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step on write
\[f(g(x))\]

step two replace
\[g(x)\] by
\[\frac{3x+7}{2x}\] to get
\[f(\frac{3x+7}{2x})\]

so far so good?

yup

zat ok?

is jsut an exercise in simplication

simplification

yup that is what my teacher told me

what??

then multiply top and bottom by 2x to get
\[f(g(x))=\frac{2x}{3x+7-2x}=\frac{2x}{x+7}\]\][

what happened to the ( )^2?

oops i lost you

are we going to replace \[f(x)=\frac{x+1}{x^2-1}\] by
\[f(x)=\frac{1}{x-1}\]?

yes because i think it is easier

ok in that case there is no square in the denominator

ooooohhh now i know

if we make that change in f, then \[f(g(x))=f(\frac{3x+7}{2x})\] yes?

if so then everything i wrote above works yes?

if not let me know

isn't it it should first be 1/(3x+7)- 1?

am i correct?

no.

why?

\[g(x)=\frac{3x+7}{2x}\]

not
\[3x+7\]

so when we replace x in f(x) by \[\frac{3x+7}{2x}\] we get \[\frac{1}{\frac{3x+7}{2x}-1}\]

yup i forgot to type 2x sorry for that

please continue

ok good

now you can simplify this compound fraction by multiplying top and bottom by 2x

k?

what happened with the 2x below 3x+7?

i multiplied top and bottom by 2x to get rid of it. canceled from the denominator

\[\frac{3x+7}{2x}\times 2x=3x+7\]

ohhh ok

clear or no?

wait ill do it on a paper ^_^

how did it become 2x/2x+7?

ok let us start with
\[\frac{1}{\frac{3x+7}{2x}-1}\] ok?

ok

multiply top and bottom by 2x. what do you get?

\[1\times 2x=2x\] that is your numerator

ok

\[\frac{3x+7}{2x}\times 2x = 3x+7\] that is why you did it to get rid of the pesky 2x in the bottom

\[-1\times 2x=-2x\]

ok

of course you have to multiply the whole denominator by 2x, not just part of it.

so you numerator is 2x

your denominator is 3x+7-2x

which give a denominator of x+7

ok

so our "final answer" is \[f(g(x))=\frac{2x}{x+7}\]

i just got confused i forgot that 3x and -2x can be combined

ah but clear now yes?

yes! thanks^_^

welcome

ah but clear now yes?