Question

(4xy/2-x+2x/2y)-(2xy/2+3x+2x/2y

Answer 1

(4xy/2-x+2x/2y)-(2xy/2+3x+2x/2y) ;
(2xy-3x/2y)-(xy+5x/2y) ;
2xy-3x/2y-xy-5x/2y ;
xy-3x/2y-5x/2y ;
2xy^2-3x-5x/2y ;
2xy^2-8x/2y ;
2(xy^2-4x)/2y ;
xy^2-4x/y .

Answer 2

To simplify the given expression:

(4xy/(2-x)) + (2x/(2y)) - (2xy/(2+3x)) - (2x/(2y))

First, let's simplify each fraction individually.

The first fraction (4xy/(2-x)) can be rewritten as (4xy/(-x+2)).

The second fraction (2x/(2y)) remains unchanged.

The third fraction (2xy/(2+3x)) can be simplified to (2xy/(3x+2)).

The fourth fraction (2x/(2y)) remains unchanged.

Now, we can combine like terms by finding a common denominator for each pair of fractions.

For the first and third fractions, the common denominator is (-x+2)(3x+2).

For the second and fourth fractions, the common denominator is 2y.

After finding the common denominators, the expression becomes:

[(4xy*(3x+2))/(2y*(-x+2))] + [(2x*(3x+2))/(2y*(3x+2))] - [(2xy*(2y))/(2y*(3x+2))] - [(2x*(-x+2))/(2y*(-x+2))]

Now, we can simplify further by canceling out common factors:

[(12x^2y + 8xy)/(2y*(-x+2))] + [(6x^2 + 4x)/(2y*(3x+2))] - [(4xy)/(3x+2)] - [(2x*(-x+2))/(2y*(-x+2))]

Simplifying the expression gives us the final result.