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Formula
Solve the equation
Answer
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Graph
$y = \dfrac { 4 } { 3 } \left ( x - 3 \right )$
$y = \dfrac { 3 } { 2 } - \dfrac { 1 - x } { 2 }$
$x$-intercept
$\left ( 3 , 0 \right )$
$y$-intercept
$\left ( 0 , - 4 \right )$
$x$-intercept
$\left ( - 2 , 0 \right )$
$y$-intercept
$\left ( 0 , 1 \right )$
$\dfrac{ 4 }{ 3 } \left( x-3 \right) = \dfrac{ 3 }{ 2 } - \dfrac{ 1-x }{ 2 }$
$x = 6$
$ $ Solve a solution to $ x$
$\color{#FF6800}{ \dfrac { 4 } { 3 } } \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right ) = \dfrac { 3 } { 2 } - \dfrac { 1 - x } { 2 }$
$ $ Multiply each term in parentheses by $ \dfrac { 4 } { 3 }$
$\color{#FF6800}{ \dfrac { 4 } { 3 } } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 4 } { 3 } } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right ) = \dfrac { 3 } { 2 } - \dfrac { 1 - x } { 2 }$
$\color{#FF6800}{ \dfrac { 4 } { 3 } } \color{#FF6800}{ x } + \dfrac { 4 } { 3 } \times \left ( - 3 \right ) = \dfrac { 3 } { 2 } - \dfrac { 1 - x } { 2 }$
$ $ Calculate the multiplication expression $ $
$\color{#FF6800}{ \dfrac { 4 x } { 3 } } + \dfrac { 4 } { 3 } \times \left ( - 3 \right ) = \dfrac { 3 } { 2 } - \dfrac { 1 - x } { 2 }$
$\dfrac { 4 x } { 3 } + \color{#FF6800}{ \dfrac { 4 } { 3 } } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right ) = \dfrac { 3 } { 2 } - \dfrac { 1 - x } { 2 }$
$ $ Calculate the product of rational numbers $ $
$\dfrac { 4 x } { 3 } \color{#FF6800}{ - } \color{#FF6800}{ 4 } = \dfrac { 3 } { 2 } - \dfrac { 1 - x } { 2 }$
$\color{#FF6800}{ \dfrac { 4 x } { 3 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } = \color{#FF6800}{ \dfrac { 3 } { 2 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 - x } { 2 } }$
$ $ Organize the expression $ $
$\color{#FF6800}{ \dfrac { 4 x } { 3 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } = \color{#FF6800}{ \dfrac { 2 + x } { 2 } }$
$\dfrac { 4 x } { 3 } - 4 = \dfrac { \color{#FF6800}{ 2 } \color{#FF6800}{ + } \color{#FF6800}{ x } } { 2 }$
$ $ Organize the expression $ $
$\dfrac { 4 x } { 3 } - 4 = \dfrac { \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } } { 2 }$
$\color{#FF6800}{ \dfrac { 4 x } { 3 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } = \color{#FF6800}{ \dfrac { x + 2 } { 2 } }$
$ $ Multiply both sides by the least common multiple for the denominators to eliminate the fraction $ $
$\color{#FF6800}{ 2 } \left ( \color{#FF6800}{ 4 } \color{#FF6800}{ x } \right ) \color{#FF6800}{ - } \color{#FF6800}{ 24 } = \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 6 }$
$\color{#FF6800}{ 2 } \left ( \color{#FF6800}{ 4 } \color{#FF6800}{ x } \right ) - 24 = 3 x + 6$
$ $ Get rid of unnecessary parentheses $ $
$\color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 4 } \color{#FF6800}{ x } - 24 = 3 x + 6$
$\color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 4 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 24 } = \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 6 }$
$ $ Organize the expression $ $
$\color{#FF6800}{ 8 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ x } = \color{#FF6800}{ 6 } \color{#FF6800}{ + } \color{#FF6800}{ 24 }$
$\color{#FF6800}{ 8 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ x } = 6 + 24$
$ $ Organize the expression $ $
$\color{#FF6800}{ 5 } \color{#FF6800}{ x } = 6 + 24$
$5 x = \color{#FF6800}{ 6 } \color{#FF6800}{ + } \color{#FF6800}{ 24 }$
$ $ Add $ 6 $ and $ 24$
$5 x = \color{#FF6800}{ 30 }$
$\color{#FF6800}{ 5 } \color{#FF6800}{ x } = \color{#FF6800}{ 30 }$
$ $ Divide both sides by the same number $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ 6 }$
$ $ 그래프 보기 $ $
Graph
Solution search results
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Calculus
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Calculus
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search-thumbnail-What is the inverse of $g\left(x\right)=\sqrt{x-1} $ $Og\left(x\right)=\sqrt{x+1} $ $+1$ $g^{'}\left(x\right)=5-$ $y_{9}$ $g\sqrt{ef\left(x\left(nght\right)} =\left(f_{tac\left(x+1\right)0sq\pi \left(x\right)}$ $Og^{'}\left(x\right)=x^{2}+1$ $○g\left(x\right)=x^{2}-1$ « Previous Ne
Other
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search-thumbnail-What is the inverse of $g\left(x\right)=\sqrt{x-1} $ $Og\left(x\right)=\sqrt{x+1} $ $+1$ $g^{'}\left(x\right)=5-$ $y_{9}$ $g\sqrt{ef\left(x\left(nght\right)} =\left(f_{tac\left(x+1\right)0sq\pi \left(x\right)}$ $Og^{'}\left(x\right)=x^{2}+1$ $○g\left(x\right)=x^{2}-1$ « Previous Ne
Other
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Trigonometry
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Other
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