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Solve the equation
Answer
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$3w-3 = 4w+11$
$w = - 14$
$ $ Solve a solution to $ w$
$3 w - 3 = \color{#FF6800}{ 4 } \color{#FF6800}{ w } + 11$
$ $ Move the variable to the left-hand side and change the symbol $ $
$3 w - 3 \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ w } = 11$
$3 w \color{#FF6800}{ - } \color{#FF6800}{ 3 } - 4 w = 11$
$ $ Move the constant to the right side and change the sign $ $
$3 w - 4 w = 11 \color{#FF6800}{ + } \color{#FF6800}{ 3 }$
$\color{#FF6800}{ 3 } \color{#FF6800}{ w } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ w } = 11 + 3$
$ $ Organize the expression $ $
$\color{#FF6800}{ - } \color{#FF6800}{ w } = 11 + 3$
$- w = \color{#FF6800}{ 11 } \color{#FF6800}{ + } \color{#FF6800}{ 3 }$
$ $ Add $ 11 $ and $ 3$
$- w = \color{#FF6800}{ 14 }$
$\color{#FF6800}{ - } \color{#FF6800}{ w } = \color{#FF6800}{ 14 }$
$ $ Change the sign of both sides of the equation $ $
$\color{#FF6800}{ w } = \color{#FF6800}{ - } \color{#FF6800}{ 14 }$
Solution search results
search-thumbnail-$ \begin{cases} \dfrac {t-5} {3}+\dfrac {w+1} {2}=1 \\ \dfrac {t-1} {5}-\dfrac {+2} {4}=2 \end{cases} $ $\left($
10th-13th grade
Algebra
search-thumbnail-$h=\dfrac {4} {5}w-$ $1$ $\dfrac {12} {5}$ 
$0w=4w+12$ 
$①$ $w=\dfrac {5} {4}n-3$ 
$○w=4m+\dfrac {12} {5}$ 
$\square m=\dfrac {5} {4}h$ $12$
7th-9th grade
Algebra
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