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Solve the equation

Answer

Graph

$y = \dfrac { 1 } { 2 } x - 1$

$x$Intercept

$\left ( 2 , 0 \right )$

$y$Intercept

$\left ( 0 , - 1 \right )$

$x = 2 y + 2$

$ $ Solve a solution to $ x$

$y = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 2 } } } \color{#FF6800}{ x } - 1$

$ $ Calculate the multiplication expression $ $

$y = \color{#FF6800}{ \dfrac { \color{#FF6800}{ x } } { \color{#FF6800}{ 2 } } } - 1$

$\color{#FF6800}{ y } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ x } } { \color{#FF6800}{ 2 } } } \color{#FF6800}{ - } \color{#FF6800}{ 1 }$

$ $ Multiply both sides by the least common multiple for the denominators to eliminate the fraction $ $

$\color{#FF6800}{ 2 } \color{#FF6800}{ y } = \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 2 }$

$2 y = \color{#FF6800}{ x } - 2$

$ $ Move $ x $ term to the left side and change the sign $ $

$2 y - x = - 2$

$\color{#FF6800}{ 2 } \color{#FF6800}{ y } - x = - 2$

$ $ Move the rest of the expression except $ x $ term to the right side and replace the sign $ $

$- x = - 2 \color{#FF6800}{ - } \left ( \color{#FF6800}{ 2 } \color{#FF6800}{ y } \right )$

$- x = \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ - } \left ( \color{#FF6800}{ 2 } \color{#FF6800}{ y } \right )$

$ $ Organize the expression $ $

$- x = \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ y } \color{#FF6800}{ - } \color{#FF6800}{ 2 }$

$\color{#FF6800}{ - } \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ y } \color{#FF6800}{ - } \color{#FF6800}{ 2 }$

$ $ Change the sign of both sides of the equation $ $

$x = 2 y + 2$

Solution search results

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}

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In the following problem $a,b,$ b,and c represent REAL NUMBERS. The derivative of $g\left(x\right)=log _{a}\left(bx\right)+cx^{2}is$ $○$ $g^{1}\left(x\right)=\right)dfrac\left(b\right)log _{-}a\left(bx\right)\right)\left(N1n$ $a\right)+2cx1\right)$ $○$ $\left(g\left(x\right)=\right)dtracb$ $loga\left(bx\right)+2c\times \right)Nln$ a}\) $○$ $g^{'}\left(x\right)=\left(dfraC\left(a\right)log _{-}a\left(bx\right)\right)\left(ln$ $\right)+2c\times 1\right)\right)$ $○$ $g^{1}\left(x\right)=\left(dfrac\left(b$ $log _{-}a\left(bx\right)\right)\left(ln$ $a1\right)$ $○$ $\left(\left(g^{1}\left(x\right)=b$ $log _{-}a\left(bx\right)+2cx1\right)$ $○$ $g\left(x\right)=\left(dfrac\left(b$ $log _{-}a\left(bx\right)\right)\left(1ln$ $a|+2c1\right)$

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$search-thumbnail-Solve the : 0<θ<90^{°}$

$11.$ Question $11$ Solve the $:$ $folloMlng'$ $0<θ<90^{°}$ $\left(1\right)$ $2sin^{2}θ=1\right)$ $\left(rac\left(3\right)\left(2\right)\right)$ $\left(11\right)$ $3tan^{2}θ+2=3$ $\left(111\right)cos^{2}θ$ $11rac\left(1\right)\left(4\right)\right)=$ $c\left(1\right)\left(4\right)\right)=11113c\left(1\right)\left(2\right)\right)$

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Can you answer this? $20$ $25$ $18$ $\left($ $\left(A\right)$ $A\right)2$ $21frac\left(5\right)\left(9\right)$ \) $\left(B\right)$ $B\right)$ $1\left(211$ $\left(C\right)$ $1\left(21$ $21+rac\left(7\right)+9\right)$ \) $\left(D\right)$ $1\left(2\right)$ 2\frac{8}{9} $ac\left(8\right)\left(9\right)$ \) $9:18PM\sqrt{} $

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$search-thumbnail-1f(u=(rac(x^{n}3+y^{n}3)(x+y))$

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