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Solve the system of equations

Answer

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 the system of equations

$\begin{cases} 3 x - 4 y = - 6 \\ 2 x + 4 y = 16 \end{cases}$

$ $ 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$

$\color{#FF6800}{ 2 } \left ( \color{#FF6800}{ \dfrac { 4 } { 3 } } \color{#FF6800}{ y } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \right ) \color{#FF6800}{ + } \color{#FF6800}{ 4 } \color{#FF6800}{ y } = \color{#FF6800}{ 16 }$

$ $ Solve a solution to $ y$

$\color{#FF6800}{ y } = \color{#FF6800}{ 3 }$

$ $ 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 $ $

$\color{#FF6800}{ x } = \color{#FF6800}{ 2 } , \color{#FF6800}{ y } = \color{#FF6800}{ 3 }$

$ $ Check if it is the solution to the system of equations $ $

$\begin{cases} \color{#FF6800}{ 3 } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } = \color{#FF6800}{ - } \color{#FF6800}{ 6 } \\ \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } \color{#FF6800}{ + } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } = \color{#FF6800}{ 16 } \end{cases}$

$ $ Simplify the equality $ $

$\begin{cases} \color{#FF6800}{ - } \color{#FF6800}{ 6 } = \color{#FF6800}{ - } \color{#FF6800}{ 6 } \\ \color{#FF6800}{ 16 } = \color{#FF6800}{ 16 } \end{cases}$

$ $ Since it is true in both equations, it is the solution of the system of equations $ $

$\color{#FF6800}{ x } = \color{#FF6800}{ 2 } , \color{#FF6800}{ y } = \color{#FF6800}{ 3 }$

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