What is the simplified form of the quantity 1 over x minus 1 over y all over 1 over x plus 1 over y ?

y plus x, over y minus x

y minus x, over y plus x

x minus y, over x plus y

x plus y, over x minus y

@mathstudent55

Question

What is the simplified form of the quantity 1 over x minus 1 over y all over 1 over x plus 1 over y ?

 y plus x, over y minus x

 y minus x, over y plus x

 x minus y, over x plus y

 x plus y, over x minus y

@mathstudent55

mathstudent55 · Answer

I'll use the equation editor to make sure I get the problem right.

mathstudent55 · Answer

Is this it?

$ \huge\dfrac{\frac{1}{x} - \frac{1}{y}}{\frac{1}{x} + \frac{1}{y}} $

anonymous · Answer

Yes!

mathstudent55 · Answer

If so, here's an easy way to simplify this fraction.
In the numerator there are two denominators, x and y.
In the denominator, there are also two denominators, x and y.
What is the LCD of x and y?

anonymous · Answer

My guess is x-y/x+y

mathstudent55 · Answer

I don't guess. I figure it out.

anonymous · Answer

Lol, I'm just telling you of my guess from when I looked at the problem.

mathstudent55 · Answer

You may very well be correct. Let's find out if your hunch is correct.

mathstudent55 · Answer

Ok, what is the LCD of x and y?

anonymous · Answer

Alright! Would it be 1?

mathstudent55 · Answer

If you had just 2 fractions like these that you needed to add, you'd need a common denominator.

$\dfrac{2}{x} + \dfrac{3}{y} $

What would you use as a common denominator?
Since x and y have no factors in common, you multiply them together to get xy.
The least common denominator (LCD) of x and y is xy.

mathstudent55 · Answer

Now back to your problem. You have denominators of x and y.
The LCD of the denominators is xy.
Let's multiply the numerator and denominator of the main fraction by xy.

mathstudent55 · Answer

$ \huge  =\dfrac{xy( \frac{1}{x} - \frac{1}{y} ) }{xy(\frac{1}{x} + \frac{1}{y})} $

Now we distribute the xy in the numeratoir and the denominator:

$ \huge = \dfrac{xy( \frac{1}{x} ) - xy(\frac{1}{y} ) }{xy(\frac{1}{x} )+ xy(\frac{1}{y})} $

Now we cancel out the x's and y's that are both in the numerator and denominator:

$ \huge = \dfrac{\cancel{x}y( \frac{1}{\cancel{x}} ) - x\cancel{y}(\frac{1}{\cancel{y}} ) }{\cancel{x}y(\frac{1}{\cancel{x}} )+ x\cancel{y}(\frac{1}{\cancel{y}})} $

$\huge =\dfrac{y - x}{y + x} $

mathstudent55 · Answer

Now compare the answer above with your guess.
The conclusion is be careful with guesses.

mathstudent55 · Answer

Ok, gtg. Bye.

anonymous · Answer

Yes, now I can see it. Thank you! @mathstudent55

mathstudent55 · Answer

You're welcome.