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Formula
Solve the system of equations
Graph
$- 2 x + y = - 1$
$- 3 x + 2 y = 1$
$x$Intercept
$\left ( \dfrac { 1 } { 2 } , 0 \right )$
$y$Intercept
$\left ( 0 , - 1 \right )$
$x$Intercept
$\left ( - \dfrac { 1 } { 3 } , 0 \right )$
$y$Intercept
$\left ( 0 , \dfrac { 1 } { 2 } \right )$
$\begin{cases} -2x+y = -1 \\-3x+2y = 1 \end{cases}$
$x = 3 , y = 5$
Solve quadratic equations using the square root
$\begin{cases} - 2 x + y = - 1 \\ - 3 x + 2 y = 1 \end{cases}$
 Solve a solution to $y$
$\begin{cases} \color{#FF6800}{ y } = \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \\ - 3 x + 2 y = 1 \end{cases}$
$\begin{cases} \color{#FF6800}{ y } = \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \\ \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \color{#FF6800}{ y } = \color{#FF6800}{ 1 } \end{cases}$
 Substitute the given $y$ value into the equation $- 3 x + 2 y = 1$
$\color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \left ( \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) = \color{#FF6800}{ 1 }$
$\color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \left ( \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) = \color{#FF6800}{ 1 }$
 Solve a solution to $x$
$\color{#FF6800}{ x } = \color{#FF6800}{ 3 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ 3 }$
 Substitute the given $x$ value into the equation $y = 2 x - 1$
$\color{#FF6800}{ y } = \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } \color{#FF6800}{ - } \color{#FF6800}{ 1 }$
$\color{#FF6800}{ y } = \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } \color{#FF6800}{ - } \color{#FF6800}{ 1 }$
 Organize the expression 
$\color{#FF6800}{ y } = \color{#FF6800}{ 5 }$
$\color{#FF6800}{ y } = \color{#FF6800}{ 5 }$
 The possible solutions are as follows 
$\color{#FF6800}{ x } = \color{#FF6800}{ 3 } , \color{#FF6800}{ y } = \color{#FF6800}{ 5 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ 3 } , \color{#FF6800}{ y } = \color{#FF6800}{ 5 }$
 Check if it is the solution to the system of equations 
$\begin{cases} \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } \color{#FF6800}{ + } \color{#FF6800}{ 5 } = \color{#FF6800}{ - } \color{#FF6800}{ 1 } \\ \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 5 } = \color{#FF6800}{ 1 } \end{cases}$
$\begin{cases} \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } \color{#FF6800}{ + } \color{#FF6800}{ 5 } = \color{#FF6800}{ - } \color{#FF6800}{ 1 } \\ \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 5 } = \color{#FF6800}{ 1 } \end{cases}$
 Simplify the equality 
$\begin{cases} \color{#FF6800}{ - } \color{#FF6800}{ 1 } = \color{#FF6800}{ - } \color{#FF6800}{ 1 } \\ \color{#FF6800}{ 1 } = \color{#FF6800}{ 1 } \end{cases}$
$\begin{cases} \color{#FF6800}{ - } \color{#FF6800}{ 1 } = \color{#FF6800}{ - } \color{#FF6800}{ 1 } \\ \color{#FF6800}{ 1 } = \color{#FF6800}{ 1 } \end{cases}$
 Since it is true in both equations, it is the solution of the system of equations 
$\color{#FF6800}{ x } = \color{#FF6800}{ 3 } , \color{#FF6800}{ y } = \color{#FF6800}{ 5 }$
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