Question

bx/a -ay/b -a+b=0
bx+ay+2ab=0

Answer 1

solving first given eq.
bx/a - ay/b-a+b=0
(b2x-a2y)/ab=a-b
b2x - a2y = a2b - ab2 .....(i)

Now the second
bx+ay+2ab=0
bx+ay= -2ab
[multiplying with a the whole equation]
abx+a2y= -2a2b .......(ii)

Adding eq. (i) and (ii), we get
b2x + abx -a2y + a2y = a2b -2a2b - ab2
bx(b+a) = -a2b - ab2
bx(a+b) = -ab(a+b)
bx = -ab
x = -a
putting the value in eq.(ii)
ab(-a) + a2y = -2a2b
-a2b + a2y = -2a2b
a2y = -2a2b + a2b
a2y = -a2b
y = -b
hope it will make you understand...!!

Answer 2

The value of x and y are -a and -b respectively.

Solution:

Solving initial given equation.

$\frac{b x}{a}-\frac{a y}{b}-a+b=0$

$\frac{b^{2} x-a^{2} y}{a b}=a-b$

$b^{2} x-a^{2} y=a^{2} b-a b^{2} \rightarrow(i)$

Now the second equation,

bx+ay+2ab = 0

bx+ay = -2ab

Multiplying (a) with the above equation,

$a b x+a^{2} y=-2 a^{2} b \rightarrow(i i)$

Adding equation (i) and (ii), we get,

$b^{2} x+a b x-a^{2} y+a^{2} y=a^{2} b-2 a^{2} b-a b^{2}$

$b^{2} x+a b x=-a^{2} b-a b^{2}$

$b x(b+a)=-a b(a+b)$

bx = -ab

x = -a

Putting the value in equation (ii),

$a b(-a)+a^{2} y=-2 a^{2} b$

$-a^{2} b+a^{2} y=-2 a^{2} b$

$a^{2} y=-2 a^{2} b+a^{2} b$

$a^{2} y=-a^{2} b$

y=-b