|4y + 1| < |2y - 3| 


I don't really understand this so could you give me full working out please

 Apr 6, 2021

Like I understand that you have to find the intervals, but I'm not sure how to solve the inequalities for them...
Three cases: (at least I think)


\(y<\frac{-1}{4} \)     and.    \(-\frac{1}{4} ≤ y < \frac{3}{2}\)     and     \(y ≥ \frac{3}{2}\)


p.s. why are only some of them like "≤ and ≥" but others aren't? 

it's just very confusing to me...

 Apr 6, 2021

> means greater than    so 6>6 is not true because 6 is not bigger than 6


\(\ge\)    means greater than or equal too.      so  \(6\ge6\)    is true because 6 is equal to 6

 Apr 6, 2021

The easiest way to tackle this is to square both sides.


\(|4y+1|<|2y-3|\\ so\\ |4y+1|^2<|2y-3|^2\\ (4y+1)^2<(2y-3)^2\\ ...\\ 3y^2+5y-2<0 \)


Solve this for  =0

Then it is a concave up parabola so your solution will be the middle bit.


Added:  When I say it is a concave up parabola I mean   y=3x^2+5x-2  is a concave up parabola.

This is the most confusing bit.

 Apr 6, 2021
edited by Melody  Apr 6, 2021

Hi, thanks for answering but I'm not sure that's how my teacher wants us to do it...

She says we have to put this on a number line and write out the cases, then solve it.

Not sure if you'll see this post again but it's fine; I'll just ask in class.

Thank you so much though :)

Logarhythm  Apr 6, 2021

That is ok.

The way I have done it is by far the easiest way 


I can help you with the way your teacher wants as well.

Let me know if you still want help.

Send me a private message as I may not see your post.

Melody  Apr 7, 2021

It's okay I figured it out :)
Thank you so much for answering nonetheless

Logarhythm  Apr 7, 2021

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