solve the following a(x+y)+b(x-y)-(a²-ab+b²)=0 and a(x+y)-b(x-y)-(a²+ab+b²)=0​

solve the following a(x+y)+b(x-y)-(a²-ab+b²)=0 and a(x+y)-b(x-y)-(a²+ab+b²)=0​

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1 thought on “solve the following a(x+y)+b(x-y)-(a²-ab+b²)=0 and a(x+y)-b(x-y)-(a²+ab+b²)=0​”

  1. Step-by-step explanation:

    Taking x+y=u and x−y=v the given system of equations becomes

    au+bu−(a

    2

    −ab+b

    2

    )=0

    au−bv−(a

    2

    +ab+b

    2

    )=0

    By cross-multiplication, we have

    b×−(a

    2

    +ab+b

    2

    )−(−b)×−(a

    2

    −ab+b

    2

    )

    u

    =

    a×−(a

    2

    +ab+b

    2

    )+a(a

    2

    −ab+b

    2

    )

    −v

    =

    a×−b−a×b

    1

    −b(a

    2

    +ab+b

    2

    )−−b(a

    2

    −ab+b

    2

    )

    u

    =

    −a(a

    2

    +ab+b

    2

    )+a(a

    2

    −ab+b

    2

    )

    −v

    =

    −ab−ab

    1

    −b(a

    2

    +ab+b

    2

    +a

    2

    −ab+b

    2

    )

    u

    =

    −a(a

    2

    +ab+b

    2

    −a

    2

    +ab−b

    2

    )

    −v

    =

    −2ab

    1

    −2b(a

    2

    +b

    2

    )

    u

    =

    −a(2ab)

    −v

    =

    −2ab

    1

    ⇒u=

    −2ab

    −2b(a

    2

    +b

    2

    )

    ,v=

    −2ab

    2a

    2

    b

    ⇒u=

    a

    a

    2

    +b

    2

    ,v=−a

    Now, u=

    a

    a

    2

    +b

    2

    ⇒x+y=

    a

    a

    2

    +b

    2

    .(i)

    and, v=−a⇒x−y=−a ..(ii)

    Adding equations (i) and (ii), we get

    2x=

    a

    a

    2

    +b

    2

    −a⇒2x=

    a

    a

    2

    +b

    2

    −a

    2

    ⇒2x=

    a

    b

    2

    ⇒x=

    2a

    b

    2

    Substitutiing equation (ii) from equation (i), we get

    2y=

    a

    a

    2

    +b

    2

    +a⇒2y=

    a

    a

    2

    +b

    2

    +a

    2

    ⇒y=

    2a

    2a

    2

    +b

    2

    Hence, the solution of the given system of equations is x=

    2a

    b

    2

    ,y=

    2a

    2a

    2

    +b

    2

    .

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