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Formula
Solve the system of equations
Answer
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Graph
$3 x - 4 y = - 6$
$2 x + 4 y = 16$
$x$-intercept
$\left ( - 2 , 0 \right )$
$y$-intercept
$\left ( 0 , \dfrac { 3 } { 2 } \right )$
$x$-intercept
$\left ( 8 , 0 \right )$
$y$-intercept
$\left ( 0 , 4 \right )$
$\begin{cases} 3x-4y = -6 \\2x+4y = 16 \end{cases}$
$x = 2 , y = 3$
Solve quadratic equations using the square root
$\begin{cases} 3 x - 4 y = - 6 \\ 2 x + 4 y = 16 \end{cases}$
$ $ Solve a solution to $ x$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 4 } { 3 } } \color{#FF6800}{ y } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \\ 2 x + 4 y = 16 \end{cases}$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 4 } { 3 } } \color{#FF6800}{ y } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \\ \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 4 } \color{#FF6800}{ y } = \color{#FF6800}{ 16 } \end{cases}$
$ $ Substitute the given $ x $ value into the equation $ 2 x + 4 y = 16$
$\color{#FF6800}{ 2 } \left ( \color{#FF6800}{ \dfrac { 4 } { 3 } } \color{#FF6800}{ y } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \right ) \color{#FF6800}{ + } \color{#FF6800}{ 4 } \color{#FF6800}{ y } = \color{#FF6800}{ 16 }$
$\color{#FF6800}{ 2 } \left ( \color{#FF6800}{ \dfrac { 4 } { 3 } } \color{#FF6800}{ y } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \right ) \color{#FF6800}{ + } \color{#FF6800}{ 4 } \color{#FF6800}{ y } = \color{#FF6800}{ 16 }$
$ $ Solve a solution to $ y$
$\color{#FF6800}{ y } = \color{#FF6800}{ 3 }$
$\color{#FF6800}{ y } = \color{#FF6800}{ 3 }$
$ $ Substitute the given $ y $ value into the equation $ x = \dfrac { 4 } { 3 } y - 2$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 4 } { 3 } } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } \color{#FF6800}{ - } \color{#FF6800}{ 2 }$
$x = \color{#FF6800}{ \dfrac { 4 } { 3 } } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } - 2$
$ $ Calculate the product of rational numbers $ $
$x = \color{#FF6800}{ 4 } - 2$
$x = \color{#FF6800}{ 4 } \color{#FF6800}{ - } \color{#FF6800}{ 2 }$
$ $ Subtract $ 2 $ from $ 4$
$x = \color{#FF6800}{ 2 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ 2 }$
$ $ The possible solutions are as follows $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ 2 } , \color{#FF6800}{ y } = \color{#FF6800}{ 3 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ 2 } , \color{#FF6800}{ y } = \color{#FF6800}{ 3 }$
$ $ Check if it is the solution to the system of equations $ $
$\begin{cases} \color{#FF6800}{ 3 } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } = \color{#FF6800}{ - } \color{#FF6800}{ 6 } \\ \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } \color{#FF6800}{ + } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } = \color{#FF6800}{ 16 } \end{cases}$
$\begin{cases} \color{#FF6800}{ 3 } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } = \color{#FF6800}{ - } \color{#FF6800}{ 6 } \\ \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } \color{#FF6800}{ + } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } = \color{#FF6800}{ 16 } \end{cases}$
$ $ Simplify the equality $ $
$\begin{cases} \color{#FF6800}{ - } \color{#FF6800}{ 6 } = \color{#FF6800}{ - } \color{#FF6800}{ 6 } \\ \color{#FF6800}{ 16 } = \color{#FF6800}{ 16 } \end{cases}$
$\begin{cases} \color{#FF6800}{ - } \color{#FF6800}{ 6 } = \color{#FF6800}{ - } \color{#FF6800}{ 6 } \\ \color{#FF6800}{ 16 } = \color{#FF6800}{ 16 } \end{cases}$
$ $ Since it is true in both equations, it is the solution of the system of equations $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ 2 } , \color{#FF6800}{ y } = \color{#FF6800}{ 3 }$
$\begin{cases} x = 2 \\ y = 3 \end{cases}$
Solve quadratic equations using the square root
$\begin{cases} \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ 6 } \\ \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 4 } \color{#FF6800}{ y } = \color{#FF6800}{ 16 } \end{cases}$
$ $ Solve the system of linear equations for $ x , y $
$\begin{cases} \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ 6 } \\ \color{#FF6800}{ 20 } \color{#FF6800}{ y } = \color{#FF6800}{ 60 } \end{cases}$
$\begin{cases} \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ 6 } \\ \color{#FF6800}{ 20 } \color{#FF6800}{ y } = \color{#FF6800}{ 60 } \end{cases}$
$ $ Solve the system of linear equations for $ x , y $
$\begin{cases} \color{#FF6800}{ 60 } \color{#FF6800}{ x } = \color{#FF6800}{ 120 } \\ \color{#FF6800}{ 20 } \color{#FF6800}{ y } = \color{#FF6800}{ 60 } \end{cases}$
$\begin{cases} \color{#FF6800}{ 60 } \color{#FF6800}{ x } = \color{#FF6800}{ 120 } \\ 20 y = 60 \end{cases}$
$ $ Solve a solution to $ x$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ 2 } \\ 20 y = 60 \end{cases}$
$\begin{cases} x = 2 \\ \color{#FF6800}{ 20 } \color{#FF6800}{ y } = \color{#FF6800}{ 60 } \end{cases}$
$ $ Solve a solution to $ y$
$\begin{cases} x = 2 \\ \color{#FF6800}{ y } = \color{#FF6800}{ 3 } \end{cases}$
$ $ 그래프 보기 $ $
Graph
Solution search results
search-thumbnail-Let's Extend To decode a name of a mathematician, find the system of the equations below then write the corresponding answer on the blank. y veb $1. \begin{cases} x-y=1 \\ 2x+y=5 \end{cases} $ $7. \begin{cases} x+4y=-10 \\ x-4y=6 \end{cases} $ $c$ R. DI A $2. \begin{cases} x+y=1 \\ x+2y=0 \end{cases} $ $8. \begin{cases} 3x-y=7 \\ x+2y=0 \end{cases} $ www 3. $ \begin{cases} x+3y=-6 \\ 2x-y=-5 \end{cases} $ $9. \begin{cases} 2x-y=-5 \\ x+4y=-7 \end{cases} $ E T. $4$ $ \begin{cases} 2x-y=-7 \\ x+2y=-1 \end{cases} $ $5. \begin{cases} x-2y=1 \\ x+y=4 \end{cases} $ 6. $ \begin{cases} 2x+3y=4 \\ x-y=-3 \end{cases} $ $\bar{1} $ $\bar{2} $ $\bar{3} $ $\bar{4} $ $\bar{5} $ $\bar{6} $ $\bar{7} $ 8 9.
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