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Formula
Solve the system of equations
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$5 x + 7 y = - 1$
$- 3 x + 4 y = - 24$
$x$-intercept
$\left ( - \dfrac { 1 } { 5 } , 0 \right )$
$y$-intercept
$\left ( 0 , - \dfrac { 1 } { 7 } \right )$
$x$-intercept
$\left ( 8 , 0 \right )$
$y$-intercept
$\left ( 0 , - 6 \right )$
$\begin{cases} 5x+7y = -1 \\-3x+4y = -24 \end{cases}$
$x = 4 , y = - 3$
Solve quadratic equations using the square root
$\begin{cases} 5 x + 7 y = - 1 \\ - 3 x + 4 y = - 24 \end{cases}$
 Solve a solution to $x$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 7 } { 5 } } \color{#FF6800}{ y } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 5 } } \\ - 3 x + 4 y = - 24 \end{cases}$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 7 } { 5 } } \color{#FF6800}{ y } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 5 } } \\ \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 4 } \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ 24 } \end{cases}$
 Substitute the given $x$ value into the equation $- 3 x + 4 y = - 24$
$\color{#FF6800}{ - } \color{#FF6800}{ 3 } \left ( \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 7 } { 5 } } \color{#FF6800}{ y } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 5 } } \right ) \color{#FF6800}{ + } \color{#FF6800}{ 4 } \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ 24 }$
$\color{#FF6800}{ - } \color{#FF6800}{ 3 } \left ( \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 7 } { 5 } } \color{#FF6800}{ y } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 5 } } \right ) \color{#FF6800}{ + } \color{#FF6800}{ 4 } \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ 24 }$
 Solve a solution to $y$
$\color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ 3 }$
$\color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ 3 }$
 Substitute the given $y$ value into the equation $x = - \dfrac { 7 } { 5 } y - \dfrac { 1 } { 5 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 7 } { 5 } } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right ) \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 5 } }$
$x = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 7 } { 5 } } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right ) - \dfrac { 1 } { 5 }$
 Calculate the product of rational numbers 
$x = \color{#FF6800}{ \dfrac { 21 } { 5 } } - \dfrac { 1 } { 5 }$
$x = \color{#FF6800}{ \dfrac { 21 } { 5 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 5 } }$
 Find the difference between the two fractions $\dfrac { 21 } { 5 }$ and $- \dfrac { 1 } { 5 }$
$x = \color{#FF6800}{ 4 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ 4 }$
 The possible solutions are as follows 
$\color{#FF6800}{ x } = \color{#FF6800}{ 4 } , \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ 3 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ 4 } , \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ 3 }$
 Check if it is the solution to the system of equations 
$\begin{cases} \color{#FF6800}{ 5 } \color{#FF6800}{ \times } \color{#FF6800}{ 4 } \color{#FF6800}{ + } \color{#FF6800}{ 7 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right ) = \color{#FF6800}{ - } \color{#FF6800}{ 1 } \\ \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ \times } \color{#FF6800}{ 4 } \color{#FF6800}{ + } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right ) = \color{#FF6800}{ - } \color{#FF6800}{ 24 } \end{cases}$
$\begin{cases} \color{#FF6800}{ 5 } \color{#FF6800}{ \times } \color{#FF6800}{ 4 } \color{#FF6800}{ + } \color{#FF6800}{ 7 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right ) = \color{#FF6800}{ - } \color{#FF6800}{ 1 } \\ \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ \times } \color{#FF6800}{ 4 } \color{#FF6800}{ + } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right ) = \color{#FF6800}{ - } \color{#FF6800}{ 24 } \end{cases}$
 Simplify the equality 
$\begin{cases} \color{#FF6800}{ - } \color{#FF6800}{ 1 } = \color{#FF6800}{ - } \color{#FF6800}{ 1 } \\ \color{#FF6800}{ - } \color{#FF6800}{ 24 } = \color{#FF6800}{ - } \color{#FF6800}{ 24 } \end{cases}$
$\begin{cases} \color{#FF6800}{ - } \color{#FF6800}{ 1 } = \color{#FF6800}{ - } \color{#FF6800}{ 1 } \\ \color{#FF6800}{ - } \color{#FF6800}{ 24 } = \color{#FF6800}{ - } \color{#FF6800}{ 24 } \end{cases}$
 Since it is true in both equations, it is the solution of the system of equations 
$\color{#FF6800}{ x } = \color{#FF6800}{ 4 } , \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ 3 }$
$\begin{cases} x = 4 \\ y = - 3 \end{cases}$
Solve quadratic equations using the square root
$\begin{cases} \color{#FF6800}{ 5 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 7 } \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ 1 } \\ \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 4 } \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ 24 } \end{cases}$
 Solve the system of linear equations for $x , y$
$\begin{cases} \color{#FF6800}{ 5 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 7 } \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ 1 } \\ \color{#FF6800}{ 41 } \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ 123 } \end{cases}$
$\begin{cases} \color{#FF6800}{ 5 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 7 } \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ 1 } \\ \color{#FF6800}{ 41 } \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ 123 } \end{cases}$
 Solve the system of linear equations for $x , y$
$\begin{cases} \color{#FF6800}{ 205 } \color{#FF6800}{ x } = \color{#FF6800}{ 820 } \\ \color{#FF6800}{ 41 } \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ 123 } \end{cases}$
$\begin{cases} \color{#FF6800}{ 205 } \color{#FF6800}{ x } = \color{#FF6800}{ 820 } \\ 41 y = - 123 \end{cases}$
 Solve a solution to $x$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ 4 } \\ 41 y = - 123 \end{cases}$
$\begin{cases} x = 4 \\ \color{#FF6800}{ 41 } \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ 123 } \end{cases}$
 Solve a solution to $y$
$\begin{cases} x = 4 \\ \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ 3 } \end{cases}$
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