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Formula
Solve the system of equations
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$2 x + 5 y = 5$
$- 3 x + 7 y = 36$
$x$Intercept
$\left ( \dfrac { 5 } { 2 } , 0 \right )$
$y$Intercept
$\left ( 0 , 1 \right )$
$x$Intercept
$\left ( - 12 , 0 \right )$
$y$Intercept
$\left ( 0 , \dfrac { 36 } { 7 } \right )$
$\begin{cases} 2x+5y = 5 \\-3x+7y = 36 \end{cases}$
$x = - 5 , y = 3$
Solve quadratic equations using the square root
$\begin{cases} 2 x + 5 y = 5 \\ - 3 x + 7 y = 36 \end{cases}$
 Solve a solution to $x$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 2 } } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 5 } { 2 } } \\ - 3 x + 7 y = 36 \end{cases}$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 2 } } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 5 } { 2 } } \\ \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 7 } \color{#FF6800}{ y } = \color{#FF6800}{ 36 } \end{cases}$
 Substitute the given $x$ value into the equation $- 3 x + 7 y = 36$
$\color{#FF6800}{ - } \color{#FF6800}{ 3 } \left ( \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 2 } } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 5 } { 2 } } \right ) \color{#FF6800}{ + } \color{#FF6800}{ 7 } \color{#FF6800}{ y } = \color{#FF6800}{ 36 }$
$\color{#FF6800}{ - } \color{#FF6800}{ 3 } \left ( \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 2 } } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 5 } { 2 } } \right ) \color{#FF6800}{ + } \color{#FF6800}{ 7 } \color{#FF6800}{ y } = \color{#FF6800}{ 36 }$
 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 $x = - \dfrac { 5 } { 2 } y + \dfrac { 5 } { 2 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 2 } } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 5 } { 2 } }$
$x = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 2 } } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } + \dfrac { 5 } { 2 }$
 Calculate the product of rational numbers 
$x = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 15 } { 2 } } + \dfrac { 5 } { 2 }$
$x = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 15 } { 2 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 5 } { 2 } }$
 Find the sum or difference of the fractions 
$x = \color{#FF6800}{ - } \color{#FF6800}{ 5 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 5 }$
 The possible solutions are as follows 
$\color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 5 } , \color{#FF6800}{ y } = \color{#FF6800}{ 3 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 5 } , \color{#FF6800}{ y } = \color{#FF6800}{ 3 }$
 Check if it is the solution to the system of equations 
$\begin{cases} \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right ) \color{#FF6800}{ + } \color{#FF6800}{ 5 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } = \color{#FF6800}{ 5 } \\ \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right ) \color{#FF6800}{ + } \color{#FF6800}{ 7 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } = \color{#FF6800}{ 36 } \end{cases}$
$\begin{cases} \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right ) \color{#FF6800}{ + } \color{#FF6800}{ 5 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } = \color{#FF6800}{ 5 } \\ \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right ) \color{#FF6800}{ + } \color{#FF6800}{ 7 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } = \color{#FF6800}{ 36 } \end{cases}$
 Simplify the equality 
$\begin{cases} \color{#FF6800}{ 5 } = \color{#FF6800}{ 5 } \\ \color{#FF6800}{ 36 } = \color{#FF6800}{ 36 } \end{cases}$
$\begin{cases} \color{#FF6800}{ 5 } = \color{#FF6800}{ 5 } \\ \color{#FF6800}{ 36 } = \color{#FF6800}{ 36 } \end{cases}$
 Since it is true in both equations, it is the solution of the system of equations 
$\color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 5 } , \color{#FF6800}{ y } = \color{#FF6800}{ 3 }$
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