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Solve the system of equations
Answer
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$\dfrac { x + 1 } { 2 } + \dfrac { y - 1 } { 3 } = \dfrac { 3 } { 2 }$
$\dfrac { x + 3 } { 5 } - \dfrac { y + 2 } { 2 } = - \dfrac { 1 } { 2 }$
$x$Intercept
$\left ( \dfrac { 8 } { 3 } , 0 \right )$
$y$Intercept
$\left ( 0 , 4 \right )$
$x$Intercept
$\left ( - \dfrac { 1 } { 2 } , 0 \right )$
$y$Intercept
$\left ( 0 , \dfrac { 1 } { 5 } \right )$
$\begin{cases} \dfrac{ x+1 }{ 2 } + \dfrac{ y-1 }{ 3 } = \dfrac{ 3 }{ 2 } \\ \dfrac{ x+3 }{ 5 } - \dfrac{ y+2 }{ 2 } = - \dfrac{ 1 }{ 2 } \end{cases}$
$x = 2 , y = 1$
Solve the system of equations
$\begin{cases} \color{#FF6800}{ \dfrac { x + 1 } { 2 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { y - 1 } { 3 } } = \dfrac { 3 } { 2 } \\ \dfrac { x + 3 } { 5 } - \dfrac { y + 2 } { 2 } = - \dfrac { 1 } { 2 } \end{cases}$
$ $ Write all numerators above the least common denominator $ $
$\begin{cases} \color{#FF6800}{ \dfrac { 3 x + 3 + 2 y - 2 } { 6 } } = \dfrac { 3 } { 2 } \\ \dfrac { x + 3 } { 5 } - \dfrac { y + 2 } { 2 } = - \dfrac { 1 } { 2 } \end{cases}$
$\begin{cases} \dfrac { 3 x + \color{#FF6800}{ 3 } + 2 y \color{#FF6800}{ - } \color{#FF6800}{ 2 } } { 6 } = \dfrac { 3 } { 2 } \\ \dfrac { x + 3 } { 5 } - \dfrac { y + 2 } { 2 } = - \dfrac { 1 } { 2 } \end{cases}$
$ $ Subtract $ 2 $ from $ 3$
$\begin{cases} \dfrac { 3 x + \color{#FF6800}{ 1 } + 2 y } { 6 } = \dfrac { 3 } { 2 } \\ \dfrac { x + 3 } { 5 } - \dfrac { y + 2 } { 2 } = - \dfrac { 1 } { 2 } \end{cases}$
$\begin{cases} \dfrac { 3 x + 1 + 2 y } { 6 } = \dfrac { 3 } { 2 } \\ \color{#FF6800}{ \dfrac { x + 3 } { 5 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { y + 2 } { 2 } } = - \dfrac { 1 } { 2 } \end{cases}$
$ $ Write all numerators above the least common denominator $ $
$\begin{cases} \dfrac { 3 x + 1 + 2 y } { 6 } = \dfrac { 3 } { 2 } \\ \color{#FF6800}{ \dfrac { 2 x + 6 - 5 y - 10 } { 10 } } = - \dfrac { 1 } { 2 } \end{cases}$
$\begin{cases} \dfrac { 3 x + 1 + 2 y } { 6 } = \dfrac { 3 } { 2 } \\ \dfrac { 2 x + \color{#FF6800}{ 6 } - 5 y \color{#FF6800}{ - } \color{#FF6800}{ 10 } } { 10 } = - \dfrac { 1 } { 2 } \end{cases}$
$ $ Subtract $ 10 $ from $ 6$
$\begin{cases} \dfrac { 3 x + 1 + 2 y } { 6 } = \dfrac { 3 } { 2 } \\ \dfrac { 2 x \color{#FF6800}{ - } \color{#FF6800}{ 4 } - 5 y } { 10 } = - \dfrac { 1 } { 2 } \end{cases}$
$\begin{cases} \dfrac { 3 x + 1 + 2 y } { 6 } = \dfrac { 3 } { 2 } \\ \dfrac { 2 x - 4 - 5 y } { 10 } = - \dfrac { 1 } { 2 } \end{cases}$
$ $ Solve a solution to $ x$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 } { 3 } } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 8 } { 3 } } \\ \dfrac { 2 x - 4 - 5 y } { 10 } = - \dfrac { 1 } { 2 } \end{cases}$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 } { 3 } } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 8 } { 3 } } \\ \color{#FF6800}{ \dfrac { 2 x - 4 - 5 y } { 10 } } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 2 } } \end{cases}$
$ $ Substitute the given $ x $ value into the equation $ \dfrac { 2 x - 4 - 5 y } { 10 } = - \dfrac { 1 } { 2 }$
$\color{#FF6800}{ \dfrac { 2 \left ( - \dfrac { 2 } { 3 } y + \dfrac { 8 } { 3 } \right ) - 4 - 5 y } { 10 } } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 2 } }$
$\color{#FF6800}{ \dfrac { 2 \left ( - \dfrac { 2 } { 3 } y + \dfrac { 8 } { 3 } \right ) - 4 - 5 y } { 10 } } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 2 } }$
$ $ Solve a solution to $ y$
$\color{#FF6800}{ y } = \color{#FF6800}{ 1 }$
$\color{#FF6800}{ y } = \color{#FF6800}{ 1 }$
$ $ Substitute the given $ y $ value into the equation $ x = - \dfrac { 2 } { 3 } y + \dfrac { 8 } { 3 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 } { 3 } } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 8 } { 3 } }$
$\color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 } { 3 } } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 8 } { 3 } }$
$ $ Organize the expression $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ 2 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ 2 }$
$ $ The possible solutions are as follows $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ 2 } , \color{#FF6800}{ y } = \color{#FF6800}{ 1 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ 2 } , \color{#FF6800}{ y } = \color{#FF6800}{ 1 }$
$ $ Check if it is the solution to the system of equations $ $
$\begin{cases} \color{#FF6800}{ \dfrac { 2 + 1 } { 2 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 1 - 1 } { 3 } } = \color{#FF6800}{ \dfrac { 3 } { 2 } } \\ \color{#FF6800}{ \dfrac { 2 + 3 } { 5 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 + 2 } { 2 } } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 2 } } \end{cases}$
$\begin{cases} \color{#FF6800}{ \dfrac { 2 + 1 } { 2 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 1 - 1 } { 3 } } = \color{#FF6800}{ \dfrac { 3 } { 2 } } \\ \color{#FF6800}{ \dfrac { 2 + 3 } { 5 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 + 2 } { 2 } } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 2 } } \end{cases}$
$ $ Simplify the equality $ $
$\begin{cases} \color{#FF6800}{ \dfrac { 3 } { 2 } } = \color{#FF6800}{ \dfrac { 3 } { 2 } } \\ \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 2 } } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 2 } } \end{cases}$
$\begin{cases} \color{#FF6800}{ \dfrac { 3 } { 2 } } = \color{#FF6800}{ \dfrac { 3 } { 2 } } \\ \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 2 } } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 2 } } \end{cases}$
$ $ Since it is true in both equations, it is the solution of the system of equations $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ 2 } , \color{#FF6800}{ y } = \color{#FF6800}{ 1 }$
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