Question

Simplify...
(x-y/x+y)+(x+y/x-y)

Answer 1

Hint: Multiply the first fraction by \(\large\frac{x-y}{x-y}\). Multiply the second fraction by \(\large\frac{x+y}{x+y}\)

Answer 2

how do I do that?

Answer 3

Like this:

\((\large\frac{x-y}{x+y} \dot\ \frac{x-y}{x-y}) + \large(\frac{x+y}{x-y} \dot\ \frac{x+y}{x+y})\)

Answer 4

In the numerator, when you multiply, you will end up with a square binomial. In the denominator, you will end up with difference of squares.

A square binomial will look like this in general: \((x - a)^2\)
Difference of squares will look like this after multiplication: \(a^2 - b^2\)

Answer 5

Those are general cases of course

Answer 6

oh ok

Answer 7

so, 2(x²+y²) /x²-y²?

Answer 8

Basically, anytime you multiply something by itself, you square it.

\(a \times a = (a)^2\)
\(b \times b = (b)^2\)
\(2x \times 2x = (2x)^2\)
\(x + 2 \times x + 2 = (x+2)^2\)

See a pattern there?

Answer 9

Well, you are close, but not exactly. Your numerator is slightly off.

Answer 10

ok

Answer 11

I'm going to correct myself. The general form for the binomial is \((x + a)^2 \) But in the first fraction, y is negative not positive so you'll end up with two different expressions for each numerator, but they will still be binomials.

Answer 12

ok, so is it (2x^2-2y^2) for the numerator?

Answer 13

no, wait, 2(x^2 + y^2) / (x^2 - y^2)?

Answer 14

I don't know where you are getting the 2 from. You should have gotten

\(\frac{(x-y)^2}{x^2-y^2} + \frac{(x+y)^2}{x^2-y^2}\) after multiplication. When you combine the fractions, you won't be able to combine the numerator since the signs in the parentheses are different. Do you see that?

Answer 15

\((x-y)^2 + (x+y)^2 \ne 2(x^2 + y^2)\)

Answer 16

You cannot combine \((x-y)^2 + (x+y)^2\) unless you expand the numerator.

Answer 17

I would have combined the fractions and left it in this form:

\(\large\frac{(x-y)^2 + (x+y)^2}{x^2-y^2}\)

Answer 18

oh ok

Answer 19

You can expand the numerator if you want, but be careful

Answer 20

alright, thank you

Answer 21

Actually, after expansion, I see that I get the same thing you got earlier \(2(x^2+y^2)\) for the numerator, so I take that back.

Answer 22

I didn't know that you expanded the numerator

Answer 23

What I mean to say is, earlier I stated that \((x-y)^2 + (x+y)^2 \ne 2(x^2 + y^2)\) but they actually are equal so your fraction

\(\frac{2(x^2 + y^2)}{x^2 - y^2}\) is also a correct form.

Answer 24

so \((x-y)^2 + (x+y)^2 = 2(x^2 + y^2)\) is indeed true. Sorry for the confusion.