Question

solve √2x-√3y=0 and √5x+√2y=0​

Answer 1

Answer:

Step-by-step explanation:

Hi friends...

here is the solution

√2x-√3y =0

multiplying by √2

4x- √6y = 0 ....................eq.(1)

√5x+√2y =0

multiplying by √2

√15x +√6y = 0 ...............eq.(2)

from eq.(1)+(2),

4x +√15x = 0

=> x = 0

putting the value of x in the eq(1)

(4×0) - (√6y) =0

=> y = 0

Hence the answer is x = 0 & y = 0

Answer 2

Here, the given pair of linear eq. is

$\sqrt{2} x - \sqrt{3} y = 0...(i)$

$\sqrt{5} x + \sqrt{2} y = 0...(ii)$

multiply (1) by $\sqrt{2}$ and (2) by $\sqrt{3}$ to make the coefficient y equal, we get,

$2x - \sqrt{6} y = 0.....(iii) \: \: ( \therefore \sqrt{2} \times \sqrt{2} = ( \cancel{√ }{2)}^{ \cancel{ 2} } = 2)$

$\sqrt{15} x + \sqrt{6} y = 0.....(iv) \: \: ( \therefore \sqrt{5} \times \sqrt{3} = \sqrt{5 \times 3} = \sqrt{15} \: and \: \sqrt{2} \times \sqrt{3} = \sqrt{2 \times 3} = \sqrt{6} )$

adding eq (iii) and (iv), we get,

$(2 + \sqrt{15} )x = 0$

$= > x = 0$

substitute x=0 in eq (ii), we get,

$\sqrt{2} y = 0$

$= > y = 0$

hence, the required solution of the given pair of linear eq. is x = 0 and y = 0