Question

1. x/a+y/b = a+b

x/a²+y/b² =2

Answer 1

Given (x/a) + (y/b) = a + b
Multiply by (1/a) both sides, we get
(x/a2) + (y/ab) = 1 + (b/a) → (1)
Given other linear equation as (x/a2) + (y/b2) = 2  → (2)
Subtract (2) from (1), we get
(x/2) + (y/ab) = 1 + (b/a)
(x/a2) + (y/b2) = 2 
----------------------------------
(y/ab) − (y/b2) = (b/a) − 1
⇒ y[(b − a)/ab2] = (b − a)/a
∴ y = b2
Put y = b2 in (x/a2) + (y/b2) = 2 
⇒ (x/a2) + (b2/b2) = 2 
⇒ (x/a2) + 1 = 2 
⇒ (x/a2)  = 1
∴ x = a2

Hope it will help you .....
If you found my answer helpful than pls mark it as brainlist

Answer 2

Hey Sista!!!!!!!


GIVEN :-

$= \frac{x}{a} + \frac{y}{b} = a + b \: \: \: \: ....(eq _{1}) \\ \\ = > \frac{x}{ {a}^{2} } + \frac{y}{ {b}^{2} } = 2 \: \: \: ...(eq _{2})$
_____________________

Now Eq(1) multiple by "1/a"

$= > \frac{x}{ {a}^{2} } + \frac{y}{ab} = \frac{a + b}{a} ...(eq _{3})$
____________________

Now

by [Eq(3)-Eq(2)] we get :-

$= > \frac{y}{ab} - \frac{y}{ {b}^{2} } = \frac{a + b}{a} - 2 \\ \\ = > y( \frac{(b - a)}{a {b}^{2} } ) = \frac{(b - a)}{a} \\ \\ = > y = {b}^{2} \: \: \: \:$
_____________[ANSWER]
__________________

Plug the value of "y" in Eq(1)

$= > \frac{x}{a} + \frac{ {b}^{2} }{b} = a + b \\ \\ = > x = {a}^{2}$
____________[ANSWER]

======================================

Hence the answer is

=> X = a²
=> y = b²

==============