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Solve the inequality
Answer
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Graph
$- 0.7 + \dfrac { 1 } { 5 } \left ( 2 - x \right ) \leq - 0.3 \left ( x - 2 \right )$
$- 0.7 + \dfrac { 1 } { 5 } \left ( 2 - x \right ) \leq - 0.3 \left ( x - 2 \right )$
Solution of inequality
$x \leq 9$
$-0.7+ \dfrac{ 1 }{ 5 } \left( 2-x \right) \leq -0.3 \left( x-2 \right)$
$x \leq 9$
$ $ Solve a solution to $ x$
$- 0.7 + \color{#FF6800}{ \dfrac { 1 } { 5 } } \left ( \color{#FF6800}{ 2 } \color{#FF6800}{ - } \color{#FF6800}{ x } \right ) \leq - 0.3 \left ( x - 2 \right )$
$ $ Multiply each term in parentheses by $ \dfrac { 1 } { 5 }$
$- 0.7 + \color{#FF6800}{ \dfrac { 1 } { 5 } } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 5 } } \color{#FF6800}{ x } \leq - 0.3 \left ( x - 2 \right )$
$- 0.7 + \dfrac { 1 } { 5 } \times 2 - \dfrac { 1 } { 5 } x \leq \color{#FF6800}{ - } \color{#FF6800}{ 0.3 } \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \right )$
$ $ Multiply each term in parentheses by $ - 0.3$
$- 0.7 + \dfrac { 1 } { 5 } \times 2 - \dfrac { 1 } { 5 } x \leq \color{#FF6800}{ - } \color{#FF6800}{ 0.3 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 0.3 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 2 } \right )$
$\color{#FF6800}{ - } \color{#FF6800}{ 0.7 } + \dfrac { 1 } { 5 } \times 2 - \dfrac { 1 } { 5 } x \leq - 0.3 x - 0.3 \times \left ( - 2 \right )$
$ $ Convert decimals to fractions $ $
$\color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 7 } { 10 } } + \dfrac { 1 } { 5 } \times 2 - \dfrac { 1 } { 5 } x \leq - 0.3 x - 0.3 \times \left ( - 2 \right )$
$- \dfrac { 7 } { 10 } + \color{#FF6800}{ \dfrac { 1 } { 5 } } \color{#FF6800}{ \times } \color{#FF6800}{ 2 } - \dfrac { 1 } { 5 } x \leq - 0.3 x - 0.3 \times \left ( - 2 \right )$
$ $ Calculate the product of rational numbers $ $
$- \dfrac { 7 } { 10 } + \color{#FF6800}{ \dfrac { 2 } { 5 } } - \dfrac { 1 } { 5 } x \leq - 0.3 x - 0.3 \times \left ( - 2 \right )$
$- \dfrac { 7 } { 10 } + \dfrac { 2 } { 5 } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 5 } } \color{#FF6800}{ x } \leq - 0.3 x - 0.3 \times \left ( - 2 \right )$
$ $ Calculate the multiplication expression $ $
$- \dfrac { 7 } { 10 } + \dfrac { 2 } { 5 } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { x } { 5 } } \leq - 0.3 x - 0.3 \times \left ( - 2 \right )$
$\color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 7 } { 10 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 2 } { 5 } } - \dfrac { x } { 5 } \leq - 0.3 x - 0.3 \times \left ( - 2 \right )$
$ $ Find the sum or difference of the fractions $ $
$\color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 10 } } - \dfrac { x } { 5 } \leq - 0.3 x - 0.3 \times \left ( - 2 \right )$
$- \dfrac { 3 } { 10 } - \dfrac { x } { 5 } \leq \color{#FF6800}{ - } \color{#FF6800}{ 0.3 } \color{#FF6800}{ x } - 0.3 \times \left ( - 2 \right )$
$ $ Calculate the multiplication expression $ $
$- \dfrac { 3 } { 10 } - \dfrac { x } { 5 } \leq \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 x } { 10 } } - 0.3 \times \left ( - 2 \right )$
$- \dfrac { 3 } { 10 } - \dfrac { x } { 5 } \leq - \dfrac { 3 x } { 10 } \color{#FF6800}{ - } \color{#FF6800}{ 0.3 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 2 } \right )$
$ $ Multiply $ - 0.3 $ and $ - 2$
$- \dfrac { 3 } { 10 } - \dfrac { x } { 5 } \leq - \dfrac { 3 x } { 10 } + \color{#FF6800}{ 0.6 }$
$- \dfrac { 3 } { 10 } - \dfrac { x } { 5 } \leq - \dfrac { 3 x } { 10 } + \color{#FF6800}{ 0.6 }$
$ $ Convert decimals to fractions $ $
$- \dfrac { 3 } { 10 } - \dfrac { x } { 5 } \leq - \dfrac { 3 x } { 10 } + \color{#FF6800}{ \dfrac { 3 } { 5 } }$
$\color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 } { 10 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { x } { 5 } } \leq \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 x } { 10 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 3 } { 5 } }$
$ $ Multiply both sides by the least common multiple for the denominators to eliminate the fraction $ $
$\color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ - } \left ( \color{#FF6800}{ 2 } \color{#FF6800}{ x } \right ) \leq \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 6 }$
$- 3 \color{#FF6800}{ - } \left ( \color{#FF6800}{ 2 } \color{#FF6800}{ x } \right ) \leq - 3 x + 6$
$ $ Get rid of unnecessary parentheses $ $
$- 3 \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \leq - 3 x + 6$
$- 3 - 2 x \leq \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ x } + 6$
$ $ Move the variable to the left-hand side and change the symbol $ $
$- 3 - 2 x \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ x } \leq 6$
$\color{#FF6800}{ - } \color{#FF6800}{ 3 } - 2 x + 3 x \leq 6$
$ $ Move the constant to the right side and change the sign $ $
$- 2 x + 3 x \leq 6 \color{#FF6800}{ + } \color{#FF6800}{ 3 }$
$\color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ x } \leq 6 + 3$
$ $ Organize the expression $ $
$\color{#FF6800}{ x } \leq 6 + 3$
$x \leq \color{#FF6800}{ 6 } \color{#FF6800}{ + } \color{#FF6800}{ 3 }$
$ $ Add $ 6 $ and $ 3$
$x \leq \color{#FF6800}{ 9 }$
$ $ 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
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