Question

bx/a -ay/b + a +b = 0,
bx-ay + 2ab = 0.
​

Answer 1

• Given equations :-

$\dfrac{bx}{a}$ - $\dfrac{ay}{b}$ + a + b = 0

bx - ay + 2ab = 0

We have to find the value of a and b.

Now..

$\dfrac{bx}{a}$ - $\dfrac{ay}{b}$ + a + b = 0

$\dfrac{ {b}^{2} x \: - \: {a}^{2}y }{ab}$ = - a - b

b²x - a²y = - a²b - ab² ___(eq. 1)

Also

bx - ay + 2ab = 0 __(eq. A)

Multiply the eq. A with b

b²x - aby + 2ab² = 0

b²x - aby = - 2ab² __(eq. 2)

Now.. Subtract eq. 1 from eq. 2

b²x - a²y - (b²x - aby) = - a²b - ab² - (- 2ab²)

b²x - a²y - b²x + aby = - a²b - ab² + 2ab²

- a²y + aby = - a²b + ab²

ay (- a + b) = ab (-a + b)

(- a + b) cancels on both sides.. we left with

ay = ab

Also a cancels on both sides.. So

y = b

Put value of y in eq. 2

b²x - ab (b) = - 2ab²

b²x - ab² = - 2ab²

b²x = - 2ab² + ab²

b²x = - ab²

b² cancels on both sides .. we left with

x = - a