qanda-logo
search-icon
Symbol
apple-logo
google-play-logo

Calculator search results

Formula
Solve the system of equations
Answer
circle-check-icon
expand-arrow-icon
expand-arrow-icon
expand-arrow-icon
expand-arrow-icon
expand-arrow-icon
Graph
$\dfrac { x - 3 y } { 4 } = - 3 - y$
$- 2 \left ( x - 1 \right ) = 3 \left ( x - y \right ) + 2$
$x$Intercept
$\left ( - 12 , 0 \right )$
$y$Intercept
$\left ( 0 , - 12 \right )$
$x$Intercept
$\left ( 0 , 0 \right )$
$y$Intercept
$\left ( 0 , 0 \right )$
$\begin{cases} \dfrac{ x-3y }{ 4 } = -3-y \\-2 \left( x-1 \right) = 3 \left( x-y \right) +2 \end{cases}$
$x = - \dfrac { 9 } { 2 } , y = - \dfrac { 15 } { 2 }$
Solve the system of equations
$\begin{cases} \color{#FF6800}{ \dfrac { x - 3 y } { 4 } } = \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ - } \color{#FF6800}{ y } \\ \color{#FF6800}{ - } \color{#FF6800}{ 2 } \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) = \color{#FF6800}{ 3 } \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ y } \right ) \color{#FF6800}{ + } \color{#FF6800}{ 2 } \end{cases}$
$ $ Organize the expression $ $
$\begin{cases} \color{#FF6800}{ \dfrac { x - 3 y } { 4 } } = \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ - } \color{#FF6800}{ y } \\ \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } = \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \end{cases}$
$\begin{cases} \dfrac { x - 3 y } { 4 } = - 3 - y \\ - 2 x + 2 = 3 x - 3 y + 2 \end{cases}$
$ $ Solve a solution to $ y$
$\begin{cases} \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 12 } \\ - 2 x + 2 = 3 x - 3 y + 2 \end{cases}$
$\begin{cases} \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 12 } \\ \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } = \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \end{cases}$
$ $ Substitute the given $ y $ value into the equation $ - 2 x + 2 = 3 x - 3 y + 2$
$\color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } = \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \left ( \color{#FF6800}{ - } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 12 } \right ) \color{#FF6800}{ + } \color{#FF6800}{ 2 }$
$\color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } = \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \left ( \color{#FF6800}{ - } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 12 } \right ) \color{#FF6800}{ + } \color{#FF6800}{ 2 }$
$ $ Solve a solution to $ x$
$\color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 9 } { 2 } }$
$\color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 9 } { 2 } }$
$ $ Substitute the given $ x $ value into the equation $ y = - x - 12$
$\color{#FF6800}{ y } = \color{#FF6800}{ - } \left ( \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 9 } { 2 } } \right ) \color{#FF6800}{ - } \color{#FF6800}{ 12 }$
$\color{#FF6800}{ y } = \color{#FF6800}{ - } \left ( \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 9 } { 2 } } \right ) \color{#FF6800}{ - } \color{#FF6800}{ 12 }$
$ $ Organize the expression $ $
$\color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 15 } { 2 } }$
$\color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 15 } { 2 } }$
$ $ The possible solutions are as follows $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 9 } { 2 } } , \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 15 } { 2 } }$
$\color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 9 } { 2 } } , \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 15 } { 2 } }$
$ $ Check if it is the solution to the system of equations $ $
$\begin{cases} \color{#FF6800}{ \dfrac { - \dfrac { 9 } { 2 } - 3 \times \left ( - \dfrac { 15 } { 2 } \right ) } { 4 } } = \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ - } \left ( \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 15 } { 2 } } \right ) \\ \color{#FF6800}{ - } \color{#FF6800}{ 2 } \left ( \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 9 } { 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) = \color{#FF6800}{ 3 } \left ( \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 9 } { 2 } } \color{#FF6800}{ - } \left ( \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 15 } { 2 } } \right ) \right ) \color{#FF6800}{ + } \color{#FF6800}{ 2 } \end{cases}$
$\begin{cases} \color{#FF6800}{ \dfrac { - \dfrac { 9 } { 2 } - 3 \times \left ( - \dfrac { 15 } { 2 } \right ) } { 4 } } = \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ - } \left ( \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 15 } { 2 } } \right ) \\ \color{#FF6800}{ - } \color{#FF6800}{ 2 } \left ( \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 9 } { 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) = \color{#FF6800}{ 3 } \left ( \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 9 } { 2 } } \color{#FF6800}{ - } \left ( \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 15 } { 2 } } \right ) \right ) \color{#FF6800}{ + } \color{#FF6800}{ 2 } \end{cases}$
$ $ Simplify the equality $ $
$\begin{cases} \color{#FF6800}{ \dfrac { 9 } { 2 } } = \color{#FF6800}{ \dfrac { 9 } { 2 } } \\ \color{#FF6800}{ 11 } = \color{#FF6800}{ 11 } \end{cases}$
$\begin{cases} \color{#FF6800}{ \dfrac { 9 } { 2 } } = \color{#FF6800}{ \dfrac { 9 } { 2 } } \\ \color{#FF6800}{ 11 } = \color{#FF6800}{ 11 } \end{cases}$
$ $ Since it is true in both equations, it is the solution of the system of equations $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 9 } { 2 } } , \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 15 } { 2 } }$
Solution search results
Have you found the solution you wanted?
Try again
Try more features at Qanda!
Search by problem image
Ask 1:1 question to TOP class teachers
AI recommend problems and video lecture
apple-logo
google-play-logo