Question

Find the value of x and y ​

Answer 1

Given

$\sf\to \dfrac{2\sqrt{3}-\sqrt{5}}{4\sqrt{3}-3\sqrt{5}} = x-\sqrt{15} y$

To find

The value of x and y

Now Take

$\sf\to \dfrac{2\sqrt{3}-\sqrt{5}}{4\sqrt{3}-3\sqrt{5}}$

Now Use Rationalization Method

$\sf\to \dfrac{2\sqrt{3}-\sqrt{5}}{4\sqrt{3}-3\sqrt{5}}\times\dfrac{4\sqrt{3}+3\sqrt{5}}{4\sqrt{3}+3\sqrt{5}}$

Using this identities

$\sf\to(a-b)(a+b)=a^2-b^2$

We get

$\sf\to \dfrac{(2\sqrt{3}-\sqrt{5})(4\sqrt{3}+3\sqrt{5})}{(4\sqrt{3}-3\sqrt{5})(4\sqrt{3}+3\sqrt{5})}$

$\sf\to \dfrac{(2\sqrt{3}-\sqrt{5})(4\sqrt{3}+3\sqrt{5})}{(4\sqrt{3})^2-(3\sqrt{5})^2}$

$\sf\to \dfrac{8\times3+2\sqrt{3}\times3\sqrt{5}-\sqrt{5}\times4\sqrt{3}-3\times5}{16\times3-9\times5}$

$\sf\to\dfrac{24+6\sqrt{15}-4\sqrt{15} -15}{48-45}$

$\sf\to\dfrac{9+2\sqrt{15}}{3}$

Now compare with

$\sf\to x-\sqrt{15} y$

we get

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

Answer 2

Question :-

$\dashrightarrow$ If $\large{\bf{\frac{2 \sqrt{3} \: - \: \sqrt{5} }{4 \sqrt{3} \: - \: 3 \sqrt{5} } = \times - \sqrt{15} y}}$

$\large\underline{\bf{Solving :}}$

To find :-

• Find the value of x and y.

$\dashrightarrow$ $\large{\bf{\frac{2 \sqrt{3} \: - \: \sqrt{5} }{4 \sqrt{3} \: - \: 3 \sqrt{5} }}}$

• Using Rationalization method

$\dashrightarrow$ $\large{\bf{ \frac{2 \sqrt{3} \: - \: \sqrt{5} }{4 \sqrt{3} \: - \: 3 \sqrt{5} } \times \frac{4 \sqrt{3} \: + \: 3 \sqrt{5} }{4 \sqrt{3} \: + \: 3 \sqrt{5} }}}$

• Using this identities

$\dashrightarrow$ $\large{\sf{(a - b)(a + b) = {a}^{2} - {b}^{2}}}$

• Now we get

$\dashrightarrow$ $\large{\bf{ \frac{(2 \sqrt{3} \: - \: \sqrt{5})(4 \sqrt{3} \: + \: 3 \sqrt{5} ) }{(4 \sqrt{3} \: - \: 3 \sqrt{5} )(4 \sqrt{3} \: + \: 3 \sqrt{5} ) } }}$

$\dashrightarrow$ $\large{\bf{\frac{(2 \sqrt{3} \: - \: \sqrt{5} )(4 \sqrt{3} \: + \: 3 \sqrt{5}) }{(4 \sqrt{3} {)}^{2} \: - \:( 3 \sqrt{5} {)}^{2} }}}$

$\dashrightarrow$ $\large{\bf{\frac{8 \: \times \: 3 \: + \: 2 \sqrt{3} \: \times \: 3 \sqrt{5} \: - \: \sqrt{5} \: \times \: 4 \sqrt{3} \: - \: 3 \: \times \: 5}{16 \: \times \: 3 \: - \: 9 \: \times \: 5}}}$

$\dashrightarrow$ $\large{\bf{\frac{24 \: + \: 6 \sqrt{15} \: - \: 4 \sqrt{15} \: - \: 15 }{48 \: - \: 45} }}$

$\dashrightarrow$ $\large{\bf{\frac{9 \: + \: 2 \sqrt{15} }{3}}}$

• Compare with

$\dashrightarrow$ $\large{\sf{x \: - \: \sqrt{15y}}}$

$\large\dag$ Hence,

$\dashrightarrow$ ${\underline{\boxed{\pink{\bf{x = 3 \: and \: y = \frac{ - 2}{3} }}}}}$★