Question

pleaseeee helps
solve and check: b-4/b-2 = b-2/b+2 + 1/b-2

Answer 1

\[\frac{ b-4 }{ b-2 } =\frac{ b-2 }{ b+2 } + \frac{ 1 }{ b-2 }\]

Answer 2

@Compassionate

Answer 3

Are you trying to solve for b?

Answer 4

yeah i think so

Answer 5

First find the greatest common factor. Do you know what that is?

Answer 6

umm would that (b-2)(b+2)?

Answer 7

would that be*

Answer 8

Correct. Now, multiply that by each numerator, but exclude the term if it is already in the numerator.

Answer 9

so
instead of
(b-2)(b+2) times (b-2) do just (b+2) times (b-2)?

Answer 10

\(\frac{ b-4 }{ b-2 } =\frac{ b-2 }{ b+2 } + \frac{ 1 }{ b-2 }\)

\(\frac{ b-4(b+2)}{ b-2 } =\frac{ b-2 }{ b+2 } + \frac{ 1 }{ b-2 }\) Notice that we multiplied it by b + 2, and not b - 2, this is because b - 2 is in its denominator.

Answer 11

You will do this for every term.

Answer 12

b^2 + 6b +8 = b^2 - 4b + 4 + b +2

Answer 13

Now that you have it eliminated, solve like a normal equation

Answer 14

b^2 + 6b + 8 = b^2 -3b + 6

Answer 15

Continue.

Answer 16

subtract b^2 both sides?

Answer 17

Nope. Subtract 6.

Answer 18

b^2 = b^2 + 9b + 2

Answer 19

b^2 + 6b + 8 = b^2 -3b + 6

b^2 + 6b = b^2 - 3b + 6
b^2 = b^2 - 9b + 6

Answer 20

whered the 8 go?

Answer 21

b^2 + 6b + 8 = b^2 -3b + 6

I messed up. We should've subtracted 6, which wouldn've maken it 2.

Answer 22

okay which gives us
b^2 + 6b + 2 = b^2-3b

Answer 23

@Compassionate

Answer 24

Now subtract the 2
Add the 3b
Divide..

Answer 25

b^2 + 9b = b^2 - 2
divide the nine?
b^2 + b = b^2/9 - 2/9
?
@Compassionate