Solve absolute rational inequalities

[@ManjulaMathew-wb3zn]

Square to remove absolute value symbol. (3-2x)^2 >9(x-1)^2 5x^2-6x

[@Fuata-j9c]

How can the two sides be negative on 0and1

[@זאבגלברד]

The function y=( - 2x+3 ) / (x - 1) is a "move" of some k/x ... If you do the devision, you see it. So it is simple to draw the function ! and find where the line y=3 intersects

[@mustafaquershi]

At 1 function is undefined. I don't know how you draw graph and give interval

[@clementkatumanani1696]

Nice work ❤

[@horsepower33]

more confused than before.

[@MARK-MAN]

😂

[@stunzeed6721]

frl

[@misheckphiri-io9lx]

You're not alone 😂😂😂,I am also confused

[@Dharmarajan-ct5ld]

Use distance concept interpretation of MODULUS. Distance of 3 from 2x > distance of 3 from 3x. It is just 3 steps

[@cnchannel5839]

Say how it is bro

[@Dharmarajan-ct5ld]

@@cnchannel5839 reply given under comments , now. Best wishes

[@Dharmarajan-ct5ld]

At 4.21, "" x>0, x-1>0"" appears questionable !!

[@hqs9585]

At the beginning it does, but she is just using the negative to accept domain instead of traditional positive, in my opinion NO NEED to do that, just be consistent with positive sign in the domain determination. In short she just made a mistake up to that point but rectified by saying negative domain acceptable.

[@rdas1]

Agree. This 2nd term x/(x-1) < 0 is an undefined one as there is no domain value of x that satisfies and the term is not = to "0" when x = 0. Of course, x can't be 1. In other words, we could also take x < 0 and x - 1 < 0. This approach confuses the audience. Above mentioned approach by Manjula Mathew is short and sweet

[@selmashikongo9649]

Is this not modulus?

[@Dharmarajan-ct5ld]

Dear Sh. chintakunta Navya, Reference your reply to my comment: x can't be 1, is obvious. Shift |x-1| to right side and take 3 inside modulus. Note: |a-t|=|b-t| for a """'constant t"""" means, either a=b or t is average of a and b . Analyse.!!!! Number line!!! Moreover, it appears author has committed a mistake with ""x/(x-1)

[@claytonjansen589]

adorable

[@juliajohaness2688]

Why do we have to equate the equations to 0?

[@kathymckenzie9164]

Julia, doing so gives the roots of each part of the rational expression, which divides the number line into regions we must inspect in our sign analysis.

[@juliajohaness2688]

@@kathymckenzie9164 thank you so much

[@zapzawesomeice2970]

This problem is flawed. Whats wrong with it?