Solve for x and y:\(\frac{5}{x+1}+\frac{2}{y-1}\) = \(\frac{1}2,\fra

1.	Solve for x and y:\(\frac{5}{x+1}+\frac{2}{y-1}\) = \(\frac{1}2,\frac{10}{x+1}-\frac{2}{y-1}\)= \(\frac{5}2\),where x ≠ 1, y ≠ 1.
Answer» The given equations are \(\frac{5}{x+1}+\frac{2}{y-1}\) = \(\frac{1}2\)..........(i) \(\frac{10}{x+1}-\frac{2}{y-1}\)= \(\frac{5}2\).........(ii) Substituting 1/x+1 = u and 1/y−1 = v, we get: 5u - 2v = 1/2 ……..(iii) 10u + 2v = 5/2 …….(iv) On adding (iii) and (iv), we get: 15u = 3 ⇒u = 3/15 = 1/5 ⇒ 1/x+1 = 1/5 ⇒ x + 1 = 5 ⇒ x = 4 On substituting u = 1/5 in (iii), we get 5 × 1/5 - 2v = 1/2 ⇒ 1 – 2v = 1/2 ⇒2v = 1/2 ⇒v = 1/4 ⇒ 1/y−1 = 1/4 ⇒ y – 1 = 4 ⇒ y = 5 Hence, the required solution is x = 4 and y = 5.

Solve for x and y:\(\frac{5}{x+1}+\frac{2}{y-1}\) = \(\frac{1}2,\frac{10}{x+1}-\frac{2}{y-1}\)= \(\frac{5}2\),where x ≠ 1, y ≠ 1.

Answer»

The given equations are

\(\frac{5}{x+1}+\frac{2}{y-1}\) = \(\frac{1}2\)..........(i)

\(\frac{10}{x+1}-\frac{2}{y-1}\)= \(\frac{5}2\).........(ii)

Substituting 1/x+1 = u and 1/y−1 = v, we get:

5u - 2v = 1/2 ……..(iii)

10u + 2v = 5/2 …….(iv)

On adding (iii) and (iv), we get:

15u = 3

⇒u = 3/15 = 1/5

⇒ 1/x+1 = 1/5 ⇒ x + 1 = 5 ⇒ x = 4

On substituting u = 1/5 in (iii), we get

5 × 1/5 - 2v = 1/2 ⇒ 1 – 2v = 1/2

⇒2v = 1/2 ⇒v = 1/4

⇒ 1/y−1 = 1/4 ⇒ y – 1 = 4 ⇒ y = 5

Hence, the required solution is x = 4 and y = 5.