Solve for x and y:\(\frac{bx}{a}-\frac{ay}b+a + b =0\),bx – ay + 2ab

1.	Solve for x and y:\(\frac{bx}{a}-\frac{ay}b+a + b =0\),bx – ay + 2ab = 0
Answer» The given equations are: bx/a - ay/b + a + b = 0 By taking LCM, we get: b²x – a²y = - a²b – b²a …..(i) and bx – ay + 2ab = 0 bx – ay = -2ab …….(ii) On multiplying (ii) by a, we get: abx – a²y = - 2a²b ……(iii) On subtracting (i) from (iii), we get: abx – b²x = 2a²b + a²b + b²a = - a²b + b²a ⇒x(ab – b²) = -ab(a – b) ⇒x(a – b)b = -ab(a – b) ∴x = −ab(a−b)/(a−b)b = -a On substituting x = -a in (i), we get: b² (-a) – a² y = -a² b – b² a ⇒ -b²a – a²y = -a²b – b²a ⇒ -a² y = -a² b ⇒ y = b Hence, the solution is x = -a and y = b.

Solve for x and y:\(\frac{bx}{a}-\frac{ay}b+a + b =0\),bx – ay + 2ab = 0

Answer»

The given equations are:

bx/a - ay/b + a + b = 0

By taking LCM, we get:

b²x – a²y = - a²b – b²a …..(i)

and bx – ay + 2ab = 0

bx – ay = -2ab …….(ii)

On multiplying (ii) by a, we get:

abx – a²y = - 2a²b ……(iii)

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

abx – b²x = 2a²b + a²b + b²a = - a²b + b²a

⇒x(ab – b²) = -ab(a – b)

⇒x(a – b)b = -ab(a – b)

∴x = −ab(a−b)/(a−b)b = -a

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

b² (-a) – a² y = -a² b – b² a

⇒ -b²a – a²y = -a²b – b²a

⇒ -a² y = -a² b

⇒ y = b

Hence, the solution is x = -a and y = b.