​ Solve the following systems of equations: 1/(3x + y) + 1/(3x - y) = 3/4 1/{2(3x+y)} - 1/{2(3x - y)} = - 1/8 ​

Question

Solve the following systems of equations:

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

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

Answer 1

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

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

Multiplying eq2 by 2 and adding to eq1

⇒ 2/(3x + y) = 3/4 - 2/8

⇒ 3x + y = 4 ---- (3)

Multiplying eq2 by 2 and subtracting from eq1

⇒ 2/(3x – y ) = 1

⇒ 3x – y = 2 ----- (4)

Adding (3) and (4)

⇒ 6x = 6

⇒ x = 1

Thus,

y = 3 – 2 = 1