a(x+y)+b (x-y)=a^2-ab+b^2;a(x+y)-b(x-y)=a^2+ab+b^2

1.	a(x+y)+b (x-y)=a^2-ab+b^2;a(x+y)-b(x-y)=a^2+ab+b^2
Answer» 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=> x = b/2 substitute value of x in (iv)y= a - b/2

Answer»

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=> x = b/2

substitute value of x in (iv)y= a - b/2

Discussion