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Formula
Solve the equation
Answer
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Graph
$y = 3 x - 40$
$y = - x$
$x$-intercept
$\left ( \dfrac { 40 } { 3 } , 0 \right )$
$y$-intercept
$\left ( 0 , - 40 \right )$
$x$-intercept
$\left ( 0 , 0 \right )$
$y$-intercept
$\left ( 0 , 0 \right )$
$3x-40 = -x$
$x = 10$
$ $ Solve a solution to $ x$
$3 x - 40 = \color{#FF6800}{ - } \color{#FF6800}{ x }$
$ $ Move the variable to the left-hand side and change the symbol $ $
$3 x - 40 \color{#FF6800}{ + } \color{#FF6800}{ x } = 0$
$3 x \color{#FF6800}{ - } \color{#FF6800}{ 40 } + x = 0$
$ $ Move the constant to the right side and change the sign $ $
$3 x + x = \color{#FF6800}{ 40 }$
$\color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ x } = 40$
$ $ Organize the expression $ $
$\color{#FF6800}{ 4 } \color{#FF6800}{ x } = 40$
$\color{#FF6800}{ 4 } \color{#FF6800}{ x } = \color{#FF6800}{ 40 }$
$ $ Divide both sides by the same number $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ 10 }$
$ $ 그래프 보기 $ $
Graph
Solution search results
search-thumbnail-$\dfrac {x+2} {6}$ $-\left(\dfrac {11-x} {3}-\dfrac {1} {4}\right)=\dfrac {3x-4} {9}$
7th-9th grade
Algebra
search-thumbnail-$1.$ $2x^{2}-x=15$ 
$2$ $10x^{2}=2x$ 
$3$ $-3x-40=-x^{2}$ 
$C$
7th-9th grade
Other
search-thumbnail-Solve $+0r$ the follaing variabde. 
$\left(1$ $3x-40=-2x-10$ 
$②$ $-15-5x=-10x+10$
7th-9th grade
Other
search-thumbnail-$=$ $25$ $\dfrac {x+2} {6}-\left(\dfrac {11-x} {3}-\dfrac {1} {4}\right)=\dfrac {3x-4} {12}$
10th-13th grade
Other
search-thumbnail-$25.$ $\dfrac {x+2} {6}-\left(\dfrac {11-x} {3}-\dfrac {1} {4}\right)=\dfrac {3x-4} {12}$
1st-6th grade
Calculus
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