Question

What about this one?

Answer 1

I would get the lcd of all the fractions first

Answer 2

the lcd would be : j + 4?

Answer 3

without even solving this I can give you a restriction of this and no it would be j(j+2)

Answer 4

i would multiply j trough the problem?
and get:

j(j+4)/ j(j+2) = 2 x j ?

Answer 5

you can automatically cancel b and d because they would make your fraction undefined

Answer 6

I cant read that ill enter it into equation form

Answer 7

ok

Answer 8

\[\frac{ j(j+4) }{ j(j+2) }=\frac{ 2j(j+2) }{ j(j+2) }-\frac{ 1(j+2) }{ j(j+2) }\]
do you know how to solve from here

Answer 9

if im right j + 4 would be left after i get rid of like terms

Answer 10

no. now you multiply both sides by j(j+2) to get rid of the fractions. then you have everything in the numerator left then you distribute everything. finish with a quadratic equation and factor it

Answer 11

and do you remember about the restrictions I stated

Answer 12

i got this from my lesson

Answer 13

your lesson is correct. but you would have j(j+4) on the left

Answer 14

so i would have (j+4) = 2 - (j+2)/ j right?

Answer 15

noooooo lol sec ill put the equation in

Answer 16

* j(j+4)=(j+2)/j

Answer 17

\[j(j+4)=2j(j+2)-j+2\]

Answer 18

you have to consider that 2 is equal to 2/1. its a fraction as well that needs the lcd of

Answer 19

im lost how did you get 2j(j+2) i think i understand how you got - j + 2

Answer 20

because you have a constant of 2. so to get the lcd there you multiply the numerator and denominator by j(j+2) giving you 2j(j+2)
because previously all there was in the denominator was 1

Answer 21

oh so the lcd is j(j+2)

Answer 22

did you miss the equation I typed out with all the terms with their lcds?

Answer 23

what is the lcd? thats were im lost now i thought it was j(j+2)

Answer 24

the lcd is j(j+2)

Answer 25

are you guys ok?

Answer 26

im ok lol im no teacher though obviously. hard over type messages

Answer 27

ok so on the right i would have : j(j+2) x j+4/ j+2 and i would cross out both x+2 thus getting j x (j+4) right?

Answer 28

that would be on the left. but yes that is correct

Answer 29

ok now im caught up so now what would i do?

Answer 30

get the lcd on the right side of the equation. did you miss all the equations I typed out? or do you not understand what I did.

Answer 31

i just didnt understand what you did i was trying to figure out where the extra j came from

Answer 32

because the term on the far rights denominator is j. you cant just add 2 to that one. you have to think as j and j+2 as totally different terms

Answer 33

ok, so after that i would: have j(j+4) =2j(j+2) − j+2 now i have to make it equal to 0 but first :
(j^2 +4j) = (2j^2 + 4j) - j + 2 where do i go from here do i go ahead and subtract?

Answer 34

just combine like terms by inverse operations across the = sign. then you just factor the quadratic equation

Answer 35

so i would get :


j^2 - 2j^2 + 4j + 4j = -j + 2

j^2 + 8j = -j + 2?

Answer 36

you have less terms on the left side. make it
\[0=j^2-j+2\]

Answer 37

subtract j^2 from 2j^2
then subtract 4j from 3j

Answer 38

The most wise FIRST thing to do would have been to discard j = -2 and j = 0.

Answer 39

as I stated before.. there were 2 answers that caused it to be undefined.

Answer 40

ok so i would get :

j^2 - 4j - 3 = j^2 - j my question is is where did you get the 3j from

Answer 41

\[j^2+4j=2j^2+4j-j+2\]
cobine like terms
\[j^2+4j=2j^2+3j+2\]
subtract j^2 from both sides
\[4j=j^2+3j+2\]
subtract 4j from both sides
\[j^2-j+2\]
then factor this to get your two possible answers

Answer 42

oh ok i see,

i would have to set it up to use the foil method (j+2) and (j- 1)

and make them equal to 0

so:

j + 2 = 0 and j - 1 = 0
j = 0 - 2 and j = 0 - 1
j = -2 and j = -1

so now i would have to check it with the original equation?

Answer 43

you have your signs backwards and you did some math wrong there lol.

Answer 44

i did wait i had said -,+ = +,- so i would not get that messed up and i did it anyway i have it backwards right?

Answer 45

it would be
\[(j+1)(j-2)\]
remember your middle term is negative

Answer 46

so i would have a -1 and +2 as my set

Answer 47

hang on I typod

Answer 48

lol

Answer 49

j^2+4j = 2j^2 + j - j + 2

-1^2 + 4 x -1 = (2 x - 1)^2 + -1 -(-1) + 2 ? for -1?

Answer 50

1 - 4 = 4 -1 +1 + 2? would not equal tha same as on the other side

Answer 51

i got it im using the wrong one

Answer 52

if you plug -1 and 2 in it does

Answer 53

can you help me with my last 2?

Answer 54

possibly ill be going to bed soon though. make a new deal the length of this is lagging me

Answer 55

ok