+0  
 
+2
89
6
avatar+527 

|4y + 1| < |2y - 3| 

 

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

 Apr 6, 2021
 #1
avatar+527 
+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...
 

 Apr 6, 2021
 #2
avatar+113190 
+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

 Apr 6, 2021
 #3
avatar+113190 
+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.

 Apr 6, 2021
edited by Melody  Apr 6, 2021
 #4
avatar+527 
+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
avatar+113190 
+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
avatar+527 
+1

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

Logarhythm  Apr 7, 2021

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