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Formula
Solve the system of equations
Graph
$x - y = 3$
$x + y = 5$
$x$-intercept
$\left ( 3 , 0 \right )$
$y$-intercept
$\left ( 0 , - 3 \right )$
$x$-intercept
$\left ( 5 , 0 \right )$
$y$-intercept
$\left ( 0 , 5 \right )$
$\begin{cases} x-y = 3 \\x+y = 5 \end{cases}$
$x = 4 , y = 1$
Solve quadratic equations using the square root
$\begin{cases} x - y = 3 \\ x + y = 5 \end{cases}$
 Solve a solution to $x$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \\ x + y = 5 \end{cases}$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \\ \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ y } = \color{#FF6800}{ 5 } \end{cases}$
 Substitute the given $x$ value into the equation $x + y = 5$
$\left ( \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right ) \color{#FF6800}{ + } \color{#FF6800}{ y } = \color{#FF6800}{ 5 }$
$\left ( \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right ) \color{#FF6800}{ + } \color{#FF6800}{ y } = \color{#FF6800}{ 5 }$
 Solve a solution to $y$
$\color{#FF6800}{ y } = \color{#FF6800}{ 1 }$
$\color{#FF6800}{ y } = \color{#FF6800}{ 1 }$
 Substitute the given $y$ value into the equation $x = y + 3$
$\color{#FF6800}{ x } = \color{#FF6800}{ 1 } \color{#FF6800}{ + } \color{#FF6800}{ 3 }$
$x = \color{#FF6800}{ 1 } \color{#FF6800}{ + } \color{#FF6800}{ 3 }$
 Add $1$ and $3$
$x = \color{#FF6800}{ 4 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ 4 }$
 The possible solutions are as follows 
$\color{#FF6800}{ x } = \color{#FF6800}{ 4 } , \color{#FF6800}{ y } = \color{#FF6800}{ 1 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ 4 } , \color{#FF6800}{ y } = \color{#FF6800}{ 1 }$
 Check if it is the solution to the system of equations 
$\begin{cases} \color{#FF6800}{ 4 } \color{#FF6800}{ - } \color{#FF6800}{ 1 } = \color{#FF6800}{ 3 } \\ \color{#FF6800}{ 4 } \color{#FF6800}{ + } \color{#FF6800}{ 1 } = \color{#FF6800}{ 5 } \end{cases}$
$\begin{cases} \color{#FF6800}{ 4 } \color{#FF6800}{ - } \color{#FF6800}{ 1 } = \color{#FF6800}{ 3 } \\ \color{#FF6800}{ 4 } \color{#FF6800}{ + } \color{#FF6800}{ 1 } = \color{#FF6800}{ 5 } \end{cases}$
 Simplify the equality 
$\begin{cases} \color{#FF6800}{ 3 } = \color{#FF6800}{ 3 } \\ \color{#FF6800}{ 5 } = \color{#FF6800}{ 5 } \end{cases}$
$\begin{cases} \color{#FF6800}{ 3 } = \color{#FF6800}{ 3 } \\ \color{#FF6800}{ 5 } = \color{#FF6800}{ 5 } \end{cases}$
 Since it is true in both equations, it is the solution of the system of equations 
$\color{#FF6800}{ x } = \color{#FF6800}{ 4 } , \color{#FF6800}{ y } = \color{#FF6800}{ 1 }$
$\begin{cases} x = 4 \\ y = 1 \end{cases}$
Solve quadratic equations using the square root
$\begin{cases} \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ y } = \color{#FF6800}{ 3 } \\ \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ y } = \color{#FF6800}{ 5 } \end{cases}$
 Solve the system of linear equations for $x , y$
$\begin{cases} \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ y } = \color{#FF6800}{ 3 } \\ \color{#FF6800}{ 2 } \color{#FF6800}{ y } = \color{#FF6800}{ 2 } \end{cases}$
$\begin{cases} \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ y } = \color{#FF6800}{ 3 } \\ \color{#FF6800}{ 2 } \color{#FF6800}{ y } = \color{#FF6800}{ 2 } \end{cases}$
 Solve the system of linear equations for $x , y$
$\begin{cases} \color{#FF6800}{ 2 } \color{#FF6800}{ x } = \color{#FF6800}{ 8 } \\ \color{#FF6800}{ 2 } \color{#FF6800}{ y } = \color{#FF6800}{ 2 } \end{cases}$
$\begin{cases} \color{#FF6800}{ 2 } \color{#FF6800}{ x } = \color{#FF6800}{ 8 } \\ 2 y = 2 \end{cases}$
 Solve a solution to $x$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ 4 } \\ 2 y = 2 \end{cases}$
$\begin{cases} x = 4 \\ \color{#FF6800}{ 2 } \color{#FF6800}{ y } = \color{#FF6800}{ 2 } \end{cases}$
 Solve a solution to $y$
$\begin{cases} x = 4 \\ \color{#FF6800}{ y } = \color{#FF6800}{ 1 } \end{cases}$
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