Question

need some help with this one
y/(y+2) - 1/(y-2) = 8/(y^2 - 4)

Answer 1

y = 5

combine fractions on left into single fraction
then set numerators equal to each other
solve for y

Answer 2

i did that, didn't work

Answer 3

did you get

y(y-2) -(y+2) = 8

Answer 4

no, y-1/(y+2)(y-2) = 8/(y+2)(y-2)

Answer 5

\[y/(y+2) - 1/(y-2)-8/(y+2)(y-2)=0\]
\[y/(y+2) - 1/(y-2)- 8/(y^2 - 4)=0\]
\[y/(y+2) - 1/(y-2) -8/(y+2)(y-2)=0\]
\[y(y-2)/(y+2)(y-2)-(y+2)/(y-2)(y+2)-8/(y-2)(y+2)=0\]
\[(y^2-2y)/(y^2-4)-(y+2)/(y^2-4)-8/(y^2-4)=0\]
\[(y^2-2y-y-2-8)/(y^2-4)=0\]
\[(y^2-3y-10)/(y^2-4)=0\]
\[((y-5)(y+2))/(y+2)(y-2)=0\]
\[(y-5)/(y-2)=0\]
\[y=5\]

Answer 6

Wrote out all of the steps so you could follow it more easily o.o

Answer 7

need common denominator to combine fractions

multiply 1st fraction by (y-2)/(y-2)
multiply 2nd fraction by (y+2)/(y+2)
this way they have same denominator

Answer 8

Hunus, I don't understand how you came up with y(y-2)/(y+2)(y-2) - (y-2)/(y-2)(y+2) - 8/(y-2)(y+2) so I'm stuck right there. Dumbcow, I already made the fractions have a common denominator. What about the other side of the equation though? Whatever is done to one side must be done to the other.

Answer 9

I made a common denominator by multiplying the fractions by
\[(y-2)/(y-2) \]
\[(y+2)/(y+2)\]
respectively

Answer 10

okay, now i'm stuck 2 steps down
you have \[(y^{2}-2y-y-2-8)/(y^2-4)=0\]
why isn't it\[(y ^{2}-2y-y+2-8)/(y^2-4)=0\]?

Answer 11

Because you're subtracting (y+2)
-(y+2)=-y-2

Answer 12

because the neg sign must be distributed

-(y+2) = -y -2

Answer 13

okay

Answer 14

:D

Answer 15

okay so you have (y-5)/(y-2)=0, how do you get, from that, that y=5?

Answer 16

nothing can be done with that

Answer 17

Multiply everything through by y-2.

Answer 18

How will that work? The fractions won't have common denominators.

Answer 19

what?
\[\frac{y-5}{y-2} = 0\]
so if you multiply by y-2, you get
\[y-5 = 0(y-2) = 0\]

Answer 20

:D okay, sorry about that, i was starting over
thanks a lot