Short expression $\frac{x^2-y^2}{x(x-y)}+\frac{x^2-y^2}{x(x+y)}$

Question

I tried to short this expression: $$\frac{x^2-y^2}{x(x-y)}+\frac{x^2-y^2}{x(x+y)}$$

The result should be $2$ but I get: \begin{align} &\frac{x^2-y^2}{x^2-xy}+\frac{x^2-y^2}{x^2+xy}\\ =& \frac{y^2}{yx}+\frac{-y^2}{yx}\\ =&0 \end{align}

What did I do wrong?

Answer 1

You did not find a common denominator before adding the fractions since $x^2-xy\neq x^2+xy$. Moreover, you cannot add fractions 'top to bottom', that is $$ \frac{2}{3}+\frac{1}{4} \neq \frac{3}{7} $$ this is why we find the common denominator. Try finding the common denominator first, it will be ....

$x(x-y)(x+y)$

and then adding your fractions.

Answer 2

Don't know how you got from $\frac{x²−y²}{x²−yx}$ to $\frac{y^2}{xy}$,

but fractions just don't work that way. To add those, you need a common denominator. Here you could use $x(x-y)(x+y)$, but it's a lot faster to note that $x^2-y^2=(x+y)(x-y)$ and then simplify both fractions so that they get common denominator $x$.

Answer 3

Indeed, $$\frac{x^2-y^2}{x(x-y)}+\frac{x^2-y^2}{x(x+y)}=\frac{x^2-y^2}{x}\left(\frac{1}{x-y}+\frac{1}{x+y}\right)=\frac{x^2-y^2}{x}\left(\frac{x+y+x-y}{(x+y)(x-y)}\right)\\=\frac{(x-y)(x+y)}{x}\left(\frac{2x}{(x+y)(x-y)}\right)=2,~~~x\neq0,~~x\neq\pm y$$

Answer 4

Start simplifying: $$ \frac{x^2-y^2}{x(x-y)}+\frac{x^2-y^2}{x(x+y)}= \frac{(x-y)(x+y)}{x(x-y)}+\frac{(x-y)(x+y)}{x(x+y)}= \frac{x+y}{x}+\frac{x-y}{x}=\dots $$