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Formula
Solve the equation
Answer
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$y = \dfrac { 2 x + 3 } { 6 } - \dfrac { 3 } { 4 } x$
$y = \dfrac { 1 } { 4 }$
$x$-intercept
$\left ( \dfrac { 6 } { 5 } , 0 \right )$
$y$-intercept
$\left ( 0 , \dfrac { 1 } { 2 } \right )$
$\dfrac{ 2x+3 }{ 6 } - \dfrac{ 3 }{ 4 } x = \dfrac{ 1 }{ 4 }$
$x = \dfrac { 3 } { 5 }$
$ $ Solve a solution to $ x$
$\dfrac { 2 x + 3 } { 6 } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 4 } } \color{#FF6800}{ x } = \dfrac { 1 } { 4 }$
$ $ Calculate the multiplication expression $ $
$\dfrac { 2 x + 3 } { 6 } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 x } { 4 } } = \dfrac { 1 } { 4 }$
$\color{#FF6800}{ \dfrac { 2 x + 3 } { 6 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 x } { 4 } } = \color{#FF6800}{ \dfrac { 1 } { 4 } }$
$ $ Multiply both sides by the least common multiple for the denominators to eliminate the fraction $ $
$\color{#FF6800}{ 2 } \left ( \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right ) \color{#FF6800}{ - } \left ( \color{#FF6800}{ 3 } \left ( \color{#FF6800}{ 3 } \color{#FF6800}{ x } \right ) \right ) = \color{#FF6800}{ 3 }$
$\color{#FF6800}{ 2 } \left ( \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right ) \color{#FF6800}{ - } \left ( \color{#FF6800}{ 3 } \left ( \color{#FF6800}{ 3 } \color{#FF6800}{ x } \right ) \right ) = \color{#FF6800}{ 3 }$
$ $ Organize the expression $ $
$\color{#FF6800}{ 4 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 6 } \color{#FF6800}{ - } \left ( \color{#FF6800}{ 3 } \left ( \color{#FF6800}{ 3 } \color{#FF6800}{ x } \right ) \right ) = \color{#FF6800}{ 3 }$
$4 x + 6 - \left ( \color{#FF6800}{ 3 } \left ( \color{#FF6800}{ 3 } \color{#FF6800}{ x } \right ) \right ) = 3$
$ $ Get rid of unnecessary parentheses $ $
$4 x + 6 - \left ( \color{#FF6800}{ 3 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } \color{#FF6800}{ x } \right ) = 3$
$\color{#FF6800}{ 4 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 6 } \color{#FF6800}{ - } \left ( \color{#FF6800}{ 3 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } \color{#FF6800}{ x } \right ) = \color{#FF6800}{ 3 }$
$ $ Organize the expression $ $
$\color{#FF6800}{ 5 } \color{#FF6800}{ x } = \color{#FF6800}{ 3 }$
$\color{#FF6800}{ 5 } \color{#FF6800}{ x } = \color{#FF6800}{ 3 }$
$ $ Divide both sides by the same number $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 3 } { 5 } }$
$ $ 그래프 보기 $ $
Graph
Solution search results
search-thumbnail-$.$ $\left(m1\right)$ $\dfrac {3} {4}x=27$ 
$\left(ix\right)$ $7x-13=12-5x-5$ 
$\left(xi\right)\dfrac {2x+3} {x+3}=\dfrac {3} {2}$
10th-13th grade
Other
search-thumbnail-$11.$ Question $11$ 
Solve the $:$ $folloMlng'$ $0<θ<90^{°}$ 
$\left(1\right)$ $2sin^{2}θ=1\right)$ $\left(rac\left(3\right)\left(2\right)\right)$ 
$\left(11\right)$ $3tan^{2}θ+2=3$ 
$\left(111\right)cos^{2}θ$ $11rac\left(1\right)\left(4\right)\right)=$ 
$c\left(1\right)\left(4\right)\right)=11113c\left(1\right)\left(2\right)\right)$
10th-13th grade
Trigonometry
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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