+592 |4y + 1| < |2y - 3|
I don't really understand this so could you give me full working out please
+592 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...
+118608 > 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
+118608 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.
+592 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 :)
+118608 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.
+592 It's okay I figured it out :)
Thank you so much for answering nonetheless
