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
Answer:
Step-by-step explanation:
Solution:
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Given:
a(x+y) + b(x-y) = a² - ab + b² ..(i).,
a(x+y) - b(x-y) = a² + ab - b² ...(ii)
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To find,
The values of x and y.
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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),
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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)
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Adding (iii) & (iv),
We get,
=> (x + y) + (x - y) a + b- a
=> 2x = b
=> ∴
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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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