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Formula
Solve the system of equations
Graph
$3 x - 2 y = - 2$
$5 x + 8 y = - 60$
$x$Intercept
$\left ( - \dfrac { 2 } { 3 } , 0 \right )$
$y$Intercept
$\left ( 0 , 1 \right )$
$x$Intercept
$\left ( - 12 , 0 \right )$
$y$Intercept
$\left ( 0 , - \dfrac { 15 } { 2 } \right )$
$\begin{cases} 3x-2y = -2 \\5x+8y = -60 \end{cases}$
$x = - 4 , y = - 5$
Solve quadratic equations using the square root
$\begin{cases} 3 x - 2 y = - 2 \\ 5 x + 8 y = - 60 \end{cases}$
 Solve a solution to $x$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 2 } { 3 } } \color{#FF6800}{ y } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 } { 3 } } \\ 5 x + 8 y = - 60 \end{cases}$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 2 } { 3 } } \color{#FF6800}{ y } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 } { 3 } } \\ \color{#FF6800}{ 5 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 8 } \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ 60 } \end{cases}$
 Substitute the given $x$ value into the equation $5 x + 8 y = - 60$
$\color{#FF6800}{ 5 } \left ( \color{#FF6800}{ \dfrac { 2 } { 3 } } \color{#FF6800}{ y } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 } { 3 } } \right ) \color{#FF6800}{ + } \color{#FF6800}{ 8 } \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ 60 }$
$\color{#FF6800}{ 5 } \left ( \color{#FF6800}{ \dfrac { 2 } { 3 } } \color{#FF6800}{ y } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 } { 3 } } \right ) \color{#FF6800}{ + } \color{#FF6800}{ 8 } \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ 60 }$
 Solve a solution to $y$
$\color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ 5 }$
$\color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ 5 }$
 Substitute the given $y$ value into the equation $x = \dfrac { 2 } { 3 } y - \dfrac { 2 } { 3 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 2 } { 3 } } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right ) \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 } { 3 } }$
$x = \color{#FF6800}{ \dfrac { 2 } { 3 } } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right ) - \dfrac { 2 } { 3 }$
 Calculate the product of rational numbers 
$x = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 10 } { 3 } } - \dfrac { 2 } { 3 }$
$x = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 10 } { 3 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 } { 3 } }$
 Find the sum or difference of the fractions 
$x = \color{#FF6800}{ - } \color{#FF6800}{ 4 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 4 }$
 The possible solutions are as follows 
$\color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 4 } , \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ 5 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 4 } , \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ 5 }$
 Check if it is the solution to the system of equations 
$\begin{cases} \color{#FF6800}{ 3 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 4 } \right ) \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right ) = \color{#FF6800}{ - } \color{#FF6800}{ 2 } \\ \color{#FF6800}{ 5 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 4 } \right ) \color{#FF6800}{ + } \color{#FF6800}{ 8 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right ) = \color{#FF6800}{ - } \color{#FF6800}{ 60 } \end{cases}$
$\begin{cases} \color{#FF6800}{ 3 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 4 } \right ) \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right ) = \color{#FF6800}{ - } \color{#FF6800}{ 2 } \\ \color{#FF6800}{ 5 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 4 } \right ) \color{#FF6800}{ + } \color{#FF6800}{ 8 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right ) = \color{#FF6800}{ - } \color{#FF6800}{ 60 } \end{cases}$
 Simplify the equality 
$\begin{cases} \color{#FF6800}{ - } \color{#FF6800}{ 2 } = \color{#FF6800}{ - } \color{#FF6800}{ 2 } \\ \color{#FF6800}{ - } \color{#FF6800}{ 60 } = \color{#FF6800}{ - } \color{#FF6800}{ 60 } \end{cases}$
$\begin{cases} \color{#FF6800}{ - } \color{#FF6800}{ 2 } = \color{#FF6800}{ - } \color{#FF6800}{ 2 } \\ \color{#FF6800}{ - } \color{#FF6800}{ 60 } = \color{#FF6800}{ - } \color{#FF6800}{ 60 } \end{cases}$
 Since it is true in both equations, it is the solution of the system of equations 
$\color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 4 } , \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ 5 }$
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