Solve the following system of equations:2/x + 3/y = 2; 4/x – 9/y =

1.	Solve the following system of equations:2/x + 3/y = 2; 4/x – 9/y = -1
Answer» Let 1/√x = u and 1/√y = v, So, the given equations becomes 2u + 3v = 2…………(i) 4u – 9v = -1 ……….(ii) Multiplying (ii) by 3 and Adding equation (i) and (ii) x 3 we get, 6u + 9v + 4u – 9v = 6 – 1 ⇒ 10u = 5 ⇒ u = \(\frac{1}{2}\) Substituting u = \(\frac{1}{2}\) in (i), we find v 2(\(\frac{1}{2}\)) + 3v = 2 ⇒ 3v = 2 – 1 ⇒ v = \(\frac{1}{3}\) Since, 1/√x = u we get x = 1/u² ⇒ x = 1/(\(\frac{1}{2}\))² = 4 And, 1/√y = v y = 1/v² ⇒ y = 1/(\(\frac{1}{3}\))² = 9 Hence, the solution is x = 4 and y = 9.

Solve the following system of equations:2/x + 3/y = 2; 4/x – 9/y = -1

Answer»

Let 1/√x = u and 1/√y = v,

So, the given equations becomes

2u + 3v = 2…………(i)

4u – 9v = -1 ……….(ii)

Multiplying (ii) by 3 and Adding equation (i) and (ii) x 3 we get,

6u + 9v + 4u – 9v = 6 – 1

⇒ 10u = 5

⇒ u = \(\frac{1}{2}\)

Substituting u = \(\frac{1}{2}\) in (i), we find v

2(\(\frac{1}{2}\)) + 3v = 2

⇒ 3v = 2 – 1

⇒ v = \(\frac{1}{3}\)

Since, 1/√x = u we get x = 1/u²

⇒ x = 1/(\(\frac{1}{2}\))² = 4

And, 1/√y = v y = 1/v²

⇒ y = 1/(\(\frac{1}{3}\))² = 9

Hence, the solution is x = 4 and y = 9.