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Formula
Solve the system of equations
Graph
$6 x - 5 y = - 9$
$4 x + 3 y = 13$
$x$Intercept
$\left ( - \dfrac { 3 } { 2 } , 0 \right )$
$y$Intercept
$\left ( 0 , \dfrac { 9 } { 5 } \right )$
$x$Intercept
$\left ( \dfrac { 13 } { 4 } , 0 \right )$
$y$Intercept
$\left ( 0 , \dfrac { 13 } { 3 } \right )$
$\begin{cases} 6x-5y = -9 \\4x+3y = 13 \end{cases}$
$x = 1 , y = 3$
Solve quadratic equations using the square root
$\begin{cases} 6 x - 5 y = - 9 \\ 4 x + 3 y = 13 \end{cases}$
 Solve a solution to $x$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 5 } { 6 } } \color{#FF6800}{ y } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 2 } } \\ 4 x + 3 y = 13 \end{cases}$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 5 } { 6 } } \color{#FF6800}{ y } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 2 } } \\ \color{#FF6800}{ 4 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ y } = \color{#FF6800}{ 13 } \end{cases}$
 Substitute the given $x$ value into the equation $4 x + 3 y = 13$
$\color{#FF6800}{ 4 } \left ( \color{#FF6800}{ \dfrac { 5 } { 6 } } \color{#FF6800}{ y } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 2 } } \right ) \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ y } = \color{#FF6800}{ 13 }$
$\color{#FF6800}{ 4 } \left ( \color{#FF6800}{ \dfrac { 5 } { 6 } } \color{#FF6800}{ y } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 2 } } \right ) \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ y } = \color{#FF6800}{ 13 }$
 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 $x = \dfrac { 5 } { 6 } y - \dfrac { 3 } { 2 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 5 } { 6 } } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 2 } }$
$x = \color{#FF6800}{ \dfrac { 5 } { 6 } } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } - \dfrac { 3 } { 2 }$
 Calculate the product of rational numbers 
$x = \color{#FF6800}{ \dfrac { 5 } { 2 } } - \dfrac { 3 } { 2 }$
$x = \color{#FF6800}{ \dfrac { 5 } { 2 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 2 } }$
 Find the difference between the two fractions $\dfrac { 5 } { 2 }$ and $- \dfrac { 3 } { 2 }$
$x = \color{#FF6800}{ 1 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ 1 }$
 The possible solutions are as follows 
$\color{#FF6800}{ x } = \color{#FF6800}{ 1 } , \color{#FF6800}{ y } = \color{#FF6800}{ 3 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ 1 } , \color{#FF6800}{ y } = \color{#FF6800}{ 3 }$
 Check if it is the solution to the system of equations 
$\begin{cases} \color{#FF6800}{ 6 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } = \color{#FF6800}{ - } \color{#FF6800}{ 9 } \\ \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } = \color{#FF6800}{ 13 } \end{cases}$
$\begin{cases} \color{#FF6800}{ 6 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } = \color{#FF6800}{ - } \color{#FF6800}{ 9 } \\ \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } = \color{#FF6800}{ 13 } \end{cases}$
 Simplify the equality 
$\begin{cases} \color{#FF6800}{ - } \color{#FF6800}{ 9 } = \color{#FF6800}{ - } \color{#FF6800}{ 9 } \\ \color{#FF6800}{ 13 } = \color{#FF6800}{ 13 } \end{cases}$
$\begin{cases} \color{#FF6800}{ - } \color{#FF6800}{ 9 } = \color{#FF6800}{ - } \color{#FF6800}{ 9 } \\ \color{#FF6800}{ 13 } = \color{#FF6800}{ 13 } \end{cases}$
 Since it is true in both equations, it is the solution of the system of equations 
$\color{#FF6800}{ x } = \color{#FF6800}{ 1 } , \color{#FF6800}{ y } = \color{#FF6800}{ 3 }$
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