|4y + 1| < |2y - 3|

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

Logarhythm Apr 6, 2021

#1**+1 **

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...

Logarhythm Apr 6, 2021

#2**+1 **

> 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

Melody Apr 6, 2021

#3**+1 **

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.

Melody Apr 6, 2021

#4**+1 **

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

#5**+1 **

That is ok.

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

BUT

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

#6