Question

Find the value of y if 2+y√3=√3-1÷√3+1

Answer 1

Answer:

$\large{\underline{\boxed{\sf y = - 1}}}$

Step-by-step explanation:

Given that ;

$\sf 2 + y \sqrt{3} = \dfrac{ \sqrt{3} - 1}{ \sqrt{3} + 1}$

Consider RHS first ;

$\sf \frac{ \sqrt{3} - 1}{ \sqrt{3} + 1 } \\ \\ \sf \frac{ \sqrt{3} - 1 }{ \sqrt{3} +1} \times \frac{ \sqrt{3} - 1}{ \sqrt{3}- 1} \\ \\ \sf \frac{( \sqrt{3} - 1) {}^{2} } {( \sqrt{3}) {}^{2} - (1) {}^{2} } \\\\ \sf \frac{3 + 1 - 2 \sqrt{3} }{3 - 1} \\ \\ \sf \frac{4 - 2 \sqrt{3} }{2} \\ \\ \sf \frac{2(2 - \sqrt{3} )}{2} \\ \\ \sf 2 - \sqrt{3}$

On comparing the RHS with LHS,

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

We get that, y = - 1.

Hence, the value of y is - 1.

Answer 2

The answer is -1.
By rationalisation
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