Solve for x and y:\(\frac{bx}a+\frac{ay}b=a^2+b^2\),x + y = 2ab

1.	Solve for x and y:\(\frac{bx}a+\frac{ay}b=a^2+b^2\),x + y = 2ab
Answer» The given equations are: \(\frac{bx}a+\frac{ay}b=a^2+b^2\) By taking LCM, we get: b²+ a²y/ab = a² + b² ⇒ b² x + a² y = (ab)a² + b² ⇒b² x + a² y = a³ b +ab³ …..(i) Also, x + y = 2ab …….(ii) On multiplying (ii) by a² , we get: a² x + a² y = 2a³ b ……(iii) On subtracting (iii) from (i), we get: (b² – a²)x = a³ b +ab³ – 2a³ b ⇒ (b² – a²)x = -a³ b +ab³ ⇒ (b² – a² )x = ab(b² – a²) ⇒(b² – a²)x = ab(b² – a²) ∴x = ab(b²− a²) (b²− a²) = ab On substituting x = ab in (i), we get: b² (ab) + a² y = a³ b + ab³ ⇒ a² y = a³ b ⇒ a³b/a² = ab Hence, the solution is x = ab and y = ab.

Answer»

The given equations are:

\(\frac{bx}a+\frac{ay}b=a^2+b^2\)

By taking LCM, we get:

b²+ a²y/ab = a² + b²

⇒ b² x + a² y = (ab)a² + b²

⇒b² x + a² y = a³ b +ab³ …..(i)

Also, x + y = 2ab …….(ii)

On multiplying (ii) by a² , we get:

a² x + a² y = 2a³ b ……(iii)

On subtracting (iii) from (i), we get:

(b² – a²)x = a³ b +ab³ – 2a³ b

⇒ (b² – a²)x = -a³ b +ab³

⇒ (b² – a² )x = ab(b² – a²)

⇒(b² – a²)x = ab(b² – a²)

∴x = ab(b²− a²) (b²− a²) = ab

On substituting x = ab in (i), we get:

b² (ab) + a² y = a³ b + ab³

⇒ a² y = a³ b

⇒ a³b/a² = ab

Hence, the solution is x = ab and y = ab.

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