How do I simplify $$1 - (x + y) \over 2 -x -y$$

I'm not sure whether it is $ 1 - x + y \over 2 - x -y$ and then $1+y \over 2 -y$

Or, if the negative applies to both values in parenthesis

and it becomes $1 - x - y \over 2 - x -y$ and then it just becomes $1\over2$

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    $\begingroup$ Try adding $\frac1{2-x-y}$ to this fraction and see what happens. What does this tell you? $\endgroup$ – Don Thousand Oct 22 '18 at 18:14
  • $\begingroup$ Why don't you just try plugging in some numbers instead of $x$ and $y$? Then you'll discover the answer yourself. $\endgroup$ – Hans Lundmark Oct 22 '18 at 20:01

When you open bracket, minus sign is applied to both the terms inside the bracket,
i.e. $$1-(x+y)=1-x-y$$
So, fraction becomes: $$\frac{1-x-y}{2-x-y}$$

Now, you are cancelling out (-x-y) from numerator and denominator, which is not permissible, since, it is not in multiplication, but rather, in subtraction.

If it were $$\frac{1{(x+y)}}{2{(x+y)}}$$ you could cancel out (x+y) and write it as $$\frac{1}{2}$$

But, right now, it just stays as $$\frac{1-x-y}{2-x-y}$$


Imagine the minus sign as $-1$. Then, it would be :

$$\frac{1-(x+y)}{2-x-y} =\frac{1 + (-1)(x+y)}{2-x-y}$$

Recall that $a(b+c) = ab + ac$, thus it would be multiplied with each variable in the parenthesis, yielding :


Now, it is wrong to simplify the $-x-y$ terms from the expression. Simplification stems from the operation of multiplication, since inverse elements cancel out. This does not happen in the case of subtraction.


It becomes $$\frac{1-x-y}{2-x-y}\ne \frac{1}{2}$$


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