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Solve the inequality
Answer
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$\dfrac { 4 - x } { 2 } \leq 3 - \dfrac { x } { 4 }$
$\dfrac { 4 - x } { 2 } \leq 3 - \dfrac { x } { 4 }$
Solution of inequality
$x \geq - 4$
$\dfrac{ 4-x }{ 2 } \leq 3- \dfrac{ x }{ 4 }$
$x \geq - 4$
$ $ Solve a solution to $ x$
$\dfrac { \color{#FF6800}{ 4 } \color{#FF6800}{ - } \color{#FF6800}{ x } } { 2 } \leq 3 - \dfrac { x } { 4 }$
$ $ Organize the expression $ $
$\dfrac { \color{#FF6800}{ - } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 4 } } { 2 } \leq 3 - \dfrac { x } { 4 }$
$\color{#FF6800}{ \dfrac { - x + 4 } { 2 } } \leq \color{#FF6800}{ 3 } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { x } { 4 } }$
$ $ Multiply both sides by the least common multiple for the denominators to eliminate the fraction $ $
$\color{#FF6800}{ 2 } \left ( \color{#FF6800}{ - } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 4 } \right ) \leq \color{#FF6800}{ - } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 12 }$
$\color{#FF6800}{ 2 } \left ( \color{#FF6800}{ - } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 4 } \right ) \leq - x + 12$
$ $ Multiply each term in parentheses by $ 2$
$\color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 4 } \leq - x + 12$
$- 2 x + \color{#FF6800}{ 2 } \color{#FF6800}{ \times } \color{#FF6800}{ 4 } \leq - x + 12$
$ $ Multiply $ 2 $ and $ 4$
$- 2 x + \color{#FF6800}{ 8 } \leq - x + 12$
$- 2 x + 8 \leq \color{#FF6800}{ - } \color{#FF6800}{ x } + 12$
$ $ Move the variable to the left-hand side and change the symbol $ $
$- 2 x + 8 \color{#FF6800}{ + } \color{#FF6800}{ x } \leq 12$
$- 2 x \color{#FF6800}{ + } \color{#FF6800}{ 8 } + x \leq 12$
$ $ Move the constant to the right side and change the sign $ $
$- 2 x + x \leq 12 \color{#FF6800}{ - } \color{#FF6800}{ 8 }$
$\color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ x } \leq 12 - 8$
$ $ Organize the expression $ $
$\color{#FF6800}{ - } \color{#FF6800}{ x } \leq 12 - 8$
$- x \leq \color{#FF6800}{ 12 } \color{#FF6800}{ - } \color{#FF6800}{ 8 }$
$ $ Subtract $ 8 $ from $ 12$
$- x \leq \color{#FF6800}{ 4 }$
$\color{#FF6800}{ - } \color{#FF6800}{ x } \leq \color{#FF6800}{ 4 }$
$ $ Change the symbol of the inequality of both sides, and reverse the symbol of the inequality to the opposite direction $ $
$\color{#FF6800}{ x } \geq \color{#FF6800}{ - } \color{#FF6800}{ 4 }$
$ $ 그래프 보기 $ $
Inequality
Solution search results
search-thumbnail-$2$ Indicate the integral part of the numbersx which satisfy the system 
of inequalities. $ \begin{cases} \dfrac {x-1} {2}-\dfrac {2x+3} {3}+\dfrac {x} {6}<2-\dfrac {x+5} {2} \\ 1-\dfrac {x+5} {8}+\dfrac {4-x} {2}<3x-\dfrac {x+1} {4} \end{cases} $
10th-13th grade
Other
search-thumbnail-$\dfrac {2-x} {2}+\dfrac {4-x} {4}$ $=$ $6-\times $ $6$
7th-9th grade
Other
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