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