Question

Solve the following simultaneous equation.

`1/(3x + y) + 1/(3x - y) = 3/4; 1/(2(3x + y))  - 1/(2(3x - y))  = -1/8`

Answer 1

Given:

`1/(3x + y) + 1/(3x - y) = 3/4; 1/(2(3x + y))  - 1/(2(3x - y))  = -1/8`

let `1/(3x + y) ​= a  "and" 1/(3x − y) = b`
Then,

a + b = `3/4`

⇒ 4a + 4b = 3     ...(I)

Similarly,

`a/2 ​− b/2 ​= (-1)/8`

⇒ `(a − b)/2 ​= (-1)/8`

⇒ `a − b = (− 2)/8` ​

⇒ `a − b = (−1​)/4`

⇒  4a − 4b = −1   ...(II)

Adding I and II we get,

4a + 4b = 3

+ 4a − 4b = −1

8a = 2

a = `2/8`​

a = `1/4`

Substituting the value of a in I we get,

4 × `1/4`​ + 4b = 3

4b = 2

b = `2/4 ​⇒ 1/2`

Then,

`1/(3x + y) ​= 1/4`​

3x + y = 4    ...(III)

`1/(3x - y) = 1/2`

3x − y = 2    ...(IV)

Adding (III) and (IV) we get,

   3x + y = 4

+ 3x − y = 2

6x = 6

x = 1

Substituting x = 1 in (III) we get,

3 × 1 + y = 4

y = 1

Thus, x = 1 and y = 1