Math, asked by amritharaobatch1, 6 months ago

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

Answers

Answered by avirajsingh26
1

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

=

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