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Formula
Solve the system of equations
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$0.6 x - 1.3 y = - 1.5$
$\dfrac { 1 } { 4 } x - \dfrac { 2 } { 9 } y = \dfrac { 1 } { 3 }$
$x$-intercept
$\left ( - \dfrac { 5 } { 2 } , 0 \right )$
$y$-intercept
$\left ( 0 , \dfrac { 15 } { 13 } \right )$
$x$-intercept
$\left ( \dfrac { 4 } { 3 } , 0 \right )$
$y$-intercept
$\left ( 0 , - \dfrac { 3 } { 2 } \right )$
$\begin{cases} 0.6x-1.3y = -1.5 \\ \dfrac{ 1 }{ 4 } x- \dfrac{ 2 }{ 9 } y = \dfrac{ 1 }{ 3 } \end{cases}$
$x = 4 , y = 3$
Solve quadratic equations using the square root
$\begin{cases} \color{#FF6800}{ 0.6 } \color{#FF6800}{ x } - 1.3 y = - 1.5 \\ \dfrac { 1 } { 4 } x - \dfrac { 2 } { 9 } y = \dfrac { 1 } { 3 } \end{cases}$
 Calculate the multiplication expression 
$\begin{cases} \color{#FF6800}{ \dfrac { 3 x } { 5 } } - 1.3 y = - 1.5 \\ \dfrac { 1 } { 4 } x - \dfrac { 2 } { 9 } y = \dfrac { 1 } { 3 } \end{cases}$
$\begin{cases} \dfrac { 3 x } { 5 } \color{#FF6800}{ - } \color{#FF6800}{ 1.3 } \color{#FF6800}{ y } = - 1.5 \\ \dfrac { 1 } { 4 } x - \dfrac { 2 } { 9 } y = \dfrac { 1 } { 3 } \end{cases}$
 Calculate the multiplication expression 
$\begin{cases} \dfrac { 3 x } { 5 } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 13 y } { 10 } } = - 1.5 \\ \dfrac { 1 } { 4 } x - \dfrac { 2 } { 9 } y = \dfrac { 1 } { 3 } \end{cases}$
$\begin{cases} \dfrac { 3 x } { 5 } - \dfrac { 13 y } { 10 } = \color{#FF6800}{ - } \color{#FF6800}{ 1.5 } \\ \dfrac { 1 } { 4 } x - \dfrac { 2 } { 9 } y = \dfrac { 1 } { 3 } \end{cases}$
 Convert decimals to fractions 
$\begin{cases} \dfrac { 3 x } { 5 } - \dfrac { 13 y } { 10 } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 2 } } \\ \dfrac { 1 } { 4 } x - \dfrac { 2 } { 9 } y = \dfrac { 1 } { 3 } \end{cases}$
$\begin{cases} \dfrac { 3 x } { 5 } - \dfrac { 13 y } { 10 } = - \dfrac { 3 } { 2 } \\ \color{#FF6800}{ \dfrac { 1 } { 4 } } \color{#FF6800}{ x } - \dfrac { 2 } { 9 } y = \dfrac { 1 } { 3 } \end{cases}$
 Calculate the multiplication expression 
$\begin{cases} \dfrac { 3 x } { 5 } - \dfrac { 13 y } { 10 } = - \dfrac { 3 } { 2 } \\ \color{#FF6800}{ \dfrac { x } { 4 } } - \dfrac { 2 } { 9 } y = \dfrac { 1 } { 3 } \end{cases}$
$\begin{cases} \dfrac { 3 x } { 5 } - \dfrac { 13 y } { 10 } = - \dfrac { 3 } { 2 } \\ \dfrac { x } { 4 } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 } { 9 } } \color{#FF6800}{ y } = \dfrac { 1 } { 3 } \end{cases}$
 Calculate the multiplication expression 
$\begin{cases} \dfrac { 3 x } { 5 } - \dfrac { 13 y } { 10 } = - \dfrac { 3 } { 2 } \\ \dfrac { x } { 4 } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 y } { 9 } } = \dfrac { 1 } { 3 } \end{cases}$
$\begin{cases} \dfrac { 3 x } { 5 } - \dfrac { 13 y } { 10 } = - \dfrac { 3 } { 2 } \\ \dfrac { x } { 4 } - \dfrac { 2 y } { 9 } = \dfrac { 1 } { 3 } \end{cases}$
 Solve a solution to $x$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 13 } { 6 } } \color{#FF6800}{ y } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 2 } } \\ \dfrac { x } { 4 } - \dfrac { 2 y } { 9 } = \dfrac { 1 } { 3 } \end{cases}$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 13 } { 6 } } \color{#FF6800}{ y } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 2 } } \\ \color{#FF6800}{ \dfrac { x } { 4 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 y } { 9 } } = \color{#FF6800}{ \dfrac { 1 } { 3 } } \end{cases}$
 Substitute the given $x$ value into the equation $\dfrac { x } { 4 } - \dfrac { 2 y } { 9 } = \dfrac { 1 } { 3 }$
$\color{#FF6800}{ \dfrac { \dfrac { 13 } { 6 } y - \dfrac { 5 } { 2 } } { 4 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 y } { 9 } } = \color{#FF6800}{ \dfrac { 1 } { 3 } }$
$\color{#FF6800}{ \dfrac { \dfrac { 13 } { 6 } y - \dfrac { 5 } { 2 } } { 4 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 y } { 9 } } = \color{#FF6800}{ \dfrac { 1 } { 3 } }$
 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 $x = \dfrac { 13 } { 6 } y - \dfrac { 5 } { 2 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 13 } { 6 } } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 2 } }$
$x = \color{#FF6800}{ \dfrac { 13 } { 6 } } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } - \dfrac { 5 } { 2 }$
 Calculate the product of rational numbers 
$x = \color{#FF6800}{ \dfrac { 13 } { 2 } } - \dfrac { 5 } { 2 }$
$x = \color{#FF6800}{ \dfrac { 13 } { 2 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 2 } }$
 Find the difference between the two fractions $\dfrac { 13 } { 2 }$ and $- \dfrac { 5 } { 2 }$
$x = \color{#FF6800}{ 4 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ 4 }$
 The possible solutions are as follows 
$\color{#FF6800}{ x } = \color{#FF6800}{ 4 } , \color{#FF6800}{ y } = \color{#FF6800}{ 3 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ 4 } , \color{#FF6800}{ y } = \color{#FF6800}{ 3 }$
 Check if it is the solution to the system of equations 
$\begin{cases} \color{#FF6800}{ 0.6 } \color{#FF6800}{ \times } \color{#FF6800}{ 4 } \color{#FF6800}{ - } \color{#FF6800}{ 1.3 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } = \color{#FF6800}{ - } \color{#FF6800}{ 1.5 } \\ \color{#FF6800}{ \dfrac { 1 } { 4 } } \color{#FF6800}{ \times } \color{#FF6800}{ 4 } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 } { 9 } } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } = \color{#FF6800}{ \dfrac { 1 } { 3 } } \end{cases}$
$\begin{cases} \color{#FF6800}{ 0.6 } \color{#FF6800}{ \times } \color{#FF6800}{ 4 } \color{#FF6800}{ - } \color{#FF6800}{ 1.3 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } = \color{#FF6800}{ - } \color{#FF6800}{ 1.5 } \\ \color{#FF6800}{ \dfrac { 1 } { 4 } } \color{#FF6800}{ \times } \color{#FF6800}{ 4 } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 } { 9 } } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } = \color{#FF6800}{ \dfrac { 1 } { 3 } } \end{cases}$
 Simplify the equality 
$\begin{cases} \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 2 } } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 2 } } \\ \color{#FF6800}{ \dfrac { 1 } { 3 } } = \color{#FF6800}{ \dfrac { 1 } { 3 } } \end{cases}$
$\begin{cases} \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 2 } } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 2 } } \\ \color{#FF6800}{ \dfrac { 1 } { 3 } } = \color{#FF6800}{ \dfrac { 1 } { 3 } } \end{cases}$
 Since it is true in both equations, it is the solution of the system of equations 
$\color{#FF6800}{ x } = \color{#FF6800}{ 4 } , \color{#FF6800}{ y } = \color{#FF6800}{ 3 }$
$\begin{cases} x = 4 \\ y = 3 \end{cases}$
Solve quadratic equations using the square root
$\begin{cases} \color{#FF6800}{ 0.6 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1.3 } \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ 1.5 } \\ \color{#FF6800}{ \dfrac { 1 } { 4 } } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 } { 9 } } \color{#FF6800}{ y } = \color{#FF6800}{ \dfrac { 1 } { 3 } } \end{cases}$
 Solve the system of linear equations for $x , y$
$\begin{cases} \color{#FF6800}{ 0.6 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1.3 } \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ 1.5 } \\ \color{#FF6800}{ \dfrac { 23 } { 120 } } \color{#FF6800}{ y } = \color{#FF6800}{ \dfrac { 23 } { 40 } } \end{cases}$
$\begin{cases} \color{#FF6800}{ 0.6 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1.3 } \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ 1.5 } \\ \color{#FF6800}{ \dfrac { 23 } { 120 } } \color{#FF6800}{ y } = \color{#FF6800}{ \dfrac { 23 } { 40 } } \end{cases}$
 Solve the system of linear equations for $x , y$
$\begin{cases} \color{#FF6800}{ \dfrac { 23 } { 200 } } \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 23 } { 50 } } \\ \color{#FF6800}{ \dfrac { 23 } { 120 } } \color{#FF6800}{ y } = \color{#FF6800}{ \dfrac { 23 } { 40 } } \end{cases}$
$\begin{cases} \color{#FF6800}{ \dfrac { 23 } { 200 } } \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 23 } { 50 } } \\ \dfrac { 23 } { 120 } y = \dfrac { 23 } { 40 } \end{cases}$
 Solve a solution to $x$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ 4 } \\ \dfrac { 23 } { 120 } y = \dfrac { 23 } { 40 } \end{cases}$
$\begin{cases} x = 4 \\ \color{#FF6800}{ \dfrac { 23 } { 120 } } \color{#FF6800}{ y } = \color{#FF6800}{ \dfrac { 23 } { 40 } } \end{cases}$
 Solve a solution to $y$
$\begin{cases} x = 4 \\ \color{#FF6800}{ y } = \color{#FF6800}{ 3 } \end{cases}$
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