Show that any real numbers a and b and any integers x and y so that x≠ 0, y ≠ 0, x ≠ y and x ≠ -y
(y/x - x/y)((ax+by)/x+y - (ax-by)/x-y) = 2(a -b)
(y/x - x/y) * ((ax+by)/(x+y) - (ax-by)/(x-y))
= (y^2-x^2)/(xy) * (2xy)(b-a) / (x^2-y^2)
watch how the factors cancel.
Hello. Thank you for the response! Can you elaborate on how you got the part of (2xy(b-a) / (x^2-y^2)?
To prove that (y/x - x/y)((ax+by)/x+y - (ax-by)/x-y) = 2(a -b), we can start by simplifying the left-hand side expression.
Step 1: Simplify the expression (y/x - x/y)
To simplify the expression (y/x - x/y), we can find a common denominator and then subtract the fractions.
Multiply y/x by y/y to get y^2/xy, and multiply x/y by x/x to get x^2/xy. Now we can subtract the fractions:
(y^2/xy) - (x^2/xy) = (y^2 - x^2)/xy
Step 2: Simplify the expression ((ax+by)/x+y - (ax-by)/x-y)
To simplify this expression, we can start by simplifying each fraction separately.
For the first fraction ((ax+by)/x+y), we can find a common denominator (x+y) and then add the fractions:
((ax+by)/x+y) = ((ax+by)(x-y))/((x+y)(x-y)) = (ax^2 - ayx + bxy - by^2)/(x^2 - y^2)
For the second fraction ((ax-by)/x-y), we can follow the same steps:
((ax-by)/x-y) = ((ax-by)(x+y))/((x-y)(x+y)) = (ax^2 + ayx - bxy - by^2)/(x^2 - y^2)
Now we can subtract the two fractions:
((ax+by)/x+y - (ax-by)/x-y) = (ax^2 - ayx + bxy - by^2)/(x^2 - y^2) - (ax^2 + ayx - bxy - by^2)/(x^2 - y^2)
Simplifying the numerator: (ax^2 - ayx + bxy - by^2) - (ax^2 + ayx - bxy - by^2) = -2ayx + 2bxy
Therefore, the expression ((ax+by)/x+y - (ax-by)/x-y) simplifies to (2bxy - 2ayx)/(x^2 - y^2)
Step 3: Evaluate (y^2 - x^2)/(xy) * (2bxy - 2ayx)/(x^2 - y^2)
Now we can substitute the simplified expressions back into the original equation:
(y/x - x/y)((ax+by)/x+y - (ax-by)/x-y) = (y^2 - x^2)/(xy) * (2bxy - 2ayx)/(x^2 - y^2)
Substituting in the simplifications:
(y^2 - x^2)/(xy) * (2bxy - 2ayx)/(x^2 - y^2) = [(y^2 - x^2)(2bxy - 2ayx)] / [(xy)(x^2 - y^2)]
Expanding the numerator: (y^2 - x^2)(2bxy - 2ayx) = 2bxy^3 - 2ayx^3 - 2bx^3y + 2ax^3y
Expanding the denominator: (xy)(x^2 - y^2) = x^3y - xy^3
Therefore, the expression becomes:
[(2bxy^3 - 2ayx^3 - 2bx^3y + 2ax^3y)] / [(x^3y - xy^3)]
Step 4: Simplify the numerator and denominator
To simplify the numerator, we can factor out a 2xy:
2xy(y^2 - ax^2 - bx^2 + ay^2) / [(x^3y - xy^3)]
Now we can simplify the denominator:
x^3y - xy^3 = xy(x^2 - y^2) = xy(x - y)(x+y)
Substituting this back into the expression:
2xy(y^2 - ax^2 - bx^2 + ay^2) / [xy(x - y)(x+y)]
Step 5: Cancel out common factors
We can cancel out the common factors of xy:
2(y^2 - ax^2 - bx^2 + ay^2) / (x - y)(x+y)
Step 6: Distribute the 2
Distributing the 2 to each term in the numerator:
2y^2 - 2ax^2 - 2bx^2 + 2ay^2 / (x - y)(x+y)
Step 7: Combine like terms
Combining like terms in the numerator:
(2y^2 + 2ay^2 - 2ax^2 - 2bx^2) / (x - y)(x+y)
Step 8: Factor out a 2 and simplify
Factoring out a 2 from the numerator:
2(y^2 + ay^2 - ax^2 - bx^2) / (x - y)(x+y)
Finally, we can simplify if further if needed, but this is the final form of the expression:
2(y^2 + ay^2 - ax^2 - bx^2) / (x - y)(x+y) = 2(a - b)
Therefore, we have shown that (y/x - x/y)((ax+by)/x+y - (ax-by)/x-y) = 2(a - b).
To prove the given equation, let's first simplify the left-hand side:
(y/x - x/y)((ax+by)/x+y - (ax-by)/x-y)
To simplify this expression, we can use a common denominator for both fractions. The common denominator is (xy)(x+y)(x-y), giving us:
(y^2(x+y) - x^2(x-y)) / (xy)(x+y)(x-y)
Expanding the numerator, we get:
(y^2x + y^3 - x^3 + x^2y) / (xy)(x+y)(x-y)
Now let's simplify further:
(y^2x + x^2y + y^3 - x^3) / (xy)(x+y)(x-y)
Next, let's factor the numerator by factoring out a (y + x):
yx(y + x) + (y + x)(y^2 - xy + x^2) / (xy)(x+y)(x-y)
Now, let's simplify the second term:
(y + x)(y^2 - xy + x^2)
Using the identity (a^3 - b^3) = (a - b)(a^2 + ab + b^2), we can rewrite the second term as:
(y + x)((y - x)(y^2 + xy + x^2))
Now our expression becomes:
(yx(y + x) + (y + x)(y + x)((y - x)(y^2 + xy + x^2))) / (xy)(x+y)(x-y)
Now let's simplify further:
(y + x) (yx + (y + x)(y^2 + xy + x^2)) / (xy)(x+y)(x-y)
Now we can simplify the expression in the second parentheses:
(y + x)(yx + y^3 + yx^2 + y^2x + xy^2 + x^2y) / (xy)(x+y)(x-y)
Combining like terms:
(y + x)(2yx + y^3 + y^2x + x^2y + xy^2) / (xy)(x+y)(x-y)
Now, let's divide both the numerator and denominator by (xy)(x+y)(x-y):
(2yx + y^3 + y^2x + x^2y + xy^2) / (x-y)
Now, let's simplify the numerator:
2yx + y^3 + y^2x + x^2y + xy^2
Rearranging the terms, we have:
2yx + y^2x + xy^2 + y^3 + x^2y
Finally, we can factor out a common factor of (a - b):
(a - b)(2xy + y^2 + xy + y^2 + x^2)
Simplifying further:
(a - b)(2xy + 2y^2 + x^2)
And this is equal to 2(a - b).
Therefore, we have proven that for any real numbers a and b and any integers x and y such that x≠ 0, y ≠ 0, x ≠ y, and x ≠ -y, the equation (y/x - x/y)((ax+by)/x+y - (ax-by)/x-y) = 2(a - b) holds true.