Question

Please koi solve Kar do please ​

Answer 1

Answer:

$\huge\colorbox{yellow}{Solution\:-}$

$Let \: \frac{1}{x + 2y} = u \: \: and \: \: \frac{5}{3x - 2y} = v$

Then, the given system of equation becomes

$\frac{u}{2} + \frac{5v}{3} = \frac{ - 3}{2}$

$= > \frac{3u + 10v}{6} = \frac{ - 3 \times 6}{3}$

$= > 3u + 10v = - 9 \: \: \: \: \: \: \: \: \: \: \: i)$

$\frac{5u}{4} - \frac{3v}{5} = \frac{61}{60}$

And,

$\frac{25u - 12v}{20} = \frac{61}{60}$

$= > 25u - 12v = \frac{61}{3} \: \: \: \: \: \: \: \: \: \: \: \: ii)$

Multiplying equation I) by 12 and equation II) by 10, we get

$36u + 120v = - 108 \: \: \: \: \: \: \: \: \: \: \: iii)$

$250u - 120v = \frac{610}{3} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: iv)$

Adding equation III) and equation IV), we get

$36u + 25v = \frac{610}{3} - 108$

$286u = \frac{610 - 324}{3}$

$286u = \frac{280}{3}$

$u = \frac{1}{3}$

Putting u = 1/3 in equation I) we get

$3 \times \frac{1}{3} + 10v = - 9$

$1 + 10v = - 9$

$10v = - 9 - 1$

$v = \frac{ - 10}{10} = - 1$

Now,

$u = \frac{1}{x + 2y}$

$\frac{1}{x + y} = \frac{1}{3}$

$x + 2y = 3$

And,

$v = \frac{1}{3x - 2y}$

$\frac{1}{3x - 2y} = - 1$

$3x - 2y = - 1$

Putting x = 1/3 in equation v) we get

$\frac{1}{2} + 2y = 3$

$2y = 3 - \frac{1}{2}$

$2y = \frac{6 - 1}{2}$

$y = \frac{5}{4}$

Hence, we conclude that

$\boxed{\pink{ \: x = \frac{1}{2}}}$

$\boxed{\pink{ \: y = \frac{5}{4}}}$

$\huge\colorbox{yellow}{Thank\:You}$

Answer 2

Answer:

Hy

Whats youre name

I'M REEN

NICE to meet you

Have a good day