Question

solve this problem ​

Answer 1

Answer:

$\boxed{\red{\bf x = 1}}$

$\boxed{\red{\bf y= 1}}$

Step-by-step explanation:

Given that -

$\sf \dfrac{1}{3x + y} + \dfrac{1}{3x - y} = \dfrac{3}{4} \: \: \: \: ....(i) \\ \\ \sf \dfrac{1}{2(3x + y)} - \dfrac{1}{2(3x - y)} = - \dfrac{1}{8} \: \: \: ....(ii)$

Let , $\sf \dfrac{1}{3x+y}=u$ and $\sf \dfrac{1}{3x - y}=v$.

Substituting the above assumptions in the equation -

Equation (i) :

⇒ $\sf \dfrac{1}{3x+y}+\dfrac{1}{3x-y}=\dfrac{3}{4}$

⇒ u + v = $\dfrac{3}{4}$

⇒ 4(u + v) = 3

⇒ 4u + 4v = 3

Equation (ii) :

⇒ $\sf \dfrac{1}{2(3x+y)}-\dfrac{1}{2(3x-y)}=\dfrac{-1}{8}$

⇒ $\sf \dfrac{u}{2}-\dfrac{v}{2}=\dfrac{-1}{8}$

⇒ $\sf \dfrac{u-v}{2}=\dfrac{-1}{8}$

⇒ u - v = $\dfrac{-1}{4}$

⇒ 4(u - v) = - 1

⇒ 4u - 4v = - 1

On adding both the equation -

⇒ 4u + 4v + 4u - 4v = 3 + ( - 1)

⇒ 8u = 2

⇒ u = $\dfrac{2}{8}$

⇒ u = $\dfrac{1}{4}$

On substituting the value of u in eqⁿ (i) -

⇒ 4u + 4v = 3

⇒ 4 * $\dfrac{1}{4}$ + 4v = 3

⇒ 1 + 4v = 3

⇒ 4v = 3 - 1

⇒ v = $\dfrac{2}{4}$

⇒ v = $\dfrac{1}{2}$

On substituting the value of u and v in our assumptions :

$\implies \sf \dfrac{1}{3x+y}=u$

$\implies \sf \dfrac{1}{3x+y}=\dfrac{1}{4}$

$\implies \sf 3x + y = 4 \: \: \:..... (iii)$

Also,

$\implies \sf \dfrac{1}{3x-y}=v$

$\implies \sf \dfrac{1}{3x-y}=\dfrac{1}{2}$

$\implies \sf 3x-y = 2 \: \: \:..... (iv)$

Again, on adding (iii) and (iv) :

⇒ 3x + y + 3x - y = 4 + 2

⇒ 6x = 6

⇒ x = 1

Substituting the value of x in (iv) :

⇒ 3x - y = 2

⇒ 3 * 1 - y = 2

⇒ y = 3 - 2

⇒ y = 1

Hence, the value of x is 1 and y is 1.

Answer 2

Step-by-step explanation:

this is the answer hope it helps