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Formula
Find the sum or difference of the fractions
Answer
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$- \dfrac{ 1 }{ 2 } - \dfrac{ 1 }{ 4 } + \dfrac{ 2 }{ 3 }$
$- \dfrac { 1 } { 12 }$
Find the sum or difference of the fractions
$- \dfrac { 1 } { \color{#FF6800}{ 2 } } - \dfrac { 1 } { \color{#FF6800}{ 4 } } + \dfrac { 2 } { \color{#FF6800}{ 3 } }$
$ $ The smallest common multiple in denominator is $ 12$
$- \dfrac { 1 } { \color{#FF6800}{ 2 } } - \dfrac { 1 } { \color{#FF6800}{ 4 } } + \dfrac { 2 } { \color{#FF6800}{ 3 } }$
$- \dfrac { 1 } { 2 } - \dfrac { 1 } { 4 } + \dfrac { 2 } { 3 }$
$ $ Multiply the denominator and the numerator so that the denominator is the smallest common multiple $ $
$- \dfrac { 1 \times \color{#FF6800}{ 6 } } { 2 \times \color{#FF6800}{ 6 } } - \dfrac { 1 \times \color{#FF6800}{ 3 } } { 4 \times \color{#FF6800}{ 3 } } + \dfrac { 2 \times \color{#FF6800}{ 4 } } { 3 \times \color{#FF6800}{ 4 } }$
$\color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 \times 6 } { 2 \times 6 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 \times 3 } { 4 \times 3 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 2 \times 4 } { 3 \times 4 } }$
$ $ Organize the expression $ $
$\color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 6 } { 12 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 12 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 8 } { 12 } }$
$\color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 6 } { 12 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 12 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 8 } { 12 } }$
$ $ Since the denominator is the same as $ 12 $ , combine the fractions into one $ $
$\color{#FF6800}{ \dfrac { - 6 - 3 + 8 } { 12 } }$
$\dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 6 } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ + } \color{#FF6800}{ 8 } } { 12 }$
$ $ Calculate the sum or the difference $ $
$\dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 1 } } { 12 }$
$\dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 1 } } { 12 }$
$ $ Move the minus sign to the front of the fraction $ $
$\color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 12 } }$
Solution search results
search-thumbnail-b. $5\dfrac {1} {2}$ $-2\dfrac {1} {3}- \begin{cases} - \\ 3\dfrac {1} {4}-\left(2\dfrac {1} {6}-4\dfrac {1} {3}+1\dfrac {2} {3}\right) \end{cases} $ 
d. $\dfrac {3} {4}+\left(\dfrac {3} {4}+\dfrac {3} {4}+\left(\dfrac {3} {4}+\dfrac {3} {4}\right)\right)$
7th-9th grade
Other
search-thumbnail-$.$ $5\dfrac {1} {2}-|2\dfrac {1} {3}- \begin{cases} - \\ 3\dfrac {1} {4}-\left(2\dfrac {1} {6}-4\dfrac {1} {3}+1\dfrac {2} {3}\right)\right) \end{cases} $
7th-9th grade
Other
search-thumbnail-$15.7\dfrac {1} {2}-2\dfrac {1} {4}+ \begin{cases} - \\ 1\dfrac {1} {4}-\dfrac {1} {2}\left(\dfrac {3} {2}-\dfrac {1} {3}-\dfrac {1} {6}\right)\right) \end{cases} $
1st-6th grade
Other
search-thumbnail-$15$ $7\dfrac {1} {2}-2\dfrac {1} {4}+ \begin{cases} - \\ 1\dfrac {1} {4}-\dfrac {1} {2}\left(\dfrac {3} {2}-\dfrac {1} {3}-\dfrac {1} {6}\right)\right) \end{cases} $
1st-6th grade
Other
search-thumbnail-
$\left(1\right)$ $\dfrac {3} {4}$ $\dfrac {1} {2}|\dfrac {2} {3}-\dfrac {3} {4}+\left(\dfrac {1} {2}-\dfrac {2} {3}+\left(\dfrac {1} {4}=\dfrac {1} {5}\right)$
7th-9th grade
Algebra
search-thumbnail-$0$ 
$15.$ $7\dfrac {1} {2}-\left(2\dfrac {1} {4}+ \begin{cases} - \\ 1\dfrac {1} {4}-\dfrac {1} {2}\left(\dfrac {3} {2}-\dfrac {1} {3}-\dfrac {1} {6}\right)\right) \end{cases} $
1st-6th grade
Algebra
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