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Formula
Solve the equation
Answer
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Graph
$y = \dfrac { 1 } { 6 } x + \dfrac { 1 } { 3 }$
$y = \dfrac { 1 } { 2 }$
$x$-intercept
$\left ( - 2 , 0 \right )$
$y$-intercept
$\left ( 0 , \dfrac { 1 } { 3 } \right )$
$\dfrac{ 1 }{ 6 } \times x+ \dfrac{ 1 }{ 3 } = \dfrac{ 1 }{ 2 }$
$x = 1$
$ $ Solve a solution to $ x$
$\color{#FF6800}{ \dfrac { 1 } { 6 } } \color{#FF6800}{ x } + \dfrac { 1 } { 3 } = \dfrac { 1 } { 2 }$
$ $ Calculate the multiplication expression $ $
$\color{#FF6800}{ \dfrac { x } { 6 } } + \dfrac { 1 } { 3 } = \dfrac { 1 } { 2 }$
$\color{#FF6800}{ \dfrac { x } { 6 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 1 } { 3 } } = \color{#FF6800}{ \dfrac { 1 } { 2 } }$
$ $ Multiply both sides by the least common multiple for the denominators to eliminate the fraction $ $
$\color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } = \color{#FF6800}{ 3 }$
$x \color{#FF6800}{ + } \color{#FF6800}{ 2 } = 3$
$ $ Move the constant to the right side and change the sign $ $
$x = 3 \color{#FF6800}{ - } \color{#FF6800}{ 2 }$
$x = \color{#FF6800}{ 3 } \color{#FF6800}{ - } \color{#FF6800}{ 2 }$
$ $ Subtract $ 2 $ from $ 3$
$x = \color{#FF6800}{ 1 }$
$ $ 그래프 보기 $ $
Graph
Solution search results
search-thumbnail-Question $4$ 
If $x$ is $6$ what $|s$ 
$\dfrac {1} {3}x$ 
$1frac\left(1\right)\left(3\right)x$
1st-6th grade
Other
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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