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Solve the system of equations
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$3 \left ( x + 2 \right ) - 2 y = 18$
$\dfrac { 2 } { 5 } x + \dfrac { 7 } { 15 } y = - \dfrac { 3 } { 5 }$
$x$Intercept
$\left ( 4 , 0 \right )$
$y$Intercept
$\left ( 0 , - 6 \right )$
$x$Intercept
$\left ( - \dfrac { 3 } { 2 } , 0 \right )$
$y$Intercept
$\left ( 0 , - \dfrac { 9 } { 7 } \right )$
$\begin{cases} 3(x+2)-2y=18 \\ \frac{2}{5}x+\frac{7}{15}y=-\frac{3}{5}\end{cases}$
$x = 2 , y = - 3$
Solve quadratic equations using the square root
$\begin{cases} \color{#FF6800}{ 3 } \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \right ) \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ y } = \color{#FF6800}{ 18 } \\ \color{#FF6800}{ \dfrac { 2 } { 5 } } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 7 } { 15 } } \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 5 } } \end{cases}$
 Organize the expression 
$\begin{cases} \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 6 } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ y } = \color{#FF6800}{ 18 } \\ \color{#FF6800}{ \dfrac { 2 } { 5 } } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 7 } { 15 } } \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 5 } } \end{cases}$
$\begin{cases} 3 x + 6 - 2 y = 18 \\ \color{#FF6800}{ \dfrac { 2 } { 5 } } \color{#FF6800}{ x } + \dfrac { 7 } { 15 } y = - \dfrac { 3 } { 5 } \end{cases}$
 Calculate the multiplication expression 
$\begin{cases} 3 x + 6 - 2 y = 18 \\ \color{#FF6800}{ \dfrac { 2 x } { 5 } } + \dfrac { 7 } { 15 } y = - \dfrac { 3 } { 5 } \end{cases}$
$\begin{cases} 3 x + 6 - 2 y = 18 \\ \dfrac { 2 x } { 5 } + \color{#FF6800}{ \dfrac { 7 } { 15 } } \color{#FF6800}{ y } = - \dfrac { 3 } { 5 } \end{cases}$
 Calculate the multiplication expression 
$\begin{cases} 3 x + 6 - 2 y = 18 \\ \dfrac { 2 x } { 5 } + \color{#FF6800}{ \dfrac { 7 y } { 15 } } = - \dfrac { 3 } { 5 } \end{cases}$
$\begin{cases} 3 x + 6 - 2 y = 18 \\ \dfrac { 2 x } { 5 } + \dfrac { 7 y } { 15 } = - \dfrac { 3 } { 5 } \end{cases}$
 Solve a solution to $x$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 2 } { 3 } } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ 4 } \\ \dfrac { 2 x } { 5 } + \dfrac { 7 y } { 15 } = - \dfrac { 3 } { 5 } \end{cases}$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 2 } { 3 } } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ 4 } \\ \color{#FF6800}{ \dfrac { 2 x } { 5 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 7 y } { 15 } } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 5 } } \end{cases}$
 Substitute the given $x$ value into the equation $\dfrac { 2 x } { 5 } + \dfrac { 7 y } { 15 } = - \dfrac { 3 } { 5 }$
$\color{#FF6800}{ \dfrac { 2 \left ( \dfrac { 2 } { 3 } y + 4 \right ) } { 5 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 7 y } { 15 } } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 5 } }$
$\color{#FF6800}{ \dfrac { 2 \left ( \dfrac { 2 } { 3 } y + 4 \right ) } { 5 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 7 y } { 15 } } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 5 } }$
 Solve a solution to $y$
$\color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ 3 }$
$\color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ 3 }$
 Substitute the given $y$ value into the equation $x = \dfrac { 2 } { 3 } y + 4$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 2 } { 3 } } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right ) \color{#FF6800}{ + } \color{#FF6800}{ 4 }$
$x = \color{#FF6800}{ \dfrac { 2 } { 3 } } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right ) + 4$
 Calculate the product of rational numbers 
$x = \color{#FF6800}{ - } \color{#FF6800}{ 2 } + 4$
$x = \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ + } \color{#FF6800}{ 4 }$
 Add $- 2$ and $4$
$x = \color{#FF6800}{ 2 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ 2 }$
 The possible solutions are as follows 
$\color{#FF6800}{ x } = \color{#FF6800}{ 2 } , \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ 3 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ 2 } , \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ 3 }$
 Check if it is the solution to the system of equations 
$\begin{cases} \color{#FF6800}{ 3 } \left ( \color{#FF6800}{ 2 } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \right ) \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right ) = \color{#FF6800}{ 18 } \\ \color{#FF6800}{ \dfrac { 2 } { 5 } } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 7 } { 15 } } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right ) = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 5 } } \end{cases}$
$\begin{cases} \color{#FF6800}{ 3 } \left ( \color{#FF6800}{ 2 } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \right ) \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right ) = \color{#FF6800}{ 18 } \\ \color{#FF6800}{ \dfrac { 2 } { 5 } } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 7 } { 15 } } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right ) = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 5 } } \end{cases}$
 Simplify the equality 
$\begin{cases} \color{#FF6800}{ 18 } = \color{#FF6800}{ 18 } \\ \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 5 } } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 5 } } \end{cases}$
$\begin{cases} \color{#FF6800}{ 18 } = \color{#FF6800}{ 18 } \\ \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 5 } } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 5 } } \end{cases}$
 Since it is true in both equations, it is the solution of the system of equations 
$\color{#FF6800}{ x } = \color{#FF6800}{ 2 } , \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ 3 }$
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