Question

$\sf\large\underline\purple{Question:-}$
Solve the value for x and y =?
(a-b)x+(a+b)y=a²-2ab+b²
(a+b)(x+y)=a²-b²
Please answer in good content
wrong answer will be deleted at the spot.​

Answer 1

$\bf{\gray{\underbrace{\blue{GIVEN:-}}}}$

1. $\bf\red{(a\:-\:b)\:x\:+\:(a\:+\:b)\:y\:=\:a^2\:-\:2ab\:+\:b^2\:}$
2. $\bf\red{(a\:+\:b)\:(x\:+\:y)\:=\:a^2\:-\:b^2\:}$

$\bf{\gray{\underbrace{\blue{TO\:FIND:-}}}}$

• The value of "x" and "y" .

$\bf{\gray{\underbrace{\blue{SOLUTION:-}}}}$

☯︎ First of all solve the first given equation .

$\pink\checkmark \: (a \: - \: b)x \: + \: (a \: + \: b)y \: = \: {a}^{2} - 2ab + {b}^{2} \\ \\ :\implies \: ax \: - \: bx \: + \: ay \: + \: by \: = \: {a}^{2} - 2ab + {b}^{2} \\ \\ :\implies \: a(x \: + \: y) - \: b(x \: - \: y) \: = \: {a^2 - 2ab + b^2} ---- (1)$

☯︎ Secondly, solve the second given equation .

$\orange\checkmark \: (a \: + \: b) \: (x \: + \: y) \: = \: {a}^{2} - {b}^{2} ---- (2) \\ \\ :\implies \: a(x \: + \: y) \: + \: b(x \: + \: y) \: = \: {a}^{2} - {b}^{2}$

$\bf\green{:\implies\:a\:(x\:+\:y)\:=\:a^2\:-\:b^2\:-\:b(x\:+\:y)\:}$

➪ Now, putting the value of "a(x + y)" in the equation (1);

$:\implies \: \Big\{ {a}^{2} - {b}^{2} - b(x \: + \: y) \Big\} \: - b(x \: - \: y) \: = \: {a}^{2} - 2ab + {b}^{2} \\ \\ :\implies \: \cancel{{a}^{2}} - {b}^{2} - bx - \cancel{by} - bx + \cancel{by} \: = \: \cancel{{a}^{2}} - 2ab + {b}^{2} \\ \\ :\implies \: - 2bx \: = \: - 2ab \: + \: 2 {b}^{2} \\ \\ :\implies \: - \cancel{2b}x \: = \: \cancel{2b}( - a \: + \: b) \\ \\ :\implies \: - x \: = \: - a \: + \: b$

$\bf\purple{:\implies\:x\:=\:(a\:-\:b)\:}$

➪ Now, putting the value of "x" in the equation (2);

$:\implies \: (a \: + \: b) \: \Big\{ (a \: - \: b) \: + \: y \Big\} \: = \: {a}^{2} - {b}^{2} \\ \\ :\implies \: (a \: - \:b) \: + \: y \: = \: \frac{(a \: - \: b) \: \cancel {(a \: + \: b)}}{\cancel {(a \: + \: b)}} \\ \\ :\implies \: y \: = \: a \: - \: b \: - \: (a \: - \: b) \\ \\ :\implies \: y \: = \: \cancel{a} \: - \: \cancel{b} \: - \: \cancel{a} \: + \: \cancel{b}$

$\bf\blue{:\implies\:y\:=\:0\:}$

$\red\therefore$ The value of x is "(a - b)" and the value of y is "0" .

Answer 2

Question :

Solve the value for x and y =?

(a-b)x+ (a+b)y = a²- 2ab + b²

(a+b) (x+y) = a²- b²

Solution :

Given :

⇒ (a-b)x + (a+b)y = a²- 2ab + b² .............. (i)

⇒ (a + b)(x + y) = a² - b²

⇒ (a + b)x + (a + b)y = a² - b².............. (ii)

Now substracting (i) from (ii) :-

⇒ (a + b)x + (a + b)y - (a - b)x - (a + b)y = a² - b² - a² + 2ab - b²

⇒ x(a + b - a + b) = 2ab - 2b²

⇒ x(2b) = 2b(a - b)

⇒ x = 2b(a - b)/2b

⇒ x = a - b

Now putting value of 'x' in equation (i) :-

⇒ (a - b)(a - b) + (a + b)y = a² - 2ab + b²

⇒ (a - b)² + (a + b)y = (a - b)²

⇒ (a + b)y = (a - b)² - (a - b)²

⇒ (a + b)y = 0

⇒ y = 0/(a + b)

⇒ y = 0

$\therefore\underline{\boxed{\bold{Required \ value \ of \ x \ = \ a - b \ }}}$

$\therefore\underline{\boxed{\bold{Required \ value \ of \ y \ = \ 0 \ }}}$