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Formula
Solve the equation
Answer
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$y = \dfrac { x + 1 } { 3 } - \dfrac { x - 3 } { 2 }$
$y = 1$
$x$-intercept
$\left ( 11 , 0 \right )$
$y$-intercept
$\left ( 0 , \dfrac { 11 } { 6 } \right )$
$\dfrac{ x+1 }{ 3 } - \dfrac{ x-3 }{ 2 } = 1$
$x = 5$
$ $ Solve a solution to $ x$
$\color{#FF6800}{ \dfrac { x + 1 } { 3 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { x - 3 } { 2 } } = 1$
$ $ Write all numerators above the least common denominator $ $
$\color{#FF6800}{ \dfrac { 2 x + 2 - 3 x + 9 } { 6 } } = 1$
$\dfrac { \color{#FF6800}{ 2 } \color{#FF6800}{ x } + 2 \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ x } + 9 } { 6 } = 1$
$ $ Calculate between similar terms $ $
$\dfrac { \color{#FF6800}{ - } \color{#FF6800}{ x } + 2 + 9 } { 6 } = 1$
$\dfrac { - x + \color{#FF6800}{ 2 } \color{#FF6800}{ + } \color{#FF6800}{ 9 } } { 6 } = 1$
$ $ Add $ 2 $ and $ 9$
$\dfrac { - x + \color{#FF6800}{ 11 } } { 6 } = 1$
$\color{#FF6800}{ \dfrac { - x + 11 } { 6 } } = \color{#FF6800}{ 1 }$
$ $ Multiply both sides by the least common multiple for the denominators to eliminate the fraction $ $
$\color{#FF6800}{ - } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 11 } = \color{#FF6800}{ 6 }$
$- x \color{#FF6800}{ + } \color{#FF6800}{ 11 } = 6$
$ $ Move the constant to the right side and change the sign $ $
$- x = 6 \color{#FF6800}{ - } \color{#FF6800}{ 11 }$
$- x = \color{#FF6800}{ 6 } \color{#FF6800}{ - } \color{#FF6800}{ 11 }$
$ $ Subtract $ 11 $ from $ 6$
$- x = \color{#FF6800}{ - } \color{#FF6800}{ 5 }$
$\color{#FF6800}{ - } \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 5 }$
$ $ Change the sign of both sides of the equation $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ - } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right )$
$x = \color{#FF6800}{ - } \left ( \color{#FF6800}{ - } 5 \right )$
$ $ Simplify Minus $ $
$x = 5$
$ $ 그래프 보기 $ $
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Solution search results
search-thumbnail-$\dfrac {x+1} {4}+\dfrac {x+2} {2}=\dfrac {x-3} {2}$ $--$ $3$ $x-1$
7th-9th grade
Geometry
search-thumbnail-$\dfrac {x+1} {4}+\dfrac {x+2} {2}=\dfrac {x-3} {2}$ $-\dfrac {x-1} {3}$
7th-9th grade
Geometry
search-thumbnail-$4$ Solve the following equations. 
$\left(1\right)$ $\dfrac {x} {6}+3\left(x+1\right)=2$ $\left(1\right)$ ) $\dfrac {x-6} {6}+\dfrac {x-3} {3}=\dfrac {x-4} {4}$ $\left(11\right)\dfrac {x+1} {3}-\dfrac {x-6} {2}=\dfrac {x} {2}$ 
$\left(1v\right)$ $\dfrac {x-1} {2}+\dfrac {x+2} {3}-\dfrac {x-3} {4}=1$ $\left(v\right)$ $\dfrac {2x+7} {5}-\dfrac {x-3} {10}-\dfrac {x+1} {15}=0$ $\left(n\right)$ $\dfrac {3} {x}=\dfrac {1} {3}$ 
(vii) $1k\dfrac {12} {x}=4$ $\left(m1\right)$ $\dfrac {4x+1} {x-3}=2$
1st-6th grade
Other
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