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Formula
Calculate the value
Answer
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$\left( \dfrac{ 1 }{ 2 } x- \dfrac{ 2 }{ 3 } y \right) - \left( \dfrac{ 3 }{ 4 } x- \dfrac{ 1 }{ 6 } y \right)$
$- \dfrac { x + 2 y } { 4 }$
Arrange the rational expression
$\left ( \color{#FF6800}{ \dfrac { 1 } { 2 } } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 } { 3 } } \color{#FF6800}{ y } \right ) \color{#FF6800}{ - } \left ( \color{#FF6800}{ \dfrac { 3 } { 4 } } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 6 } } \color{#FF6800}{ y } \right )$
$ $ Get rid of unnecessary parentheses $ $
$\color{#FF6800}{ \dfrac { 1 } { 2 } } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 } { 3 } } \color{#FF6800}{ y } \color{#FF6800}{ - } \left ( \color{#FF6800}{ \dfrac { 3 } { 4 } } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 6 } } \color{#FF6800}{ y } \right )$
$\color{#FF6800}{ \dfrac { 1 } { 2 } } \color{#FF6800}{ x } - \dfrac { 2 } { 3 } y - \left ( \dfrac { 3 } { 4 } x - \dfrac { 1 } { 6 } y \right )$
$ $ Calculate the multiplication expression $ $
$\color{#FF6800}{ \dfrac { x } { 2 } } - \dfrac { 2 } { 3 } y - \left ( \dfrac { 3 } { 4 } x - \dfrac { 1 } { 6 } y \right )$
$\dfrac { x } { 2 } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 } { 3 } } \color{#FF6800}{ y } - \left ( \dfrac { 3 } { 4 } x - \dfrac { 1 } { 6 } y \right )$
$ $ Calculate the multiplication expression $ $
$\dfrac { x } { 2 } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 y } { 3 } } - \left ( \dfrac { 3 } { 4 } x - \dfrac { 1 } { 6 } y \right )$
$\dfrac { x } { 2 } - \dfrac { 2 y } { 3 } - \left ( \color{#FF6800}{ \dfrac { 3 } { 4 } } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 6 } } \color{#FF6800}{ y } \right )$
$ $ Calculate the expression as a fraction format $ $
$\dfrac { x } { 2 } - \dfrac { 2 y } { 3 } - \color{#FF6800}{ \dfrac { 9 x - 2 y } { 12 } }$
$\color{#FF6800}{ \dfrac { x } { 2 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 2 y } { 3 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 9 x - 2 y } { 12 } }$
$ $ Calculate the expression as a fraction format $ $
$\color{#FF6800}{ - } \color{#FF6800}{ \dfrac { x + 2 y } { 4 } }$
Solution search results
search-thumbnail-The is the statement of the Mean Value Theorem from your $te\times tb00k$ $Tneoren$ $m4.5$ $Ne8n$ Value Theorem Let f be continuous over $leclose$ interval $\left(a.b\right)aod$ $i$ $eo$ $xd$ $csnlc$ $\left(ab\right)$ $mmd$ exists at least one point cE $\left(a.b\right)$ $si1dhtba$ $f^{'}\left(c\right)=\dfrac {f\left(b\right)-f\left(a\right)} {b-a}$ What is the geometric interpretation of the conclusion of the theorem? O The tangent line to the graph of \(f\left(x\right)\) at $l\left(c1\right)$ is parallel to the secant line connecting \(\left(a,f\left(a\right)\right)\) and \(\left(b,f\left(b\right)\right)\). O The tangent line to the graph of $\left(f\left(lef\left(x\left(right\right)$ at \(c\) is the secant line connecting \(\left(a,f\left(a\right)\right)\) $ana$ \(\left(b,f\left(b\right)\right)\). O The tangent line to the graph $of$ $\left(f\left(leF\left(x\left(right\right)\left(\right)at1\left(c1\right)is$ perpendicular to the secant line connecting \(\left(a,f\left(a\right)\right)\) and A(\left(b,f\left(b\right)\right)\). O The tangent line to the graph of $\left(fleH\left(x\right)nght\right)\right)\right)$ $at$ $\left(c\right)\right)ishoizonta|$ $Tnere$ is more than one tangent line to the graph $0no$ $\left(flcf\left(x\right)night\right)\left($ $at1\left(c1\right)$
Calculus
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search-thumbnail- $\left(i\right)$ $\dfrac {3} {4}x-\dfrac {2} {3}y=1$ $\dfrac {3} {8}x-\dfrac {1} {6}y$ $=1$
7th-9th grade
Algebra
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search-thumbnail-$2$ $\left(1\right)$ $\dfrac {3} {4}x-\dfrac {2} {3}y=1$ $\dfrac {3} {8}x-\dfrac {1} {6}y=1$ 1
English
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search-thumbnail-Which of the following rational numbers are equivalent? $0Ptionsy$ A \frac{5}{6}, \frac{30}{36} B $s\sqrt{rac\left(} -2\right)\left(3\right)\sqrt{1rac} \sqrt{4\right)16\right)4} $ C $s\sqrt{11aC\left(} -4\right)1-7b,\sqrt{1rac\left(16\sqrt{35\right)9} } $ D \frac{1}{2},\frac{3}{8}
7th-9th grade
Other
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