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Solve the equation
Answer
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Graph
$y = 4 \left ( 0.2 x + 1 \right )$
$y = 0.3 \left ( 4 - 2 x \right )$
$x$-intercept
$\left ( - 5 , 0 \right )$
$y$-intercept
$\left ( 0 , 4 \right )$
$x$-intercept
$\left ( 2 , 0 \right )$
$y$-intercept
$\left ( 0 , \dfrac { 6 } { 5 } \right )$
$4 \left( 0.2x+1 \right) = 0.3 \left( 4-2x \right)$
$x = - 2$
$ $ Solve a solution to $ x$
$\color{#FF6800}{ 4 } \left ( \color{#FF6800}{ 0.2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right ) = \color{#FF6800}{ 0.3 } \left ( \color{#FF6800}{ 4 } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \right )$
$ $ Organize the expression $ $
$\color{#FF6800}{ 0.8 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 4 } = \color{#FF6800}{ 0.3 } \color{#FF6800}{ \times } \color{#FF6800}{ 4 } \color{#FF6800}{ + } \color{#FF6800}{ 0.3 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 2 } \right ) \color{#FF6800}{ x }$
$\color{#FF6800}{ 0.8 } \color{#FF6800}{ x } + 4 = 0.3 \times 4 + 0.3 \times \left ( - 2 \right ) x$
$ $ Calculate the multiplication expression $ $
$\color{#FF6800}{ \dfrac { 4 x } { 5 } } + 4 = 0.3 \times 4 + 0.3 \times \left ( - 2 \right ) x$
$\dfrac { 4 x } { 5 } + 4 = \color{#FF6800}{ 0.3 } \color{#FF6800}{ \times } \color{#FF6800}{ 4 } + 0.3 \times \left ( - 2 \right ) x$
$ $ Multiply $ 0.3 $ and $ 4$
$\dfrac { 4 x } { 5 } + 4 = \color{#FF6800}{ 1.2 } + 0.3 \times \left ( - 2 \right ) x$
$\dfrac { 4 x } { 5 } + 4 = \color{#FF6800}{ 1.2 } + 0.3 \times \left ( - 2 \right ) x$
$ $ Convert decimals to fractions $ $
$\dfrac { 4 x } { 5 } + 4 = \color{#FF6800}{ \dfrac { 6 } { 5 } } + 0.3 \times \left ( - 2 \right ) x$
$\dfrac { 4 x } { 5 } + 4 = \dfrac { 6 } { 5 } + \color{#FF6800}{ 0.3 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 2 } \right ) \color{#FF6800}{ x }$
$ $ Simplify the expression $ $
$\dfrac { 4 x } { 5 } + 4 = \dfrac { 6 } { 5 } \color{#FF6800}{ - } \color{#FF6800}{ 0.6 } \color{#FF6800}{ x }$
$\dfrac { 4 x } { 5 } + 4 = \dfrac { 6 } { 5 } \color{#FF6800}{ - } \color{#FF6800}{ 0.6 } \color{#FF6800}{ x }$
$ $ Calculate the multiplication expression $ $
$\dfrac { 4 x } { 5 } + 4 = \dfrac { 6 } { 5 } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 x } { 5 } }$
$\color{#FF6800}{ \dfrac { 4 x } { 5 } } \color{#FF6800}{ + } \color{#FF6800}{ 4 } = \color{#FF6800}{ \dfrac { 6 } { 5 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 3 x } { 5 } }$
$ $ Multiply both sides by the least common multiple for the denominators to eliminate the fraction $ $
$\color{#FF6800}{ 4 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 20 } = \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 6 }$
$\color{#FF6800}{ 4 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 20 } = \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 6 }$
$ $ Organize the expression $ $
$\color{#FF6800}{ 4 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ x } = \color{#FF6800}{ 6 } \color{#FF6800}{ - } \color{#FF6800}{ 20 }$
$\color{#FF6800}{ 4 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ x } = 6 - 20$
$ $ Organize the expression $ $
$\color{#FF6800}{ 7 } \color{#FF6800}{ x } = 6 - 20$
$7 x = \color{#FF6800}{ 6 } \color{#FF6800}{ - } \color{#FF6800}{ 20 }$
$ $ Subtract $ 20 $ from $ 6$
$7 x = \color{#FF6800}{ - } \color{#FF6800}{ 14 }$
$\color{#FF6800}{ 7 } \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 14 }$
$ $ Divide both sides by the same number $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 2 }$
$ $ 그래프 보기 $ $
Graph
Solution search results
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search-thumbnail-What is the inverse of $g\left(x\right)=\sqrt{x-1} $ $Og\left(x\right)=\sqrt{x+1} $ $+1$ $g^{'}\left(x\right)=5-$ $y_{9}$ $g\sqrt{ef\left(x\left(nght\right)} =\left(f_{tac\left(x+1\right)0sq\pi \left(x\right)}$ $Og^{'}\left(x\right)=x^{2}+1$ $○g\left(x\right)=x^{2}-1$ « Previous Ne
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search-thumbnail-What is the inverse of $g\left(x\right)=\sqrt{x-1} $ $Og\left(x\right)=\sqrt{x+1} $ $+1$ $g^{'}\left(x\right)=5-$ $y_{9}$ $g\sqrt{ef\left(x\left(nght\right)} =\left(f_{tac\left(x+1\right)0sq\pi \left(x\right)}$ $Og^{'}\left(x\right)=x^{2}+1$ $○g\left(x\right)=x^{2}-1$ « Previous Ne
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