See below for the proof of the equation [tex]\frac{ax + by}{x + y} - \frac{ax - by}{x - y}= 2(a - b)[/tex]
The equation is given as:
[tex]\frac{ax + by}{x + y} - \frac{ax - by}{x - y}= 2(a - b)[/tex]
Take the LCM
[tex]\frac{(ax + by)(x - y) -(ax - by)(x + y)}{(x + y)(x - y)}= 2(a - b)[/tex]
Expand
[tex]\frac{ax^2 - axy + bxy - by^2 -ax^2 - axy + bxy + by^2}{(x + y)(x - y)}= 2(a - b)[/tex]
Evaluate the like terms
[tex]\frac{-2axy + 2bxy }{(x + y)(x - y)}= 2(a - b)[/tex]
Rewrite as:
[tex]\frac{-2axy + 2bxy }{x^2 - y^2}= 2(a - b)[/tex]
Factorize the numerator
[tex]\frac{2(a - b)(x^2 - y^2)}{x^2 - y^2}= 2(a - b)[/tex]
Divide
2(a - b)= 2(a - b)
Both sides are equal
Hence, the equation [tex]\frac{ax + by}{x + y} - \frac{ax - by}{x - y}= 2(a - b)[/tex] has been proved
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