Question

Can someone help me?
I need to prove this Identity is true
(x - y) (y - x) = (xy – x^2 - y^2 - yx)

Answer 1

Just multiply it out...

\[(x-y)(y-x) = x(y-x) -y(y-x)\]by distributive property
Now distribute again on that and you should be done!

Answer 2

Oh, well any way I could plug numbers in? Like to get 35=35

Answer 3

Somebody please help me, this is taking much longer than it should

Answer 4

no need... just expand the (x-y)(y-x) with the distributive property and it's there.

Answer 5

start with the left side of the equation.

Answer 6

it's like either use the distributive property on the left to get the right hand side of the equation
or factor the right hand side of the equation to get the left side of the equation

Answer 7

ugh man I fell for that.. nice @whpalmer4 I saw what you did there :/ you used factor by grouping dude. >_<
good one :)

Answer 8

Thank you for explaining, but if you, say, make x=# and y=#, what should I do to make them equal to eachother

Answer 9

hmmm we could try letting x and y be the same value

Answer 10

though I'm trying to do one in my head ... letting x = 1 and y = 1
the result would become
(1-1)(1-1)=((1)(1)-1^2-1^2-(1)(1))
(0)(0)=(1-1-1-1)
0=(0-1-1)
0=(-1-1)
and that's a false statement

Answer 11

I don't think that would work, because on the left it would be zero right off the bat

Answer 12

0 doesn't equal to -2. I know that.. so x and y needs to be different

Answer 13

in looking for something that gives the same number on the left to the right

Answer 14

I've been working on this for days and its really confusing

Answer 15

maybe x = 2 , and y = 3 ?

(2-3)(3-2)= ((2)(3)-2^2-3^2-(3)(2))
(-1)(0) = (6-4-9-6)
0=(2-9-6)
nooooooope

I'm assuming that there is no x,y pair and we may need to do with what @whpalmer4 suggested.

Answer 16

\[(x - y) (y - x) = (xy – x^2 - y^2 \color{red}{+} yx)\]

Answer 17

@Zarkon Why would there be a +?

Answer 18

\[(-y)(-x)=yx\]

Answer 19

DOH! Sign error that makes a big difference.

Answer 20

-___________________-!

Answer 21

a negative times a negative is positive . Now let's see if we can find an x,y pair that makes the equation equal

Answer 22

If this is what was hanging me up for days I will go into a damn screaming fit

Answer 23

also, you cant prove it with examples. you need to multiply it out like @whpalmer4 was doing

Answer 24

which is either start from the left to get the result on the right hand side of the equation or vice versa right @Zarkon ?

Answer 25

(1 - 1) (1 - 1) = (1 – 1^2 - 1^2 +1)
(0)(0) = (1-1-1+1)
(0) = (0-1+1)
0=(-1+1)
0=0
ok I'm gonna be guilty for doing this because a proof shouldn't be plug x and y but I let x = 1 and y =1 and the equation holds true. It's the sign error that threw everything out of the mix.

Answer 26

Could this come out the same if it was a bigger number like 3?

Answer 27

Yeah, those pesky - signs are the work of the devil! I'm told that in days of old, mathematicians would often write their equations so they didn't need to use them!

\[x^2 -3x + 6 = 0\]would be written \[x^2 + 6 = 3x\]and so on.

Answer 28

ALso I didn't eat dinner yet. Can't do math without energy. yeah it should be the same

Answer 29

heh, Im so happy this is finally put to rest. Sorry for my sign mistake, this damn thing kept me up till 3 am each night

Answer 30

Actually, I think the original problem is incorrectly stated.

\[(x-y)(y - x) = x(y -x) - y(y - x) = x(y-x) - y(-1)(x-y) = x(y-x) + y(x-y) \]\[= xy - x^2 + yx - y^2 = 2xy - x^2 - y^2\]

Answer 31

Why have you dragged me back to hell

Answer 32

\[y-x = -1(x-y)\]so we just have \[-1(x-y)(x-y) = -1(x-y)^2 = -1(x^2 - 2xy +y^2) = -x^2 + 2xy- y^2\]

Hey, who copied the problem?

Answer 33

wasn't me.

Answer 34

This is for an assignment where I have to create a new identity for polynomials and have to go through proving it

Answer 35

Possibly the problem was to prove that \[(x-y)(y+x)= (xy - x^2 - y^2 - yx)\]?

Notice that the right side simplifies to just \(-x^2 -y^2\) because \(xy - yx = xy - xy = 0\)

Answer 36

omg I see it now. -_-

Answer 37

OMG! THIS IS TROLL LEVEL >:O

Answer 38

like xy and yx cancel out

Answer 39

Well, anyhow, to prove an identity, just manipulate both sides of the equation as necessary to meet in the middle with the same thing on both sides...

Answer 40

I need a result where (x - y) (y - x) = (xy – x^ - y^2 + yx)
have a result of (example) 20=20

Answer 41

leaving
(x-y)(y+x) = -x^2-y^2
and then
(x-y)(y+x) = -(x+y)(x-y)
............ x+X

Answer 42

Well, you don't prove it works for all numbers by plugging in numbers, though you can prove that it DOESN'T work for all numbers by plugging in numbers.

Answer 43

This assignment is going to be the death of my grade in this class. I really should know how to do this, it really shouldn't be this hard, and theres no + in (y - x)

Answer 44

Well, I'm sorry, but the identity you posted simply is not an identity. Try plugging in \(x =1, y = 2\)

Answer 45

wait a sec, I worked it out and it was -1=-1

Answer 46

\[(x-y)(y-x) = xy - x^2 - y^2 - yx\]is your statement, yes?

\[(1-2)(2-1) = (1)(2) - (1)^2 - (2)^2 - (2)(1)\]\[-1*2 = 2-1-4-2\]\[-2=-5\]not in my universe :-)

Answer 47

no it was + yx I screwed up in my signs
and (2-1)=1
so -1*1=-1

Answer 48

so for the last time, what is the identity you are trying to prove?

Answer 49

(x - y) (y - x)
it's a real pain in the retriceand confusing

Answer 50

no, what is the entire statement? You have no = sign there

Answer 51

ah, sorry im a bit tired, its
is (x - y) (y - x) = (xy – x^ - y^2 + yx)

Answer 52

Okay, that's what I was saying earlier:\[(x-y)(y-x) = 2xy - x^2 - y^2\]your version just hasn't collected the \(xy\) terms together

Here is a proof:

\[(x-y)(y-x) = (xy - x^2 - y^2 + yx)\]we expand the left side twice with the distributive property:
\[x(y-x) - y(y-x) = xy - x^2 - y^2 + yx\]\[xy - x^2 - y^2 -(-yx) = xy - x^2 - y^2 +yx\]
\[xy - x^2 - y^2 + xy = xy - x^2 - y^2 + yx\]now we change the order of the product for the last term on the left hand side because multiplication is commutative:
\[xy - x^2 - y^2 + yx = xy - x^2 - y^2 + yx\]
Done.

Answer 53

actually, didn't even need that last step if I hadn't changed \(-(-yx)\) to \(+xy\) by habit of always going in ascending lexicographic order

Answer 54

Thank you for clearing things up! I love you all for helping ♥

Answer 55

Thank you @whpalmer4 senpai!

Answer 56

You're welcome!