OpenStudy (anonymous):

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

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (asnaseer):

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

4 years ago
OpenStudy (anonymous):

no it is y-2 just verified

4 years ago
OpenStudy (asnaseer):

then your question is incorrect

4 years ago
OpenStudy (anonymous):

what is incorrect about it?

4 years ago
OpenStudy (asnaseer):

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

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (asnaseer):

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

4 years ago
OpenStudy (anonymous):

yes sorry about the confusion

4 years ago
OpenStudy (asnaseer):

np

4 years ago
OpenStudy (asnaseer):

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

4 years ago
OpenStudy (asnaseer):

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

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (asnaseer):

perfect - now gather the terms involving 'y'

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (asnaseer):

you got it! :)

4 years ago
OpenStudy (anonymous):

thank you so much

4 years ago
OpenStudy (asnaseer):

yw :)

4 years ago
OpenStudy (anonymous):

are you able to help me with verifying the inverse now

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (asnaseer):

I don't understand what you are doing?

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (asnaseer):

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

4 years ago
OpenStudy (asnaseer):

you first need to simplify the denominator into a single fraction

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (anonymous):

for the two parts under the denominator

4 years ago
OpenStudy (asnaseer):

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

4 years ago
OpenStudy (asnaseer):

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

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (asnaseer):

no

4 years ago
OpenStudy (asnaseer):

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

4 years ago
OpenStudy (asnaseer):

try working this out first

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (asnaseer):

|dw:1377639339110:dw|

4 years ago