FF6800}{ 2.8 } \\ \color{#FF6800}{ \dfrac { 1 } { 3 } } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 1 } { 2 } } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } = \color{#FF6800}{ 2 } \end{cases}$
$ $ Simplify the equality $ $
$\begin{cases} \color{#FF6800}{ \dfrac { 14 } { 5 } } = \color{#FF6800}{ \dfrac { 14 } { 5 } } \\ \color{#FF6800}{ 2 } = \color{#FF6800}{ 2 } \end{cases}$
$\begin{cases} \color{#FF6800}{ \dfrac { 14 } { 5 } } = \color{#FF6800}{ \dfrac { 14 } { 5 } } \\ \color{#FF6800}{ 2 } = \color{#FF6800}{ 2 } \end{cases}$
$ $ Since it is true in both equations, it is the solution of the system of equations $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ 3 } , \color{#FF6800}{ y } = \color{#FF6800}{ 2 }$
$ $ 그래프 보기 $ $
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