Question

Algebraic fractions help!

So in the picture below, I have to add & subtract. But first I have to get a common denominator.

I know you can multiply them all together & get one, but I was wondering if there was a more simple way to do this.

Also, can someone help me work it out until I get the answer? It should be a number or something because of the box.. Like it shouldn't be an equation or a fraction.

Thanks a million and four!!

Answer 1

What is the LCD in this case?

Answer 2

Firstly just lets take a look at the denominators
x+y
x-y
y-x
y^2-x^2
They all look a bit alike. Can you figure out what could be the common denominator?

Answer 3

hint: a^2-b^2=(a+b)*(a-b)

Answer 4

You could use y^2=x^2?

Answer 5

Did you mean y^2 + x^2 ?

Answer 6

Whoops. No I meant y^2-x^2

Answer 7

yes, that's the LCD

Answer 8

Okay, so for the first number, you could multiply the denominator by (x-y), right? and the second number by (x+y), right? But I don't know for the third since it is reversed..

Answer 9

and LCD is not a drug, it is the least common denominator

Answer 10

you're thinking LSD...but that's beside the point

Answer 11

Hahahaha. Thanks for clearing that up :P

Answer 12

anyway...

Answer 13

can you multiply y-x with something to get x-y?

Answer 14

Negative 1?

Answer 15

& then you just have to switch around the numbers, right?

Answer 16

you can multiply the first denominator by x+y since

(x - y)(x+y)

(-y+x)(y+x)

-(y-x)(y+x)

-(y^2-x^2)

Answer 17

yeah, all the signs change

Answer 18

So

\[\Large \frac{1}{x-y}\]

then becomes

\[\Large -\frac{x+y}{y^2-x^2}\]

Answer 19

And then you would multiply the second denominator by (x-y) making it also y^2-x^2.

Answer 20

good

Answer 21

oh wait, y - x

Answer 22

(x+y)(y-x)

(y+x)(y-x)

y^2 - x^2

Answer 23

So

\[\Large -\frac{8}{x+y}\]

becomes

\[\Large -\frac{8(y-x)}{y^2-x^2}\]

\[\Large -\frac{8y-8x}{y^2-x^2}\]

Answer 24

And then for the third you would first multiply by -1, then (x+y) (essentially just multiplying by (-x-y), right?) giving you y^2-x^2. and the fourth already has the denominator. So then you just also multiply the numerators by the same thing you did the denominators and add/subtract where appropriate. Okay, guys. Thanks!

Answer 25

yes, you multiply the third fraction (top and bottom) by y+x (same thing as x+y)

to go from

\[\Large \frac{4}{y-x}\]

to

\[\Large \frac{4(y+x)}{y^2-x^2}\]

\[\Large \frac{4y+4x}{y^2-x^2}\]

Answer 26

you're welcome

Answer 27

So with that I got -16x+8y. Is that right?

Answer 28

on which part? or are you saying that's your final answer?

Answer 29

We have

\[\Large \frac{1}{x-y}-\frac{8}{x+y}+\frac{4}{y-x}-\frac{11x-5y}{y^{2}-x^{2}} \]

and after you get every denominator to the LCD, we then get

\[\Large -\frac{x+y}{y^2-x^2}-\frac{8y-8x}{y^2-x^2}+\frac{4y+4x}{y^2-x^2}-\frac{11x-5y}{y^{2}-x^{2}} \]

did you get this?

Answer 30

Yes, I did

Answer 31

ok, then combine the numerators:


-(x+y)-(8y-8x)+4y+4x-(11x-5y)

-x-y - 8y+8x + 4y+4x - 11x+5y


Tell me what you get

Answer 32

I re-did it and got -16x - 8y

Answer 33

unfortunately that's not correct

Answer 34

Okay, well just the x's are: -x + 8x + 4x + -11x. That gave me 0.
Then just they y's are: -y + -8y + 4y +5y. That gave me 0 also.
So my answer is 0?

Answer 35

you nailed it

Answer 36

I had a bunch of sign mistakes.. & Thanks for walking me through it!

Answer 37

since 0/(any number but 0) = 0, the whole thing is just 0

Answer 38

so that nasty expression is the same thing as 0

Answer 39

Oh, well yay! Haha. thanks :)

Answer 40

you're welcome