Question

find the inverse
y=(x+3/(x-2)

Answer 1

I have
\[x=\frac{ y+3 }{ y-2 }\]
\[x(y-2)=y+3\]
\[x(y-2)-y=3\]
now im stuck

Answer 2

you seem to have an error in your first line - I believe the denominator should be y+2 not y-2

Answer 3

no it is y-2 just verified

Answer 4

then your question is incorrect

Answer 5

what is incorrect about it?

Answer 6

the question state the expression as:\[y=\frac{x+3}{x+2}\]note the positive sign in the denominator

Answer 7

sorry the first question is wrong it is suppose to be x-2

Answer 8

ok, so you need to find the inverse of:\[y=\frac{x+3}{x-2}\]correct?

Answer 9

yes sorry about the confusion

Answer 10

np

Answer 11

so your steps up to:\[x(y−2)−y=3\]are correct

Answer 12

next you need to expand the expression \(x(y-2)\) by multiplying it out

Answer 13

ok so
\[xy-2x-y=3\]

Answer 14

perfect - now gather the terms involving 'y'

Answer 15

oh then I can add 2x and then factor out y right

Answer 16

you got it! :)

Answer 17

thank you so much

Answer 18

yw :)

Answer 19

are you able to help me with verifying the inverse now

Answer 20

i pluged \[f^{-1} into f\] and got this so far
\[\frac{ 2x+3 }{ x-1 }+3*\frac{ x-1 }{ 2x+3 }-\frac{ 1 }{ 2 }\]

Answer 21

I don't understand what you are doing?

Answer 22

so I am trying to verify the inevrse through function composition so i first had
\[\frac{ \frac{ 2x+3 }{ x-1 }+3 }{ \frac{ 2x+3 }{ x-1 }-2 }\]
I just flipped the bottom and changed it to multiplication

Answer 23

you flipped it incorrectly.\[\frac{1}{\frac{a}{b}-c}\ne\frac{b}{a}-\frac{1}{c}\]

Answer 24

you first need to simplify the denominator into a single fraction

Answer 25

so would i multiply the bottom by x-1 to get a common denominator

Answer 26

for the two parts under the denominator

Answer 27

i.e.\[\frac{1}{\frac{a}{b}-c}=\frac{1}{\frac{a-bc}{b}}=\frac{b}{a-bc}\]

Answer 28

so first evaluate:\[\frac{ 2x+3 }{ x-1 }-2\]

Answer 29

so it would be 2x+3-2x-2 on the bottom?

Answer 30

no

Answer 31

\[\frac{ 2x+3 }{ x-1 }-2=?\]

Answer 32

try working this out first

Answer 33

\[\frac{ 2x+3 }{ x-1 }-2(\frac{ x-1 }{ x-1 })\]
\[\frac{ 2x+3-2(x-1) }{ x-1 }\] the x-1s cancel and you have
2x+1 right?