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Formula
Solve the inequality
Answer
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Graph
$1 - \left ( 4 + 8 x \right ) \geq - 2 \left ( x - 1 \right ) + 5$
$1 - \left ( 4 + 8 x \right ) \geq - 2 \left ( x - 1 \right ) + 5$
Solution of inequality
$x \leq - \dfrac { 5 } { 3 }$
$1- \left( 4+8x \right) \geq -2 \left( x-1 \right) +5$
$x \leq - \dfrac { 5 } { 3 }$
$ $ Solve a solution to $ x$
$1 - \left ( 4 + 8 x \right ) \geq \color{#FF6800}{ - } \color{#FF6800}{ 2 } \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) + 5$
$ $ Multiply each term in parentheses by $ - 2$
$1 - \left ( 4 + 8 x \right ) \geq \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } + \color{#FF6800}{ 2 } + 5$
$1 \color{#FF6800}{ - } \left ( \color{#FF6800}{ 4 } \color{#FF6800}{ + } \color{#FF6800}{ 8 } \color{#FF6800}{ x } \right ) \geq - 2 x + 2 + 5$
$ $ Change the symbol of each term in parentheses when there is a (-) symbol in front of parentheses $ $
$1 \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ - } \color{#FF6800}{ 8 } \color{#FF6800}{ x } \geq - 2 x + 2 + 5$
$\color{#FF6800}{ 1 } \color{#FF6800}{ - } \color{#FF6800}{ 4 } - 8 x \geq - 2 x + 2 + 5$
$ $ Subtract $ 4 $ from $ 1$
$\color{#FF6800}{ - } \color{#FF6800}{ 3 } - 8 x \geq - 2 x + 2 + 5$
$- 3 - 8 x \geq - 2 x + \color{#FF6800}{ 2 } \color{#FF6800}{ + } \color{#FF6800}{ 5 }$
$ $ Add $ 2 $ and $ 5$
$- 3 - 8 x \geq - 2 x + \color{#FF6800}{ 7 }$
$\color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ - } \color{#FF6800}{ 8 } \color{#FF6800}{ x } \geq - 2 x + 7$
$ $ Organize the expression $ $
$\color{#FF6800}{ - } \color{#FF6800}{ 8 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \geq - 2 x + 7$
$- 8 x - 3 \geq \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } + 7$
$ $ Move the variable to the left-hand side and change the symbol $ $
$- 8 x - 3 \color{#FF6800}{ + } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \geq 7$
$- 8 x \color{#FF6800}{ - } \color{#FF6800}{ 3 } + 2 x \geq 7$
$ $ Move the constant to the right side and change the sign $ $
$- 8 x + 2 x \geq 7 \color{#FF6800}{ + } \color{#FF6800}{ 3 }$
$\color{#FF6800}{ - } \color{#FF6800}{ 8 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \geq 7 + 3$
$ $ Organize the expression $ $
$\color{#FF6800}{ - } \color{#FF6800}{ 6 } \color{#FF6800}{ x } \geq 7 + 3$
$- 6 x \geq \color{#FF6800}{ 7 } \color{#FF6800}{ + } \color{#FF6800}{ 3 }$
$ $ Add $ 7 $ and $ 3$
$- 6 x \geq \color{#FF6800}{ 10 }$
$\color{#FF6800}{ - } \color{#FF6800}{ 6 } \color{#FF6800}{ x } \geq \color{#FF6800}{ 10 }$
$ $ Change the symbol of the inequality of both sides, and reverse the symbol of the inequality to the opposite direction $ $
$6 x \leq - 10$
$\color{#FF6800}{ 6 } \color{#FF6800}{ x } \leq \color{#FF6800}{ - } \color{#FF6800}{ 10 }$
$ $ Divide both sides by the same number $ $
$\color{#FF6800}{ x } \leq \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 5 } { 3 } }$
$ $ 그래프 보기 $ $
Inequality
Solution search results
search-thumbnail-If the sum of two consecutive 
numbers is $45$ and one number is $X$ 
.This statement in the form of 
equation $1s:$ 
$\left(1$ Point) $\right)$ 
$○5x+1$ $1eft\left(x+1$ $r1gnt\right)=45s$ 
$○sx+1ef\left(x+2$ $r1gnt\right)=145s$ 
$sx+1x=45s$
7th-9th grade
Algebra
search-thumbnail-$s|ef\left(-1n$ $\left($ }\right)^{50}\ $\right)$ \ | | is\ equal\ to\ $S$ 
$s1S$ 
$S-1S$ 
$s2S$ 
$s50s$
7th-9th grade
Other
search-thumbnail-Given the set of ordered pairs $\left(\left(-7.0\right),\left(-6,5\right),\left(-5,-3\right),\left(-1,2\right)$ $\left(1,6\right),\left(2,-2\right)$ $\left(5,3\right)\left(7,-8\right)\right)$ 
Find f(7)fAleft(7\right) 
O a 
O b -8 
6. 
$5$
7th-9th grade
Algebra
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