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Solve the system of equations
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$7 x + 6 y = 1$
$4 x + 3 y = - 2$
$x$Intercept
$\left ( \dfrac { 1 } { 7 } , 0 \right )$
$y$Intercept
$\left ( 0 , \dfrac { 1 } { 6 } \right )$
$x$Intercept
$\left ( - \dfrac { 1 } { 2 } , 0 \right )$
$y$Intercept
$\left ( 0 , - \dfrac { 2 } { 3 } \right )$
$\begin{cases} 7x+6y = 1 \\4x+3y = -2 \end{cases}$
$x = - 5 , y = 6$
Solve the system of equations
$\begin{cases} 7 x + 6 y = 1 \\ 4 x + 3 y = - 2 \end{cases}$
$ $ Solve a solution to $ x$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 6 } { 7 } } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 1 } { 7 } } \\ 4 x + 3 y = - 2 \end{cases}$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 6 } { 7 } } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 1 } { 7 } } \\ \color{#FF6800}{ 4 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ 2 } \end{cases}$
$ $ Substitute the given $ x $ value into the equation $ 4 x + 3 y = - 2$
$\color{#FF6800}{ 4 } \left ( \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 6 } { 7 } } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 1 } { 7 } } \right ) \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ 2 }$
$\color{#FF6800}{ 4 } \left ( \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 6 } { 7 } } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 1 } { 7 } } \right ) \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ 2 }$
$ $ Solve a solution to $ y$
$\color{#FF6800}{ y } = \color{#FF6800}{ 6 }$
$\color{#FF6800}{ y } = \color{#FF6800}{ 6 }$
$ $ Substitute the given $ y $ value into the equation $ x = - \dfrac { 6 } { 7 } y + \dfrac { 1 } { 7 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 6 } { 7 } } \color{#FF6800}{ \times } \color{#FF6800}{ 6 } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 1 } { 7 } }$
$x = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 6 } { 7 } } \color{#FF6800}{ \times } \color{#FF6800}{ 6 } + \dfrac { 1 } { 7 }$
$ $ Calculate the product of rational numbers $ $
$x = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 36 } { 7 } } + \dfrac { 1 } { 7 }$
$x = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 36 } { 7 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 1 } { 7 } }$
$ $ Find the sum or difference of the fractions $ $
$x = \color{#FF6800}{ - } \color{#FF6800}{ 5 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 5 }$
$ $ The possible solutions are as follows $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 5 } , \color{#FF6800}{ y } = \color{#FF6800}{ 6 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 5 } , \color{#FF6800}{ y } = \color{#FF6800}{ 6 }$
$ $ Check if it is the solution to the system of equations $ $
$\begin{cases} \color{#FF6800}{ 7 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right ) \color{#FF6800}{ + } \color{#FF6800}{ 6 } \color{#FF6800}{ \times } \color{#FF6800}{ 6 } = \color{#FF6800}{ 1 } \\ \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right ) \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ \times } \color{#FF6800}{ 6 } = \color{#FF6800}{ - } \color{#FF6800}{ 2 } \end{cases}$
$\begin{cases} \color{#FF6800}{ 7 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right ) \color{#FF6800}{ + } \color{#FF6800}{ 6 } \color{#FF6800}{ \times } \color{#FF6800}{ 6 } = \color{#FF6800}{ 1 } \\ \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right ) \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ \times } \color{#FF6800}{ 6 } = \color{#FF6800}{ - } \color{#FF6800}{ 2 } \end{cases}$
$ $ Simplify the equality $ $
$\begin{cases} \color{#FF6800}{ 1 } = \color{#FF6800}{ 1 } \\ \color{#FF6800}{ - } \color{#FF6800}{ 2 } = \color{#FF6800}{ - } \color{#FF6800}{ 2 } \end{cases}$
$\begin{cases} \color{#FF6800}{ 1 } = \color{#FF6800}{ 1 } \\ \color{#FF6800}{ - } \color{#FF6800}{ 2 } = \color{#FF6800}{ - } \color{#FF6800}{ 2 } \end{cases}$
$ $ Since it is true in both equations, it is the solution of the system of equations $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 5 } , \color{#FF6800}{ y } = \color{#FF6800}{ 6 }$
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