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Solve the system of equations
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$0.3 x + 0.2 y = 1.1$
$\dfrac { x - 1 } { 2 } + \dfrac { y + 1 } { 6 } = \dfrac { 1 } { 3 }$
$x$Intercept
$\left ( \dfrac { 11 } { 3 } , 0 \right )$
$y$Intercept
$\left ( 0 , \dfrac { 11 } { 2 } \right )$
$x$Intercept
$\left ( \dfrac { 4 } { 3 } , 0 \right )$
$y$Intercept
$\left ( 0 , 4 \right )$
$\begin{cases} 0.3x+0.2y = 1.1 \\ \dfrac{ x-1 }{ 2 } + \dfrac{ y+1 }{ 6 } = \dfrac{ 1 }{ 3 } \end{cases}$
$x = - 1 , y = 7$
Solve the system of equations
$\begin{cases} \color{#FF6800}{ 0.3 } \color{#FF6800}{ x } + 0.2 y = 1.1 \\ \dfrac { x - 1 } { 2 } + \dfrac { y + 1 } { 6 } = \dfrac { 1 } { 3 } \end{cases}$
 Calculate the multiplication expression 
$\begin{cases} \color{#FF6800}{ \dfrac { 3 x } { 10 } } + 0.2 y = 1.1 \\ \dfrac { x - 1 } { 2 } + \dfrac { y + 1 } { 6 } = \dfrac { 1 } { 3 } \end{cases}$
$\begin{cases} \dfrac { 3 x } { 10 } + \color{#FF6800}{ 0.2 } \color{#FF6800}{ y } = 1.1 \\ \dfrac { x - 1 } { 2 } + \dfrac { y + 1 } { 6 } = \dfrac { 1 } { 3 } \end{cases}$
 Calculate the multiplication expression 
$\begin{cases} \dfrac { 3 x } { 10 } + \color{#FF6800}{ \dfrac { y } { 5 } } = 1.1 \\ \dfrac { x - 1 } { 2 } + \dfrac { y + 1 } { 6 } = \dfrac { 1 } { 3 } \end{cases}$
$\begin{cases} \dfrac { 3 x } { 10 } + \dfrac { y } { 5 } = \color{#FF6800}{ 1.1 } \\ \dfrac { x - 1 } { 2 } + \dfrac { y + 1 } { 6 } = \dfrac { 1 } { 3 } \end{cases}$
 Convert decimals to fractions 
$\begin{cases} \dfrac { 3 x } { 10 } + \dfrac { y } { 5 } = \color{#FF6800}{ \dfrac { 11 } { 10 } } \\ \dfrac { x - 1 } { 2 } + \dfrac { y + 1 } { 6 } = \dfrac { 1 } { 3 } \end{cases}$
$\begin{cases} \dfrac { 3 x } { 10 } + \dfrac { y } { 5 } = \dfrac { 11 } { 10 } \\ \color{#FF6800}{ \dfrac { x - 1 } { 2 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { y + 1 } { 6 } } = \dfrac { 1 } { 3 } \end{cases}$
 Write all numerators above the least common denominator 
$\begin{cases} \dfrac { 3 x } { 10 } + \dfrac { y } { 5 } = \dfrac { 11 } { 10 } \\ \color{#FF6800}{ \dfrac { 3 x - 3 + y + 1 } { 6 } } = \dfrac { 1 } { 3 } \end{cases}$
$\begin{cases} \dfrac { 3 x } { 10 } + \dfrac { y } { 5 } = \dfrac { 11 } { 10 } \\ \dfrac { 3 x \color{#FF6800}{ - } \color{#FF6800}{ 3 } + y \color{#FF6800}{ + } \color{#FF6800}{ 1 } } { 6 } = \dfrac { 1 } { 3 } \end{cases}$
 Add $- 3$ and $1$
$\begin{cases} \dfrac { 3 x } { 10 } + \dfrac { y } { 5 } = \dfrac { 11 } { 10 } \\ \dfrac { 3 x \color{#FF6800}{ - } \color{#FF6800}{ 2 } + y } { 6 } = \dfrac { 1 } { 3 } \end{cases}$
$\begin{cases} \dfrac { 3 x } { 10 } + \dfrac { y } { 5 } = \dfrac { 11 } { 10 } \\ \dfrac { 3 x - 2 + y } { 6 } = \dfrac { 1 } { 3 } \end{cases}$
 Solve a solution to $x$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 } { 3 } } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 11 } { 3 } } \\ \dfrac { 3 x - 2 + y } { 6 } = \dfrac { 1 } { 3 } \end{cases}$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 } { 3 } } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 11 } { 3 } } \\ \color{#FF6800}{ \dfrac { 3 x - 2 + y } { 6 } } = \color{#FF6800}{ \dfrac { 1 } { 3 } } \end{cases}$
 Substitute the given $x$ value into the equation $\dfrac { 3 x - 2 + y } { 6 } = \dfrac { 1 } { 3 }$
$\color{#FF6800}{ \dfrac { 3 \left ( - \dfrac { 2 } { 3 } y + \dfrac { 11 } { 3 } \right ) - 2 + y } { 6 } } = \color{#FF6800}{ \dfrac { 1 } { 3 } }$
$\color{#FF6800}{ \dfrac { 3 \left ( - \dfrac { 2 } { 3 } y + \dfrac { 11 } { 3 } \right ) - 2 + y } { 6 } } = \color{#FF6800}{ \dfrac { 1 } { 3 } }$
 Solve a solution to $y$
$\color{#FF6800}{ y } = \color{#FF6800}{ 7 }$
$\color{#FF6800}{ y } = \color{#FF6800}{ 7 }$
 Substitute the given $y$ value into the equation $x = - \dfrac { 2 } { 3 } y + \dfrac { 11 } { 3 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 } { 3 } } \color{#FF6800}{ \times } \color{#FF6800}{ 7 } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 11 } { 3 } }$
$x = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 } { 3 } } \color{#FF6800}{ \times } \color{#FF6800}{ 7 } + \dfrac { 11 } { 3 }$
 Calculate the product of rational numbers 
$x = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 14 } { 3 } } + \dfrac { 11 } { 3 }$
$x = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 14 } { 3 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 11 } { 3 } }$
 Find the sum or difference of the fractions 
$x = \color{#FF6800}{ - } \color{#FF6800}{ 1 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 1 }$
 The possible solutions are as follows 
$\color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 1 } , \color{#FF6800}{ y } = \color{#FF6800}{ 7 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 1 } , \color{#FF6800}{ y } = \color{#FF6800}{ 7 }$
 Check if it is the solution to the system of equations 
$\begin{cases} \color{#FF6800}{ 0.3 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) \color{#FF6800}{ + } \color{#FF6800}{ 0.2 } \color{#FF6800}{ \times } \color{#FF6800}{ 7 } = \color{#FF6800}{ 1.1 } \\ \color{#FF6800}{ \dfrac { - 1 - 1 } { 2 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 7 + 1 } { 6 } } = \color{#FF6800}{ \dfrac { 1 } { 3 } } \end{cases}$
$\begin{cases} \color{#FF6800}{ 0.3 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) \color{#FF6800}{ + } \color{#FF6800}{ 0.2 } \color{#FF6800}{ \times } \color{#FF6800}{ 7 } = \color{#FF6800}{ 1.1 } \\ \color{#FF6800}{ \dfrac { - 1 - 1 } { 2 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 7 + 1 } { 6 } } = \color{#FF6800}{ \dfrac { 1 } { 3 } } \end{cases}$
 Simplify the equality 
$\begin{cases} \color{#FF6800}{ \dfrac { 11 } { 10 } } = \color{#FF6800}{ \dfrac { 11 } { 10 } } \\ \color{#FF6800}{ \dfrac { 1 } { 3 } } = \color{#FF6800}{ \dfrac { 1 } { 3 } } \end{cases}$
$\begin{cases} \color{#FF6800}{ \dfrac { 11 } { 10 } } = \color{#FF6800}{ \dfrac { 11 } { 10 } } \\ \color{#FF6800}{ \dfrac { 1 } { 3 } } = \color{#FF6800}{ \dfrac { 1 } { 3 } } \end{cases}$
 Since it is true in both equations, it is the solution of the system of equations 
$\color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 1 } , \color{#FF6800}{ y } = \color{#FF6800}{ 7 }$
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