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Formula
Solve the system of equations
Graph
$y = 2 x - 1$
$x + 2 y = 8$
$x$-intercept
$\left ( \dfrac { 1 } { 2 } , 0 \right )$
$y$-intercept
$\left ( 0 , - 1 \right )$
$x$-intercept
$\left ( 8 , 0 \right )$
$y$-intercept
$\left ( 0 , 4 \right )$
$\begin{cases} y = 2x-1 \\x+2y = 8 \end{cases}$
$x = 2 , y = 3$
Solve quadratic equations using the square root
$\begin{cases} \color{#FF6800}{ y } = \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \\ \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \color{#FF6800}{ y } = \color{#FF6800}{ 8 } \end{cases}$
 Substitute the given $y$ value into the equation $x + 2 y = 8$
$\color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \left ( \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) = \color{#FF6800}{ 8 }$
$\color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \left ( \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) = \color{#FF6800}{ 8 }$
 Solve a solution to $x$
$\color{#FF6800}{ x } = \color{#FF6800}{ 2 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ 2 }$
 Substitute the given $x$ value into the equation $y = 2 x - 1$
$\color{#FF6800}{ y } = \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } \color{#FF6800}{ - } \color{#FF6800}{ 1 }$
$y = \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } - 1$
 Multiply $2$ and $2$
$y = \color{#FF6800}{ 4 } - 1$
$y = \color{#FF6800}{ 4 } \color{#FF6800}{ - } \color{#FF6800}{ 1 }$
 Subtract $1$ from $4$
$y = \color{#FF6800}{ 3 }$
$\color{#FF6800}{ y } = \color{#FF6800}{ 3 }$
 The possible solutions are as follows 
$\color{#FF6800}{ x } = \color{#FF6800}{ 2 } , \color{#FF6800}{ y } = \color{#FF6800}{ 3 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ 2 } , \color{#FF6800}{ y } = \color{#FF6800}{ 3 }$
 Check if it is the solution to the system of equations 
$\begin{cases} \color{#FF6800}{ 3 } = \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \\ \color{#FF6800}{ 2 } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } = \color{#FF6800}{ 8 } \end{cases}$
$\begin{cases} \color{#FF6800}{ 3 } = \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \\ \color{#FF6800}{ 2 } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } = \color{#FF6800}{ 8 } \end{cases}$
 Simplify the equality 
$\begin{cases} \color{#FF6800}{ 3 } = \color{#FF6800}{ 3 } \\ \color{#FF6800}{ 8 } = \color{#FF6800}{ 8 } \end{cases}$
$\begin{cases} \color{#FF6800}{ 3 } = \color{#FF6800}{ 3 } \\ \color{#FF6800}{ 8 } = \color{#FF6800}{ 8 } \end{cases}$
 Since it is true in both equations, it is the solution of the system of equations 
$\color{#FF6800}{ x } = \color{#FF6800}{ 2 } , \color{#FF6800}{ y } = \color{#FF6800}{ 3 }$
$\begin{cases} x = 2 \\ y = 3 \end{cases}$
Solve quadratic equations using the square root
$\begin{cases} \color{#FF6800}{ y } = \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \\ x + 2 y = 8 \end{cases}$
 Do transposition 
$\begin{cases} \color{#FF6800}{ y } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 1 } \\ x + 2 y = 8 \end{cases}$
$\begin{cases} \color{#FF6800}{ y } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 1 } \\ \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \color{#FF6800}{ y } = \color{#FF6800}{ 8 } \end{cases}$
 Solve the system of linear equations for $x , y$
$\begin{cases} \color{#FF6800}{ y } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 1 } \\ \color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ 15 } \end{cases}$
$\begin{cases} \color{#FF6800}{ y } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 1 } \\ \color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ 15 } \end{cases}$
 Solve the system of linear equations for $x , y$
$\begin{cases} \color{#FF6800}{ 10 } \color{#FF6800}{ x } = \color{#FF6800}{ 20 } \\ \color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ 15 } \end{cases}$
$\begin{cases} \color{#FF6800}{ 10 } \color{#FF6800}{ x } = \color{#FF6800}{ 20 } \\ - 5 y = - 15 \end{cases}$
 Solve a solution to $x$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ 2 } \\ - 5 y = - 15 \end{cases}$
$\begin{cases} x = 2 \\ \color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ 15 } \end{cases}$
 Solve a solution to $y$
$\begin{cases} x = 2 \\ \color{#FF6800}{ y } = \color{#FF6800}{ 3 } \end{cases}$
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