Question

plz solve this
x?y?​

Answer 1

Answer:

$x = a^{2}$

$y = b^{2}$

Step-by-step explanation:

The given equations is-

$\dfrac{x}{a} +\dfrac{y}{b} = (a + b)$           ------------ 1

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

Multiply in equation 1 by $\dfrac{1}{a}$

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

$\dfrac{x}{a^{2} } + \dfrac{y}{b^{2} } = 2$  

              Subtracting

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

$y[\dfrac{1}{ab} -\dfrac{1}{b^{2} }] = \dfrac{a+b-2b}{2}$

$y[\dfrac{b-a}{ab^{2} }] = \dfrac{b-a}{a}$

$\dfrac{y}{ab^{2} } = \dfrac{1}{a}$

$y = \dfrac{ab^{2} }{a}$

$y = b^{2}$

Put $y = b^{2}$ in equation 1

$\dfrac{x}{a} +\dfrac{y}{b} = (a + b)$

$\dfrac{x}{a} + \dfrac{b^{2} }{b} = a+b$

$\dfrac{x}{a} + b = a+b$

$\dfrac{x}{a} = a$

$x = a^{2}$

Hence, the value of x and y is $a^{2} \ and \ y^{2}$.