Question

Please answer fastly​

Answer 1

SOLUTION

$= > \frac{bx}{a} - \frac{ay}{b} + a + b = 0.........(1) \\ = > bx - ay + 2ab = 0 \\ = > bx = ay - 2ab \\ = > x = \frac{ay - 2ab}{b} .............(2)$

Substituting value of (2) in (1),

$\frac{b}{a} ( \frac{ay - 2b}{b} ) - \frac{ay}{b} + a + b = 0 \\ = > \frac{ay - 2ab}{a} - \frac{ay}{b} + a + b = 0 \\ = > \frac{a(y - 2b)}{a} - \frac{ay}{b} + a + b = 0 \\ = > y - 2b - \frac{ay}{b} + a + b = 0 \frac{b - a}{b} \\ = > \frac{y(1 - a)}{b} + a - b = 0 \\ = > \frac{y(1 - a)}{b} = b - a \\ \\ = > y = \frac{(b - a)}{1 - \frac{a}{b} } = \frac{ \frac{(b - a)}{(b - a)} }{b} = > b$

Substituting value of y = b in equation (2).

$x = \frac{a(b) - 2ab}{b} \\ = > x = \frac{ab - 2ab}{b} \\ = > x = - \frac{ab}{b} \\ = > x = - a$

Therefore,

x= -a & y= b

hope it helps ☺️