Question

find the value of x and y​

Answer 1

Given:-

$\frac{1}{(3x + y)} + \frac{1}{(3x - y)} = \frac{3}{4} \\ \\ \frac{1}{2(3x + y)} + \frac{1}{2(3x - y)} = \frac{ - 1}{8}$

$put \: \frac{1}{(3x + y)} = a \\ and \: \frac{1}{(3x - y)} = b$

we get,

$a + b = \frac{3}{4} ......(i)\\ \frac{1}{2} a - \frac{1}{2} b = \frac{ - 1}{8} .......(ii)$

multiing eq.(ii) by 2 and subtracting equations(ii) from (i)

a + b = 3/4

a - b = -1/4

- + +

__________

2b = 1

=> b = 1/2

Putting value of b in eq. (i)

a + 1/2 = 3/4

=> a = 3/4 - 1/2 = 1/4

We got,

$\: \: \: \: \: b = \frac{1}{2} \\ = > \frac{1}{(3x - y)} = \frac{1}{2} \\ = > 3x - y = 2......(iii)$

And,

$\: \: \: \: \: a = \frac{1}{4} \\ = > \frac{1}{( 3x + y)} = \frac{1}{4} \\ = > 3x + y = 4......(iv)$

Adding eq. (iii) & (iv)

3x + y = 4

3x - y = 2

_________

6x = 6

=> x =1

Putting value of x in eq. (iii)

3 + y = 4

=> y = 1

Answer 2

Answer:

answer is 1

1. Step-by-step explanation:

thanksfor this