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Formula
Solve the inequality
Answer
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Graph
$\dfrac { x } { 3 } - 0.2 \left ( x + 5 \right ) \leq 1$
$\dfrac { x } { 3 } - 0.2 \left ( x + 5 \right ) \leq 1$
Solution of inequality
$x \leq 15$
$\dfrac{ x }{ 3 } -0.2 \left( x+5 \right) \leq 1$
$x \leq 15$
$ $ Solve a solution to $ x$
$\dfrac { x } { 3 } \color{#FF6800}{ - } \color{#FF6800}{ 0.2 } \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 5 } \right ) \leq 1$
$ $ Multiply each term in parentheses by $ - 0.2$
$\dfrac { x } { 3 } \color{#FF6800}{ - } \color{#FF6800}{ 0.2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 0.2 } \color{#FF6800}{ \times } \color{#FF6800}{ 5 } \leq 1$
$\dfrac { x } { 3 } \color{#FF6800}{ - } \color{#FF6800}{ 0.2 } \color{#FF6800}{ x } - 0.2 \times 5 \leq 1$
$ $ Calculate the multiplication expression $ $
$\dfrac { x } { 3 } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { x } { 5 } } - 0.2 \times 5 \leq 1$
$\dfrac { x } { 3 } - \dfrac { x } { 5 } \color{#FF6800}{ - } \color{#FF6800}{ 0.2 } \color{#FF6800}{ \times } \color{#FF6800}{ 5 } \leq 1$
$ $ Multiply $ - 0.2 $ and $ 5$
$\dfrac { x } { 3 } - \dfrac { x } { 5 } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \leq 1$
$\color{#FF6800}{ \dfrac { x } { 3 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { x } { 5 } } - 1 \leq 1$
$ $ Write all numerators above the least common denominator $ $
$\color{#FF6800}{ \dfrac { 5 x - 3 x } { 15 } } - 1 \leq 1$
$\dfrac { 5 x - 3 x } { 15 } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \leq 1$
$ $ Convert an equation to a fraction using $ a=\dfrac{a}{1}$
$\dfrac { 5 x - 3 x } { 15 } + \color{#FF6800}{ \dfrac { - 1 } { 1 } } \leq 1$
$\color{#FF6800}{ \dfrac { 5 x - 3 x } { 15 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { - 1 } { 1 } } \leq 1$
$ $ Write all numerators above the least common denominator $ $
$\color{#FF6800}{ \dfrac { 5 x - 3 x - 15 } { 15 } } \leq 1$
$\dfrac { \color{#FF6800}{ 5 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ x } - 15 } { 15 } \leq 1$
$ $ Calculate between similar terms $ $
$\dfrac { \color{#FF6800}{ 2 } \color{#FF6800}{ x } - 15 } { 15 } \leq 1$
$\color{#FF6800}{ \dfrac { 2 x - 15 } { 15 } } \leq \color{#FF6800}{ 1 }$
$ $ Multiply both sides by the least common multiple for the denominators to eliminate the fraction $ $
$\color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 15 } \leq \color{#FF6800}{ 15 }$
$2 x \color{#FF6800}{ - } \color{#FF6800}{ 15 } \leq 15$
$ $ Move the constant to the right side and change the sign $ $
$2 x \leq 15 \color{#FF6800}{ + } \color{#FF6800}{ 15 }$
$2 x \leq \color{#FF6800}{ 15 } \color{#FF6800}{ + } \color{#FF6800}{ 15 }$
$ $ Add $ 15 $ and $ 15$
$2 x \leq \color{#FF6800}{ 30 }$
$\color{#FF6800}{ 2 } \color{#FF6800}{ x } \leq \color{#FF6800}{ 30 }$
$ $ Divide both sides by the same number $ $
$\color{#FF6800}{ x } \leq \color{#FF6800}{ 15 }$
$ $ 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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