(6x-2y)/(x^2+xy-2y^2)-(3x+2y)/(x^2+3xy-4y^2)
x^2+xy-2y^2 = (x+2y)(x-y)
x^2+3xy-4y^2 = (x+4y)(x-y)
so, the LCD = (x+2y)(x+4y)(x-y) and putting it all over that LCD, the numerator then becomes
(6x-2y)(x+4y) - (3x+2y)(x+2y)
= 6x^2 +22xy - 8y^2
- 3x^2 +8xy + 4y^2
= 3x^2 + 14xy - 12y^2
so, the final fraction is
3x^2 + 14xy - 12y^2
-------------------------
(x+2y)(x+4y)(x-y)
To simplify the expression (6x-2y)/(x^2+xy-2y^2) - (3x+2y)/(x^2+3xy-4y^2), we need to find a common denominator for the two fractions and then combine them.
1. First, let's find the common denominator for the two fractions.
The denominators are x^2+xy-2y^2 and x^2+3xy-4y^2.
2. Next, factorize the denominators.
The first denominator x^2+xy-2y^2 can be factored as (x+y)(x-2y).
The second denominator x^2+3xy-4y^2 can be factored as (x+4y)(x-y).
The factored denominators are (x+y)(x-2y) and (x+4y)(x-y) respectively.
3. Now, we can rewrite the expression with the common denominators.
(6x-2y)/(x+y)(x-2y) - (3x+2y)/(x+4y)(x-y)
4. To subtract the fractions, we need to multiply each fraction by the missing factor in the denominator of the other fraction.
For the first fraction (6x-2y)/(x+y)(x-2y), multiply the numerator and denominator by (x+4y)(x-y).
(6x-2y)(x+4y)(x-y) / (x+y)(x-2y)(x+4y)(x-y)
For the second fraction (3x+2y)/(x+4y)(x-y), multiply the numerator and denominator by (x+y)(x-2y).
(3x+2y)(x+y)(x-2y) / (x+y)(x-2y)(x+4y)(x-y)
5. Now, simplify the expression by multiplying out the numerators.
Numerator for the first fraction: (6x-2y)(x+4y)(x-y) = 6x^2 - 2xy + 24xy - 8y^2 - 6x^2 + 2xy - 8xy + 2y^2 = 12xy + 10xy - 6y^2 = 22xy - 6y^2
Numerator for the second fraction: (3x+2y)(x+y)(x-2y) = 3x^3 - 6x^2y + 2xy^2 + 3xy^2 + 6x^2y - 2xy^2 = 3x^3 + 5xy^2
6. Finally, simplify the expression by canceling out the common factors.
(22xy - 6y^2) / (x+y)(x-2y)(x+4y)(x-y) - (3x^3 + 5xy^2) / (x+y)(x-2y)(x+4y)(x-y)
Therefore, the simplified expression is (22xy - 6y^2 - 3x^3 - 5xy^2) / (x+y)(x-2y)(x+4y)(x-y).