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Formula
Solve the equation
Answer
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Graph
$y = - 5 x$
$y = x + 12$
$x$-intercept
$\left ( 0 , 0 \right )$
$y$-intercept
$\left ( 0 , 0 \right )$
$x$-intercept
$\left ( - 12 , 0 \right )$
$y$-intercept
$\left ( 0 , 12 \right )$
$-5x = x+12$
$x = - 2$
$ $ Solve a solution to $ x$
$- 5 x = \color{#FF6800}{ x } + 12$
$ $ Move the variable to the left-hand side and change the symbol $ $
$- 5 x - x = 12$
$\color{#FF6800}{ - } \color{#FF6800}{ 5 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ x } = 12$
$ $ Organize the expression $ $
$\color{#FF6800}{ - } \color{#FF6800}{ 6 } \color{#FF6800}{ x } = 12$
$\color{#FF6800}{ - } \color{#FF6800}{ 6 } \color{#FF6800}{ x } = \color{#FF6800}{ 12 }$
$ $ Change the sign of both sides of the equation $ $
$6 x = - 12$
$\color{#FF6800}{ 6 } \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 12 }$
$ $ Divide both sides by the same number $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 2 }$
$ $ 그래프 보기 $ $
Graph
Solution search results
search-thumbnail-$5\left(x-4\right)=x+16$ 
$5x-1$ $=x+16$ 
$5x-20+20=x+16+20$ 
$5x=x+36$ 
$=36$ 
$x=4$
1st-6th grade
Algebra
search-thumbnail-$63a$ Determine the sign of the following 
expressions for $\times ∈R\left(1\right)$ $x^{2}+x+1$ 
$.$ $\left(11\right)$ $x^{2}-5x+6$
10th-13th grade
Algebra
search-thumbnail-Given: $3x=x+12$ 
Prove: x = 6 STATEMENT REASON 
$3x=x+12$ 5 
$3x-x=x+12-x$ 
– – 
$2x=x+12-x$ Simplifying 
$2x=0+12$ $7$ $-$ 
$2x=12$ 8 
$2\times \left(\dfrac {1} {2}\right)=12\left(\dfrac {1} {2}\right)$ $2$ 
$x-12\left(\dfrac {1} {2}\right)$ $10$ 
$x=6$ Simplifying 
Reason $5$ * $1$ point 
$○$ Prove 
$○$ Statement 
$○$ Property 
$○$ Given
7th-9th grade
Other
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