We have

and

Where x + 2y ≠ 0 and 3x - 2y ≠ 0

Lets simplify these equations. Assuming and

⇒

Multiply it with 6, we get

3p + 10q = -9 …(i)

Also,

⇒

Multiply it with 20, we get

25p – 12q = 61/3

⇒ 75 – 36q = 61 …(ii)

To solve these equations, we need to make one of the variables (in both the equations) have same coefficient.

Multiply equation (i) by 36 and equation (ii) by 10, so that the variables p and q in both the equations have same coefficients.

Recalling equations (i) and (ii),

3p + 10q = -9 [×36

75p – 36q = 61 [×10

⇒ 858p = 286

⇒ p = 286/858 = 1/3

Substitute p = 1/3 in eq.(i)/eq.(ii), as per convenience of solving.

Thus, substituting in eq.(i), we get

3(1/3) + 10q = -9

⇒ 1 + 10q = -9

⇒ 10q = -9-1 = -10

⇒ q = -1

Thus, p=1/3 and q=-1

As p = 1/(x + 2y),

⇒

⇒ x + 2y = 3 …(iii)

And q = 1/(3x – 2y)

⇒

⇒ 2y – 3x = 1 …(iv)

Subtracting equations (iii) and (iv) to obtain x and y,

(x + 2y) – (2y – 3x) = 3 – 1

⇒ x + 2y – 2y + 3x = 2

⇒ 4x = 2

⇒ x = 1/2

Putting the value of x in equation (iv), we get

2y – 3(1/2) = 1

⇒ 4y – 3 = 2

⇒ 4y = 2 + 3 = 5

⇒ y = 5/4

Hence, we have x=1/2 and y=5/4