a(x+y) + b(x-y) – (a^2-ab+b^2) = 0 a(x+y) – b(x-y) – (a^2+ab+b^2) = 0 solve the above equations by addition and subtraction method

Question

a(x+y) + b(x-y) – (a^2-ab+b^2) = 0 a(x+y) – b(x-y) – (a^2+ab+b^2) = 0 solve the above equations by addition and subtraction method

in progress 0
Gabriella 7 days 2021-09-11T12:57:13+00:00 1 Answer 0 views 0

Answers ( )

    0
    2021-09-11T12:59:08+00:00

    Answer:

    Step-by-step explanation:

    Solution:

    _____________________________________________________________

    Given:

    a(x+y) + b(x-y) = a² – ab + b² ..(i).,

    a(x+y) – b(x-y) = a² + ab – b² …(ii)

    _____________________________________________________________

    To find,

    The values of x and y.

    _____________________________________________________________

    Adding both the equations,

    We get,

    => a(x + y) + b(x – y) + a(x + y) – b(x – y) = a² – ab + b² + a² +ab – b²

    => 2a(x + y) = 2a²

    => x + y = a …(iii),

    ____________________

    Subtracting (ii) from (i),

    => a(x + y) + b(x – y) -(a(x + y)  – b(x – y)) = a² – ab + b² – (a² + ab -b²)

    => a(x + y) +  b(x – y) – a(x – y) + b(x – y) = a² – ab + b² – a² – ab + b²

    => 2b(x – y) = -2ab + 2b²

    => 2b(x – y) = 2b² – 2ab

    => 2b(x – y) = 2b(b – a)

    => x – y = b – a ..(iv)

    _______________________

    Adding (iii) & (iv),

    We get,

    => (x + y) + (x – y) a + b- a

    => 2x = b

    => ∴

    __________________________

    Substituting value of x in (iv),

    We get,

    => x – y = b – a

    =>

    =>

    =>

    => ∴Solution:

    _____________________________________________________________

    Given:

    a(x+y) + b(x-y) = a² – ab + b² ..(i).,

    a(x+y) – b(x-y) = a² + ab – b² …(ii)

    _____________________________________________________________

    To find,

    The values of x and y.

    _____________________________________________________________

    Adding both the equations,

    We get,

    => a(x + y) + b(x – y) + a(x + y) – b(x – y) = a² – ab + b² + a² +ab – b²

    => 2a(x + y) = 2a²

    => x + y = a …(iii),

    ____________________

    Subtracting (ii) from (i),

    => a(x + y) + b(x – y) -(a(x + y)  – b(x – y)) = a² – ab + b² – (a² + ab -b²)

    => a(x + y) +  b(x – y) – a(x – y) + b(x – y) = a² – ab + b² – a² – ab + b²

    => 2b(x – y) = -2ab + 2b²

    => 2b(x – y) = 2b² – 2ab

    => 2b(x – y) = 2b(b – a)

    => x – y = b – a ..(iv)

    _______________________

    Adding (iii) & (iv),

    We get,

    => (x + y) + (x – y) a + b- a

    => 2x = b

    => ∴

    __________________________

    Substituting value of x in (iv),

    We get,

    => x – y = b – a

    =

Leave an answer

Browse
Browse

18:9+8+9*3-7:3-1*13 = ? ( )