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Solve the equation
Answer
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Graph
$y = \dfrac { 2 } { 3 } \left ( x - 3 \right ) - \dfrac { 1 } { 7 } \left ( x + 1 \right )$
$y = 1$
$x$Intercept
$\left ( \dfrac { 45 } { 11 } , 0 \right )$
$y$Intercept
$\left ( 0 , - \dfrac { 15 } { 7 } \right )$
$\dfrac{ 2 }{ 3 } \left( x-3 \right) - \dfrac{ 1 }{ 7 } \left( x+1 \right) = 1$
$x = 6$
$ $ Solve a solution to $ x$
$\color{#FF6800}{ \dfrac { 2 } { 3 } } \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right ) \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 7 } } \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right ) = \color{#FF6800}{ 1 }$
$ $ Organize the expression $ $
$\color{#FF6800}{ \dfrac { 2 } { 3 } } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 2 } { 3 } } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right ) \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 7 } } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 7 } } = \color{#FF6800}{ 1 }$
$\color{#FF6800}{ \dfrac { 2 } { 3 } } \color{#FF6800}{ x } + \dfrac { 2 } { 3 } \times \left ( - 3 \right ) - \dfrac { 1 } { 7 } x - \dfrac { 1 } { 7 } = 1$
$ $ Calculate the multiplication expression $ $
$\color{#FF6800}{ \dfrac { 2 x } { 3 } } + \dfrac { 2 } { 3 } \times \left ( - 3 \right ) - \dfrac { 1 } { 7 } x - \dfrac { 1 } { 7 } = 1$
$\dfrac { 2 x } { 3 } + \color{#FF6800}{ \dfrac { 2 } { 3 } } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right ) - \dfrac { 1 } { 7 } x - \dfrac { 1 } { 7 } = 1$
$ $ Calculate the product of rational numbers $ $
$\dfrac { 2 x } { 3 } \color{#FF6800}{ - } \color{#FF6800}{ 2 } - \dfrac { 1 } { 7 } x - \dfrac { 1 } { 7 } = 1$
$\dfrac { 2 x } { 3 } - 2 \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 7 } } \color{#FF6800}{ x } - \dfrac { 1 } { 7 } = 1$
$ $ Calculate the multiplication expression $ $
$\dfrac { 2 x } { 3 } - 2 \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { x } { 7 } } - \dfrac { 1 } { 7 } = 1$
$\color{#FF6800}{ \dfrac { 2 x } { 3 } } - 2 \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { x } { 7 } } - \dfrac { 1 } { 7 } = 1$
$ $ Write all numerators above the least common denominator $ $
$\color{#FF6800}{ \dfrac { 14 x - 3 x } { 21 } } - 2 - \dfrac { 1 } { 7 } = 1$
$\dfrac { 14 x - 3 x } { 21 } \color{#FF6800}{ - } \color{#FF6800}{ 2 } - \dfrac { 1 } { 7 } = 1$
$ $ Convert an equation to a fraction using $ a=\dfrac{a}{1}$
$\dfrac { 14 x - 3 x } { 21 } + \color{#FF6800}{ \dfrac { - 2 } { 1 } } - \dfrac { 1 } { 7 } = 1$
$\color{#FF6800}{ \dfrac { 14 x - 3 x } { 21 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { - 2 } { 1 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 7 } } = 1$
$ $ Write all numerators above the least common denominator $ $
$\color{#FF6800}{ \dfrac { 14 x - 3 x - 42 - 3 } { 21 } } = 1$
$\color{#FF6800}{ \dfrac { 14 x - 3 x - 42 - 3 } { 21 } } = \color{#FF6800}{ 1 }$
$ $ Organize the expression $ $
$\color{#FF6800}{ \dfrac { 11 x - 45 } { 21 } } = \color{#FF6800}{ 1 }$
$\color{#FF6800}{ \dfrac { 11 x - 45 } { 21 } } = \color{#FF6800}{ 1 }$
$ $ Multiply both sides by the least common multiple for the denominators to eliminate the fraction $ $
$\color{#FF6800}{ 11 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 45 } = \color{#FF6800}{ 21 }$
$\color{#FF6800}{ 11 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 45 } = \color{#FF6800}{ 21 }$
$ $ Organize the expression $ $
$\color{#FF6800}{ 11 } \color{#FF6800}{ x } = \color{#FF6800}{ 66 }$
$\color{#FF6800}{ 11 } \color{#FF6800}{ x } = \color{#FF6800}{ 66 }$
$ $ Divide both sides by the same number $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ 6 }$
$x = 6$
Solve the fractional equation
$\color{#FF6800}{ \dfrac { 2 } { 3 } } \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right ) \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 7 } } \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right ) = \color{#FF6800}{ 1 }$
$ $ Solve a solution to $ x$
$\color{#FF6800}{ x } = \color{#FF6800}{ 6 }$
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