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Formula
Solve the inequality
Answer
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Graph
$\dfrac { 1 } { 3 } x + 0.5 \left ( x - \dfrac { 1 } { 3 } \right ) \geq \dfrac { 3 } { 2 } x$
$\dfrac { 1 } { 3 } x + 0.5 \left ( x - \dfrac { 1 } { 3 } \right ) \geq \dfrac { 3 } { 2 } x$
Solution of inequality
$x \leq - \dfrac { 1 } { 4 }$
$\dfrac{ 1 }{ 3 } x+0.5 \left( x- \dfrac{ 1 }{ 3 } \right) \geq \dfrac{ 3 }{ 2 } x$
$x \leq - \dfrac { 1 } { 4 }$
$ $ Solve a solution to $ x$
$\dfrac { 1 } { 3 } x + \color{#FF6800}{ 0.5 } \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 3 } } \right ) \geq \dfrac { 3 } { 2 } x$
$ $ Multiply each term in parentheses by $ 0.5$
$\dfrac { 1 } { 3 } x + \color{#FF6800}{ 0.5 } \color{#FF6800}{ x } + \color{#FF6800}{ 0.5 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 3 } } \right ) \geq \dfrac { 3 } { 2 } x$
$\color{#FF6800}{ \dfrac { 1 } { 3 } } \color{#FF6800}{ x } + 0.5 x + 0.5 \times \left ( - \dfrac { 1 } { 3 } \right ) \geq \dfrac { 3 } { 2 } x$
$ $ Calculate the multiplication expression $ $
$\color{#FF6800}{ \dfrac { x } { 3 } } + 0.5 x + 0.5 \times \left ( - \dfrac { 1 } { 3 } \right ) \geq \dfrac { 3 } { 2 } x$
$\dfrac { x } { 3 } + \color{#FF6800}{ 0.5 } \color{#FF6800}{ x } + 0.5 \times \left ( - \dfrac { 1 } { 3 } \right ) \geq \dfrac { 3 } { 2 } x$
$ $ Calculate the multiplication expression $ $
$\dfrac { x } { 3 } + \color{#FF6800}{ \dfrac { x } { 2 } } + 0.5 \times \left ( - \dfrac { 1 } { 3 } \right ) \geq \dfrac { 3 } { 2 } x$
$\dfrac { x } { 3 } + \dfrac { x } { 2 } + \color{#FF6800}{ 0.5 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 3 } } \right ) \geq \dfrac { 3 } { 2 } x$
$ $ Calculate multiplication and division of rational numbers $ $
$\dfrac { x } { 3 } + \dfrac { x } { 2 } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 6 } } \geq \dfrac { 3 } { 2 } x$
$\color{#FF6800}{ \dfrac { x } { 3 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { x } { 2 } } - \dfrac { 1 } { 6 } \geq \dfrac { 3 } { 2 } x$
$ $ Write all numerators above the least common denominator $ $
$\color{#FF6800}{ \dfrac { 2 x + 3 x } { 6 } } - \dfrac { 1 } { 6 } \geq \dfrac { 3 } { 2 } x$
$\color{#FF6800}{ \dfrac { 2 x + 3 x } { 6 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 6 } } \geq \dfrac { 3 } { 2 } x$
$ $ Since the denominator is the same as $ 6 $ , combine the fractions into one $ $
$\color{#FF6800}{ \dfrac { 2 x + 3 x - 1 } { 6 } } \geq \dfrac { 3 } { 2 } x$
$\dfrac { \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ x } - 1 } { 6 } \geq \dfrac { 3 } { 2 } x$
$ $ Calculate between similar terms $ $
$\dfrac { \color{#FF6800}{ 5 } \color{#FF6800}{ x } - 1 } { 6 } \geq \dfrac { 3 } { 2 } x$
$\dfrac { 5 x - 1 } { 6 } \geq \color{#FF6800}{ \dfrac { 3 } { 2 } } \color{#FF6800}{ x }$
$ $ Calculate the multiplication expression $ $
$\dfrac { 5 x - 1 } { 6 } \geq \color{#FF6800}{ \dfrac { 3 x } { 2 } }$
$\color{#FF6800}{ \dfrac { 5 x - 1 } { 6 } } \geq \color{#FF6800}{ \dfrac { 3 x } { 2 } }$
$ $ Multiply both sides by the least common multiple for the denominators to eliminate the fraction $ $
$\color{#FF6800}{ 5 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \geq \color{#FF6800}{ 9 } \color{#FF6800}{ x }$
$5 x - 1 \geq \color{#FF6800}{ 9 } \color{#FF6800}{ x }$
$ $ Move the variable to the left-hand side and change the symbol $ $
$5 x - 1 \color{#FF6800}{ - } \color{#FF6800}{ 9 } \color{#FF6800}{ x } \geq 0$
$5 x \color{#FF6800}{ - } \color{#FF6800}{ 1 } - 9 x \geq 0$
$ $ Move the constant to the right side and change the sign $ $
$5 x - 9 x \geq \color{#FF6800}{ 1 }$
$\color{#FF6800}{ 5 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 9 } \color{#FF6800}{ x } \geq 1$
$ $ Organize the expression $ $
$\color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ x } \geq 1$
$\color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ x } \geq \color{#FF6800}{ 1 }$
$ $ Change the symbol of the inequality of both sides, and reverse the symbol of the inequality to the opposite direction $ $
$4 x \leq - 1$
$\color{#FF6800}{ 4 } \color{#FF6800}{ x } \leq \color{#FF6800}{ - } \color{#FF6800}{ 1 }$
$ $ Divide both sides by the same number $ $
$\color{#FF6800}{ x } \leq \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 4 } }$
$ $ 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-$11.$ Question $11$ 
Solve the $:$ $folloMlng'$ $0<θ<90^{°}$ 
$\left(1\right)$ $2sin^{2}θ=1\right)$ $\left(rac\left(3\right)\left(2\right)\right)$ 
$\left(11\right)$ $3tan^{2}θ+2=3$ 
$\left(111\right)cos^{2}θ$ $11rac\left(1\right)\left(4\right)\right)=$ 
$c\left(1\right)\left(4\right)\right)=11113c\left(1\right)\left(2\right)\right)$
10th-13th grade
Trigonometry
search-thumbnail-Which of the following rational numbers are 
equivalent? 
$0Ptionsy$ 
A \frac{5}{6}, \frac{30}{36} 
B $s\sqrt{rac\left(} -2\right)\left(3\right)\sqrt{1rac} \sqrt{4\right)16\right)4} $ 
C $s\sqrt{11aC\left(} -4\right)1-7b,\sqrt{1rac\left(16\sqrt{35\right)9} } $ 
D \frac{1}{2},\frac{3}{8}
7th-9th grade
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