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$\\ \text{Verification:-}\\ \text{For verlflcatlon, we have to put value of x}$ in glven equation\\ $\mathrm{L.H.S.}=\frac{3\left(2+\frac{6}{5}\right)-5\left(2 \times \frac{6}{5}-3\right)}{5-3 \times \frac{6}{5}}$ \\ Using BODMAS,\\ L.H.S. $=\frac{3\left(\frac{10+6}{5}\right)-5\left(\frac{12}{5}-3\right)}{5-\frac{18}{5}}$\\ $\mathrm{L.H.S.}=\frac{3\left(\frac{16}{5}\right)-5\left(\frac{12-15}{5}\right)}{\frac{25-18}{5}}$\\ $\mathrm{L.H.S.}=\frac{\left(\frac{48}{5}\right)-5\left(\frac{-3}{5}\right)}{\frac{7}{5}}$\\ $\mathrm{L.H.S.}=\frac{\frac{48}{5}+\frac{15}{5}}{\frac{7}{5}}$ \\ L.H.S. $=\frac{\frac{48+15}{5}}{\frac{7}{5}}$ \\ $\mathrm{L.H.S.}=\frac{\frac{63}{5}}{\frac{7}{5}}$ \\ $\mathrm{L.H.S.}=\frac{63}{7}$ \\ $\mathrm{L.H.S.}=9$ \\ $\mathrm{L} . \mathrm{H.S.}=\mathrm{R.H.S.}$ \\ Hence Proved$

Plz Find The Value Of X And Give The Verification A 3 2 X 5 2 X 3 5 3 X 9

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$\begin{array}{l} \Rightarrow \frac{3(2+x)-5(2 x-3)}{5-3 x}=9 \\ \Rightarrow 6+3 x-10 x+15=45-27 x \\ \Rightarrow x(27+3-10)=45-21 \\ \Rightarrow x(20)=24 \\ \Rightarrow x=\frac{24}{20} \\\\ \Rightarrow x=\frac{6}{5} \end{array}$

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$\\As , g=\frac{G M}{R^{2}}$ or $M=\frac{g R^{2}}{G}$ \\ \\This relation is true not only to the earth but for any heavenly body which is assumed to be spherical Now, $g=1.67 m s^{-2}, R=1.74 \times 10^{6} m$ \\ \\$G=6.67 \times 10^{-11} Nm ^{-2} kg ^{-2}$ \\ \\ $\therefore$ Mass of the moon, $M=\frac{1.67 \times\left(1.74 \times 10^{6}\right)^{2}}{6.67 \times 10^{-11}} kg$ $=7.58 \times 10^{22} kg$$

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