Question

एक्स अपऑन ए प्लस बी अपॉन बी इक्वल टू ए प्लस बी एंड एक्स अपॉन ए स्क्वायर प्लस वन अपॉन बी स्क्वायर इक्वल टू 2 एंड फाइंड द वैल्यू ऑफ एक्स वाई 55 ​

Answer 1

Solve for x and y :-

$\tt \:  ⟼ \dfrac{x}{a} + \dfrac{y}{b} = a + b$

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

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$\large\underline\purple{\bold{Solution :- }}$

☆ Given Ist equation

$\tt \:  ⟼ \dfrac{x}{a} + \dfrac{y}{b} = a + b$

$\tt \:  ⟼ \dfrac{bx + ay}{ab} = a + b$

$\tt \:  ⟼ bx + ay = {ba}^{2} + {ab}^{2} - - - (1)$

☆ Given second equation

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

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

$\tt \:  ⟼ {xb}^{2} + {ya}^{2} = 2 {a}^{2} {b}^{2} - - - - (2)$

☆ Now, solving equation (1) and (2) by method of elimination.

☆ Multiply equation (1) by b, we get

$\tt \:  ⟼ {xb}^{2} + aby = {a}^{2} {b}^{2} + {ab}^{3} - - - (3)$

☆ Now, Subtracting equation (2) from equation (3), we get

$\tt \:  ⟼ aby - {ya}^{2} = {ab}^{3} - {a}^{2} {b}^{2}$

$\tt \:  ⟼ ay(b - a) = {ab}^{2} (b - a)$

$\bf\implies \:y = { b}^{2} - - - (4)$

☆ Now, substituting equation (4) in equation (2), we get

$\tt \:  ⟼ {xb}^{2} + {a}^{2} {b}^{2} = {2a}^{2} {b}^{2}$

$\tt \:  ⟼ {xb}^{2} = {2a}^{2} {b}^{2} - {a}^{2} {b}^{2}$

$\tt \:  ⟼ {xb}^{2} = {a}^{2} {b}^{2}$

$\bf\implies \:x = {a}^{2} - - - (5)$

$\begin{gathered}\begin{gathered}\bf Hence -  \begin{cases} &\tt{x \: = {a}^{2} } \\ &\tt{y \: = {b}^{2} } \end{cases}\end{gathered}\end{gathered}$