\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

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.​

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Alexandra 1 month 2021-08-21T12:00:47+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-08-21T12:02:23+00:00

    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 \ }}}

    0
    2021-08-21T12:02:43+00:00

    \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” .

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