Question

if 6+√3/3-√3=x+√3y, find X & y​

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{x=6}}}$

$\green{\tt{\therefore{y=-\frac{2}{3}}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies 6 + \frac{ \sqrt{3} }{3} - \sqrt{3} = x + \sqrt{3}y \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies Value \: of \: x =? \\ \\ \tt: \implies Value \: of \: y=?$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies 6 + \frac{ \sqrt{3} }{3} - \sqrt{3} = x + \sqrt{3}y \\ \\ \tt: \implies \frac{6 \times 3 + \sqrt{3} - \sqrt{3} \times 3 }{3} = x + \sqrt{3} y \\ \\ \tt: \implies \frac{18 + \sqrt{3} - 3 \sqrt{3} }{3} = x + \sqrt{3}y \\ \\ \tt: \implies \frac{18 - 2 \sqrt{3} }{3} = x + \sqrt{3} y \\ \\ \tt: \implies \frac{18}{3} - \frac{2 \sqrt{3} }{3} = x + \sqrt{3} y \\ \\ \tt: \implies 6 - \frac{2}{ \sqrt{3} } = x + \sqrt{3} y\\ \\ \text{On \: comparing \: both \: side } \\ \\ \green{\tt: \implies x = 6} \\ \\ \tt: \implies - \frac{2}{ \sqrt{3} } = \sqrt{3} y \\ \\ \tt: \implies - 2 = \sqrt{3} \times \sqrt{3} y \\ \\ \tt: \implies - 2 = 3y \\ \\ \green{\tt: \implies y = \frac{ - 2}{3} }$

Answer 2

$\Huge{\underline{\underline {\boxed{\texttt{\red{Answer:-}}}}}}$

$\sf \: x = 6 \\ \\ \sf \: y = - \dfrac{2}{3}$

$\huge{\underline{\underline {\boxed{\texttt{\red{Explanation:-}}}}}}$

$\\ \\ \bold{\blue{\large{\boxed{\bigstar{\bf{Given:-}}}}}}</p><p>$

$\\ \\ \sf\longmapsto6 + \dfrac{ \sqrt{3} }{3} - \sqrt{3} = x + \sqrt{3} y$

$\bold{\pink{\large{\boxed{\bigstar{\bf{To find:-}}}}}}$

$\\ \\ \implies\sf\ \: Value \: of \: x \: and \: y$

$\\ \sf \:Now,$

$\\ \\ \sf\longmapsto \:6 + \dfrac{ \sqrt{3} }{3} - \sqrt{3} = x + \sqrt{3}y \:$

$\\ \\ \sf\longmapsto \: \dfrac{6 \times 3 + \sqrt{3} - 3\sqrt{3} }{3} = x + \sqrt{3}y$

$\\ \\ \sf\longmapsto \frac{18 + \sqrt{3} - 3\sqrt{3} }{3} = x + \sqrt{3}y$

$\\ \\ \sf\longmapsto \dfrac{18 - 2\sqrt{3} }{3} = x + \sqrt{3}y$

$\\ \\ \sf\longmapsto \dfrac{18}{3} - \dfrac{2 \sqrt{3} }{3} = x + \sqrt{3}y$

$\sf\longmapsto \: 6 - \frac{2}{ \sqrt{3} } = x + \sqrt{3} y$

$\sf\longmapsto\dfrac{6\sqrt{3}-2}{\sqrt{3}}=x+\sqrt{3}y$

$\sf\longmapsto6\sqrt{3}-2=\sqrt{3}x+y$

On comparing sides,

$\sf\:x=6$ and $\sf\:-2=3y$

$\sf\longmapsto\:y=\dfrac{-2}{3}$

Therefore,

$\sf\:x=6\:and \:y=\dfrac{-2}{3}$