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Solve the system of equations
Answer
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Graph
$0.3 x - 0.2 y = 0.6$
$0.4 x + 0.3 y = 2.5$
$x$Intercept
$\left ( 2 , 0 \right )$
$y$Intercept
$\left ( 0 , - 3 \right )$
$x$Intercept
$\left ( \dfrac { 25 } { 4 } , 0 \right )$
$y$Intercept
$\left ( 0 , \dfrac { 25 } { 3 } \right )$
$\begin{cases} 0.3x-0.2y = 0.6 \\0.4x+0.3y = 2.5 \end{cases}$
$x = 4 , y = 3$
Solve the system of equations
$\begin{cases} \color{#FF6800}{ 0.3 } \color{#FF6800}{ x } - 0.2 y = 0.6 \\ 0.4 x + 0.3 y = 2.5 \end{cases}$
$ $ Calculate the multiplication expression $ $
$\begin{cases} \color{#FF6800}{ \dfrac { 3 x } { 10 } } - 0.2 y = 0.6 \\ 0.4 x + 0.3 y = 2.5 \end{cases}$
$\begin{cases} \dfrac { 3 x } { 10 } \color{#FF6800}{ - } \color{#FF6800}{ 0.2 } \color{#FF6800}{ y } = 0.6 \\ 0.4 x + 0.3 y = 2.5 \end{cases}$
$ $ Calculate the multiplication expression $ $
$\begin{cases} \dfrac { 3 x } { 10 } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { y } { 5 } } = 0.6 \\ 0.4 x + 0.3 y = 2.5 \end{cases}$
$\begin{cases} \dfrac { 3 x } { 10 } - \dfrac { y } { 5 } = \color{#FF6800}{ 0.6 } \\ 0.4 x + 0.3 y = 2.5 \end{cases}$
$ $ Convert decimals to fractions $ $
$\begin{cases} \dfrac { 3 x } { 10 } - \dfrac { y } { 5 } = \color{#FF6800}{ \dfrac { 3 } { 5 } } \\ 0.4 x + 0.3 y = 2.5 \end{cases}$
$\begin{cases} \dfrac { 3 x } { 10 } - \dfrac { y } { 5 } = \dfrac { 3 } { 5 } \\ \color{#FF6800}{ 0.4 } \color{#FF6800}{ x } + 0.3 y = 2.5 \end{cases}$
$ $ Calculate the multiplication expression $ $
$\begin{cases} \dfrac { 3 x } { 10 } - \dfrac { y } { 5 } = \dfrac { 3 } { 5 } \\ \color{#FF6800}{ \dfrac { 2 x } { 5 } } + 0.3 y = 2.5 \end{cases}$
$\begin{cases} \dfrac { 3 x } { 10 } - \dfrac { y } { 5 } = \dfrac { 3 } { 5 } \\ \dfrac { 2 x } { 5 } + \color{#FF6800}{ 0.3 } \color{#FF6800}{ y } = 2.5 \end{cases}$
$ $ Calculate the multiplication expression $ $
$\begin{cases} \dfrac { 3 x } { 10 } - \dfrac { y } { 5 } = \dfrac { 3 } { 5 } \\ \dfrac { 2 x } { 5 } + \color{#FF6800}{ \dfrac { 3 y } { 10 } } = 2.5 \end{cases}$
$\begin{cases} \dfrac { 3 x } { 10 } - \dfrac { y } { 5 } = \dfrac { 3 } { 5 } \\ \dfrac { 2 x } { 5 } + \dfrac { 3 y } { 10 } = \color{#FF6800}{ 2.5 } \end{cases}$
$ $ Convert decimals to fractions $ $
$\begin{cases} \dfrac { 3 x } { 10 } - \dfrac { y } { 5 } = \dfrac { 3 } { 5 } \\ \dfrac { 2 x } { 5 } + \dfrac { 3 y } { 10 } = \color{#FF6800}{ \dfrac { 5 } { 2 } } \end{cases}$
$\begin{cases} \dfrac { 3 x } { 10 } - \dfrac { y } { 5 } = \dfrac { 3 } { 5 } \\ \dfrac { 2 x } { 5 } + \dfrac { 3 y } { 10 } = \dfrac { 5 } { 2 } \end{cases}$
$ $ Solve a solution to $ x$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 2 } { 3 } } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \\ \dfrac { 2 x } { 5 } + \dfrac { 3 y } { 10 } = \dfrac { 5 } { 2 } \end{cases}$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 2 } { 3 } } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \\ \color{#FF6800}{ \dfrac { 2 x } { 5 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 3 y } { 10 } } = \color{#FF6800}{ \dfrac { 5 } { 2 } } \end{cases}$
$ $ Substitute the given $ x $ value into the equation $ \dfrac { 2 x } { 5 } + \dfrac { 3 y } { 10 } = \dfrac { 5 } { 2 }$
$\color{#FF6800}{ \dfrac { 2 \left ( \dfrac { 2 } { 3 } y + 2 \right ) } { 5 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 3 y } { 10 } } = \color{#FF6800}{ \dfrac { 5 } { 2 } }$
$\color{#FF6800}{ \dfrac { 2 \left ( \dfrac { 2 } { 3 } y + 2 \right ) } { 5 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 3 y } { 10 } } = \color{#FF6800}{ \dfrac { 5 } { 2 } }$
$ $ Solve a solution to $ y$
$\color{#FF6800}{ y } = \color{#FF6800}{ 3 }$
$\color{#FF6800}{ y } = \color{#FF6800}{ 3 }$
$ $ Substitute the given $ y $ value into the equation $ x = \dfrac { 2 } { 3 } y + 2$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 2 } { 3 } } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } \color{#FF6800}{ + } \color{#FF6800}{ 2 }$
$x = \color{#FF6800}{ \dfrac { 2 } { 3 } } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } + 2$
$ $ Calculate the product of rational numbers $ $
$x = \color{#FF6800}{ 2 } + 2$
$x = \color{#FF6800}{ 2 } \color{#FF6800}{ + } \color{#FF6800}{ 2 }$
$ $ Add $ 2 $ and $ 2$
$x = \color{#FF6800}{ 4 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ 4 }$
$ $ The possible solutions are as follows $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ 4 } , \color{#FF6800}{ y } = \color{#FF6800}{ 3 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ 4 } , \color{#FF6800}{ y } = \color{#FF6800}{ 3 }$
$ $ Check if it is the solution to the system of equations $ $
$\begin{cases} \color{#FF6800}{ 0.3 } \color{#FF6800}{ \times } \color{#FF6800}{ 4 } \color{#FF6800}{ - } \color{#FF6800}{ 0.2 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } = \color{#FF6800}{ 0.6 } \\ \color{#FF6800}{ 0.4 } \color{#FF6800}{ \times } \color{#FF6800}{ 4 } \color{#FF6800}{ + } \color{#FF6800}{ 0.3 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } = \color{#FF6800}{ 2.5 } \end{cases}$
$\begin{cases} \color{#FF6800}{ 0.3 } \color{#FF6800}{ \times } \color{#FF6800}{ 4 } \color{#FF6800}{ - } \color{#FF6800}{ 0.2 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } = \color{#FF6800}{ 0.6 } \\ \color{#FF6800}{ 0.4 } \color{#FF6800}{ \times } \color{#FF6800}{ 4 } \color{#FF6800}{ + } \color{#FF6800}{ 0.3 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } = \color{#FF6800}{ 2.5 } \end{cases}$
$ $ Simplify the equality $ $
$\begin{cases} \color{#FF6800}{ \dfrac { 3 } { 5 } } = \color{#FF6800}{ \dfrac { 3 } { 5 } } \\ \color{#FF6800}{ \dfrac { 5 } { 2 } } = \color{#FF6800}{ \dfrac { 5 } { 2 } } \end{cases}$
$\begin{cases} \color{#FF6800}{ \dfrac { 3 } { 5 } } = \color{#FF6800}{ \dfrac { 3 } { 5 } } \\ \color{#FF6800}{ \dfrac { 5 } { 2 } } = \color{#FF6800}{ \dfrac { 5 } { 2 } } \end{cases}$
$ $ Since it is true in both equations, it is the solution of the system of equations $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ 4 } , \color{#FF6800}{ y } = \color{#FF6800}{ 3 }$
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