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Formula
Solve the equation
Answer
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Graph
$y = 0.1 \left ( x - 2 \right )$
$y = 0.03 x - 0.34$
$x$-intercept
$\left ( 2 , 0 \right )$
$y$-intercept
$\left ( 0 , - \dfrac { 1 } { 5 } \right )$
$x$-intercept
$\left ( \dfrac { 34 } { 3 } , 0 \right )$
$y$-intercept
$\left ( 0 , - \dfrac { 17 } { 50 } \right )$
$0.1 \left( x-2 \right) = 0.03x-0.34$
$x = - 2$
$ $ Solve a solution to $ x$
$\color{#FF6800}{ 0.1 } \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \right ) = 0.03 x - 0.34$
$ $ Multiply each term in parentheses by $ 0.1$
$\color{#FF6800}{ 0.1 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 0.1 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 2 } \right ) = 0.03 x - 0.34$
$\color{#FF6800}{ 0.1 } \color{#FF6800}{ x } + 0.1 \times \left ( - 2 \right ) = 0.03 x - 0.34$
$ $ Calculate the multiplication expression $ $
$\color{#FF6800}{ \dfrac { x } { 10 } } + 0.1 \times \left ( - 2 \right ) = 0.03 x - 0.34$
$\dfrac { x } { 10 } + \color{#FF6800}{ 0.1 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 2 } \right ) = 0.03 x - 0.34$
$ $ Multiply $ 0.1 $ and $ - 2$
$\dfrac { x } { 10 } \color{#FF6800}{ - } \color{#FF6800}{ 0.2 } = 0.03 x - 0.34$
$\dfrac { x } { 10 } \color{#FF6800}{ - } \color{#FF6800}{ 0.2 } = 0.03 x - 0.34$
$ $ Convert decimals to fractions $ $
$\dfrac { x } { 10 } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 5 } } = 0.03 x - 0.34$
$\dfrac { x } { 10 } - \dfrac { 1 } { 5 } = \color{#FF6800}{ 0.03 } \color{#FF6800}{ x } - 0.34$
$ $ Calculate the multiplication expression $ $
$\dfrac { x } { 10 } - \dfrac { 1 } { 5 } = \color{#FF6800}{ \dfrac { 3 x } { 100 } } - 0.34$
$\dfrac { x } { 10 } - \dfrac { 1 } { 5 } = \dfrac { 3 x } { 100 } \color{#FF6800}{ - } \color{#FF6800}{ 0.34 }$
$ $ Convert decimals to fractions $ $
$\dfrac { x } { 10 } - \dfrac { 1 } { 5 } = \dfrac { 3 x } { 100 } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 17 } { 50 } }$
$\color{#FF6800}{ \dfrac { x } { 10 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 1 } { 5 } } = \color{#FF6800}{ \dfrac { 3 x } { 100 } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 17 } { 50 } }$
$ $ Multiply both sides by the least common multiple for the denominators to eliminate the fraction $ $
$\color{#FF6800}{ 10 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 20 } = \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 34 }$
$\color{#FF6800}{ 10 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 20 } = \color{#FF6800}{ 3 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 34 }$
$ $ Organize the expression $ $
$\color{#FF6800}{ 10 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 34 } \color{#FF6800}{ + } \color{#FF6800}{ 20 }$
$\color{#FF6800}{ 10 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ x } = - 34 + 20$
$ $ Organize the expression $ $
$\color{#FF6800}{ 7 } \color{#FF6800}{ x } = - 34 + 20$
$7 x = \color{#FF6800}{ - } \color{#FF6800}{ 34 } \color{#FF6800}{ + } \color{#FF6800}{ 20 }$
$ $ Add $ - 34 $ and $ 20$
$7 x = \color{#FF6800}{ - } \color{#FF6800}{ 14 }$
$\color{#FF6800}{ 7 } \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 14 }$
$ $ 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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7th-9th grade
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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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