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Solve the system of equations
Graph
$4 x + 3 y = 5$
$3 x - 4 y = 10$
$x$Intercept
$\left ( \dfrac { 5 } { 4 } , 0 \right )$
$y$Intercept
$\left ( 0 , \dfrac { 5 } { 3 } \right )$
$x$Intercept
$\left ( \dfrac { 10 } { 3 } , 0 \right )$
$y$Intercept
$\left ( 0 , - \dfrac { 5 } { 2 } \right )$
$\begin{cases} 4x+3y = 5 \\3x-4y = 10 \end{cases}$
$x = 2 , y = - 1$
Solve the system of equations
$\begin{cases} 4 x + 3 y = 5 \\ 3 x - 4 y = 10 \end{cases}$
 Solve a solution to $x$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 4 } } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 5 } { 4 } } \\ 3 x - 4 y = 10 \end{cases}$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 4 } } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 5 } { 4 } } \\ \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ y } = \color{#FF6800}{ 10 } \end{cases}$
 Substitute the given $x$ value into the equation $3 x - 4 y = 10$
$\color{#FF6800}{ 3 } \left ( \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 4 } } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 5 } { 4 } } \right ) \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ y } = \color{#FF6800}{ 10 }$
$\color{#FF6800}{ 3 } \left ( \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 4 } } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 5 } { 4 } } \right ) \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ y } = \color{#FF6800}{ 10 }$
 Solve a solution to $y$
$\color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ 1 }$
$\color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ 1 }$
 Substitute the given $y$ value into the equation $x = - \dfrac { 3 } { 4 } y + \dfrac { 5 } { 4 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 4 } } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 5 } { 4 } }$
$x = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 4 } } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) + \dfrac { 5 } { 4 }$
 Calculate the product of rational numbers 
$x = \color{#FF6800}{ \dfrac { 3 } { 4 } } + \dfrac { 5 } { 4 }$
$x = \color{#FF6800}{ \dfrac { 3 } { 4 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 5 } { 4 } }$
 Find the sum of the fractions 
$x = \color{#FF6800}{ 2 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ 2 }$
 The possible solutions are as follows 
$\color{#FF6800}{ x } = \color{#FF6800}{ 2 } , \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ 1 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ 2 } , \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ 1 }$
 Check if it is the solution to the system of equations 
$\begin{cases} \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) = \color{#FF6800}{ 5 } \\ \color{#FF6800}{ 3 } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) = \color{#FF6800}{ 10 } \end{cases}$
$\begin{cases} \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) = \color{#FF6800}{ 5 } \\ \color{#FF6800}{ 3 } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) = \color{#FF6800}{ 10 } \end{cases}$
 Simplify the equality 
$\begin{cases} \color{#FF6800}{ 5 } = \color{#FF6800}{ 5 } \\ \color{#FF6800}{ 10 } = \color{#FF6800}{ 10 } \end{cases}$
$\begin{cases} \color{#FF6800}{ 5 } = \color{#FF6800}{ 5 } \\ \color{#FF6800}{ 10 } = \color{#FF6800}{ 10 } \end{cases}$
 Since it is true in both equations, it is the solution of the system of equations 
$\color{#FF6800}{ x } = \color{#FF6800}{ 2 } , \color{#FF6800}{ y } = \color{#FF6800}{ - } \color{#FF6800}{ 1 }$
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