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Solve the system of equations
Answer
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Graph
$3 x - 2 y = 16$
$3 y - 2 x = 6$
$x$Intercept
$\left ( \dfrac { 16 } { 3 } , 0 \right )$
$y$Intercept
$\left ( 0 , - 8 \right )$
$x$Intercept
$\left ( - 3 , 0 \right )$
$y$Intercept
$\left ( 0 , 2 \right )$
$\begin{cases} 3x-2y = 16 \\3y-2x = 6 \end{cases}$
$x = 12 , y = 10$
Solve the system of equations
$\begin{cases} 3 x - 2 y = 16 \\ 3 y - 2 x = 6 \end{cases}$
$ $ Solve a solution to $ x$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 2 } { 3 } } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 16 } { 3 } } \\ 3 y - 2 x = 6 \end{cases}$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 2 } { 3 } } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 16 } { 3 } } \\ \color{#FF6800}{ 3 } \color{#FF6800}{ y } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } = \color{#FF6800}{ 6 } \end{cases}$
$ $ Substitute the given $ x $ value into the equation $ 3 y - 2 x = 6$
$\color{#FF6800}{ 3 } \color{#FF6800}{ y } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \left ( \color{#FF6800}{ \dfrac { 2 } { 3 } } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 16 } { 3 } } \right ) = \color{#FF6800}{ 6 }$
$\color{#FF6800}{ 3 } \color{#FF6800}{ y } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \left ( \color{#FF6800}{ \dfrac { 2 } { 3 } } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 16 } { 3 } } \right ) = \color{#FF6800}{ 6 }$
$ $ Solve a solution to $ y$
$\color{#FF6800}{ y } = \color{#FF6800}{ 10 }$
$\color{#FF6800}{ y } = \color{#FF6800}{ 10 }$
$ $ Substitute the given $ y $ value into the equation $ x = \dfrac { 2 } { 3 } y + \dfrac { 16 } { 3 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 2 } { 3 } } \color{#FF6800}{ \times } \color{#FF6800}{ 10 } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 16 } { 3 } }$
$x = \color{#FF6800}{ \dfrac { 2 } { 3 } } \color{#FF6800}{ \times } \color{#FF6800}{ 10 } + \dfrac { 16 } { 3 }$
$ $ Calculate the product of rational numbers $ $
$x = \color{#FF6800}{ \dfrac { 20 } { 3 } } + \dfrac { 16 } { 3 }$
$x = \color{#FF6800}{ \dfrac { 20 } { 3 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 16 } { 3 } }$
$ $ Find the sum of the fractions $ $
$x = \color{#FF6800}{ 12 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ 12 }$
$ $ The possible solutions are as follows $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ 12 } , \color{#FF6800}{ y } = \color{#FF6800}{ 10 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ 12 } , \color{#FF6800}{ y } = \color{#FF6800}{ 10 }$
$ $ Check if it is the solution to the system of equations $ $
$\begin{cases} \color{#FF6800}{ 3 } \color{#FF6800}{ \times } \color{#FF6800}{ 12 } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 10 } = \color{#FF6800}{ 16 } \\ \color{#FF6800}{ 3 } \color{#FF6800}{ \times } \color{#FF6800}{ 10 } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 12 } = \color{#FF6800}{ 6 } \end{cases}$
$\begin{cases} \color{#FF6800}{ 3 } \color{#FF6800}{ \times } \color{#FF6800}{ 12 } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 10 } = \color{#FF6800}{ 16 } \\ \color{#FF6800}{ 3 } \color{#FF6800}{ \times } \color{#FF6800}{ 10 } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 12 } = \color{#FF6800}{ 6 } \end{cases}$
$ $ Simplify the equality $ $
$\begin{cases} \color{#FF6800}{ 16 } = \color{#FF6800}{ 16 } \\ \color{#FF6800}{ 6 } = \color{#FF6800}{ 6 } \end{cases}$
$\begin{cases} \color{#FF6800}{ 16 } = \color{#FF6800}{ 16 } \\ \color{#FF6800}{ 6 } = \color{#FF6800}{ 6 } \end{cases}$
$ $ Since it is true in both equations, it is the solution of the system of equations $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ 12 } , \color{#FF6800}{ y } = \color{#FF6800}{ 10 }$
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