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Solve the system of equations
Answer
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$3 x - 4 y = 4$
$- 2 x + 3 y = - 2$
$x$Intercept
$\left ( \dfrac { 4 } { 3 } , 0 \right )$
$y$Intercept
$\left ( 0 , - 1 \right )$
$x$Intercept
$\left ( 1 , 0 \right )$
$y$Intercept
$\left ( 0 , - \dfrac { 2 } { 3 } \right )$
$\begin{cases} 3x-4y = 4 \\-2x+3y = -2 \end{cases}$
$x = 4 , y = 2$
Solve the system of equations
$\begin{cases} 3 x - 4 y = 4 \\ - 2 x + 3 y = - 2 \end{cases}$
$ $ Solve a solution to $ x$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 4 } { 3 } } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 4 } { 3 } } \\ - 2 x + 3 y = - 2 \end{cases}$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 4 } { 3 } } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 4 } { 3 } } \\ \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ 2 } \end{cases}$
$ $ Substitute the given $ x $ value into the equation $ - 2 x + 3 y = - 2$
$\color{#FF6800}{ - } \color{#FF6800}{ 2 } \left ( \color{#FF6800}{ \dfrac { 4 } { 3 } } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 4 } { 3 } } \right ) \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ 2 }$
$\color{#FF6800}{ - } \color{#FF6800}{ 2 } \left ( \color{#FF6800}{ \dfrac { 4 } { 3 } } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 4 } { 3 } } \right ) \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ 2 }$
$ $ 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 { 4 } { 3 } y + \dfrac { 4 } { 3 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 4 } { 3 } } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 4 } { 3 } }$
$x = \color{#FF6800}{ \dfrac { 4 } { 3 } } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } + \dfrac { 4 } { 3 }$
$ $ Calculate the product of rational numbers $ $
$x = \color{#FF6800}{ \dfrac { 8 } { 3 } } + \dfrac { 4 } { 3 }$
$x = \color{#FF6800}{ \dfrac { 8 } { 3 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 4 } { 3 } }$
$ $ Find the sum of the fractions $ $
$x = \color{#FF6800}{ 4 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ 4 }$
$ $ The possible solutions are as follows $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ 4 } , \color{#FF6800}{ y } = \color{#FF6800}{ 2 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ 4 } , \color{#FF6800}{ y } = \color{#FF6800}{ 2 }$
$ $ Check if it is the solution to the system of equations $ $
$\begin{cases} \color{#FF6800}{ 3 } \color{#FF6800}{ \times } \color{#FF6800}{ 4 } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } = \color{#FF6800}{ 4 } \\ \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 4 } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } = \color{#FF6800}{ - } \color{#FF6800}{ 2 } \end{cases}$
$\begin{cases} \color{#FF6800}{ 3 } \color{#FF6800}{ \times } \color{#FF6800}{ 4 } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } = \color{#FF6800}{ 4 } \\ \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 4 } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } = \color{#FF6800}{ - } \color{#FF6800}{ 2 } \end{cases}$
$ $ Simplify the equality $ $
$\begin{cases} \color{#FF6800}{ 4 } = \color{#FF6800}{ 4 } \\ \color{#FF6800}{ - } \color{#FF6800}{ 2 } = \color{#FF6800}{ - } \color{#FF6800}{ 2 } \end{cases}$
$\begin{cases} \color{#FF6800}{ 4 } = \color{#FF6800}{ 4 } \\ \color{#FF6800}{ - } \color{#FF6800}{ 2 } = \color{#FF6800}{ - } \color{#FF6800}{ 2 } \end{cases}$
$ $ Since it is true in both equations, it is the solution of the system of equations $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ 4 } , \color{#FF6800}{ y } = \color{#FF6800}{ 2 }$
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