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