## 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

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

## Answers ( )

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

=