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Formula
Solve the system of equations
Graph
$2 x + y = 1$
$5 x - y = 15$
$x$Intercept
$\left ( \dfrac { 1 } { 2 } , 0 \right )$
$y$Intercept
$\left ( 0 , 1 \right )$
$x$Intercept
$\left ( 3 , 0 \right )$
$y$Intercept
$\left ( 0 , - 15 \right )$
$\begin{cases} 2x+y = 1 \\5x-y = 15 \end{cases}$
$x = \dfrac { 16 } { 7 } , y = - \dfrac { 25 } { 7 }$
Solve quadratic equations using the square root
$\begin{cases} 2 x + y = 1 \\ 5 x - y = 15 \end{cases}$
 Solve a solution to $y$
$\begin{cases} \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \\ 5 x - y = 15 \end{cases}$
$\begin{cases} \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \\ \color{#FF6800}{ 5 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ y } = \color{#FF6800}{ 15 } \end{cases}$
 Substitute the given $y$ value into the equation $5 x - y = 15$
$\color{#FF6800}{ 5 } \color{#FF6800}{ x } \color{#FF6800}{ - } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right ) = \color{#FF6800}{ 15 }$
$\color{#FF6800}{ 5 } \color{#FF6800}{ x } \color{#FF6800}{ - } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right ) = \color{#FF6800}{ 15 }$
 Solve a solution to $x$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 16 } { 7 } }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 16 } { 7 } }$
 Substitute the given $x$ value into the equation $y = - 2 x + 1$
$\color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ \dfrac { 16 } { 7 } } \color{#FF6800}{ + } \color{#FF6800}{ 1 }$
$y = \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ \dfrac { 16 } { 7 } } + 1$
 Calculate the product of rational numbers 
$y = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 32 } { 7 } } + 1$
$y = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 32 } { 7 } } \color{#FF6800}{ + } \color{#FF6800}{ 1 }$
 Get the subtract 
$y = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 25 } { 7 } }$
$\color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 25 } { 7 } }$
 The possible solutions are as follows 
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 16 } { 7 } } , \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 25 } { 7 } }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 16 } { 7 } } , \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 25 } { 7 } }$
 Check if it is the solution to the system of equations 
$\begin{cases} \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ \dfrac { 16 } { 7 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 25 } { 7 } } = \color{#FF6800}{ 1 } \\ \color{#FF6800}{ 5 } \color{#FF6800}{ \times } \color{#FF6800}{ \dfrac { 16 } { 7 } } \color{#FF6800}{ - } \left ( \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 25 } { 7 } } \right ) = \color{#FF6800}{ 15 } \end{cases}$
$\begin{cases} \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ \dfrac { 16 } { 7 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 25 } { 7 } } = \color{#FF6800}{ 1 } \\ \color{#FF6800}{ 5 } \color{#FF6800}{ \times } \color{#FF6800}{ \dfrac { 16 } { 7 } } \color{#FF6800}{ - } \left ( \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 25 } { 7 } } \right ) = \color{#FF6800}{ 15 } \end{cases}$
 Simplify the equality 
$\begin{cases} \color{#FF6800}{ 1 } = \color{#FF6800}{ 1 } \\ \color{#FF6800}{ 15 } = \color{#FF6800}{ 15 } \end{cases}$
$\begin{cases} \color{#FF6800}{ 1 } = \color{#FF6800}{ 1 } \\ \color{#FF6800}{ 15 } = \color{#FF6800}{ 15 } \end{cases}$
 Since it is true in both equations, it is the solution of the system of equations 
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 16 } { 7 } } , \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 25 } { 7 } }$
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