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Solve the equation

Answer

Graph

$y = \dfrac { x } { 2 } - \dfrac { 2 - x } { 6 }$

$y = 7$

$x$Intercept

$\left ( \dfrac { 1 } { 2 } , 0 \right )$

$y$Intercept

$\left ( 0 , - \dfrac { 1 } { 3 } \right )$

$x = 11$

$ $ Solve a solution to $ x$

$\dfrac { x } { 2 } - \dfrac { \color{#FF6800}{ 2 } \color{#FF6800}{ - } \color{#FF6800}{ x } } { 6 } = 7$

$ $ Organize the expression $ $

$\dfrac { x } { 2 } - \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } } { 6 } = 7$

$\color{#FF6800}{ \dfrac { \color{#FF6800}{ x } } { \color{#FF6800}{ 2 } } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } } { \color{#FF6800}{ 6 } } } = \color{#FF6800}{ 7 }$

$ $ Multiply both sides by the least common multiple for the denominators to eliminate the fraction $ $

$\color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ - } \left ( \color{#FF6800}{ - } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \right ) = \color{#FF6800}{ 42 }$

$3 x \color{#FF6800}{ - } \left ( \color{#FF6800}{ - } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \right ) = 42$

$ $ Change the symbol of each term in parentheses when there is a (-) symbol in front of parentheses $ $

$3 x + \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 } = 42$

$\color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ x } - 2 = 42$

$ $ Calculate between similar terms $ $

$\color{#FF6800}{ 4 } \color{#FF6800}{ x } - 2 = 42$

$4 x \color{#FF6800}{ - } \color{#FF6800}{ 2 } = 42$

$ $ Move the constant to the right side and change the sign $ $

$4 x = 42 \color{#FF6800}{ + } \color{#FF6800}{ 2 }$

$4 x = \color{#FF6800}{ 42 } \color{#FF6800}{ + } \color{#FF6800}{ 2 }$

$ $ Add $ 42 $ and $ 2$

$4 x = \color{#FF6800}{ 44 }$

$\color{#FF6800}{ 4 } \color{#FF6800}{ x } = \color{#FF6800}{ 44 }$

$ $ Divide both sides by the same number $ $

$\color{#FF6800}{ x } = \color{#FF6800}{ 11 }$

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