Question

solve the inequality 2-3x < |x-3|.

Answer 1

So 8x^2 -6x-5<0
x=-1/2 and x=5/4
But why is th3 answer only x=-1/2??

Answer 2

Well, you shouldn't have equals given that it's an inequality

Answer 3

However, you should have: \[
2-3x<x-3
\]As well as \[
2-3x<-(x-3)
\]And you simplify them.

Answer 4

But theres modulus. I can't put the negative there.

Answer 5

You solve modulus by breaking it down into both possible cases: the positive case and the negative case.

Answer 6

Okay I did (2-3x)^2 < (x-3)^2

Answer 7

Then?

Answer 8

Using that method is a bit problematic

Answer 9

It requires to use this method.

Answer 10

Squaring both sides of the inequality doesn't always work that way

Answer 11

The answerscheme says that gives a mark.

Answer 12

And afterall I need to find wheather its less or more than

Answer 13

Positive case:\[
2-3x<x-3\implies 5<4x \implies \frac 54 <x
\]Negative case:\[
2-3x<-(x-3)\implies -2+3x>x-3\implies 2x>-1 \implies x>-\frac 12
\]

Answer 14

Now \(-1/2 < 5/4\).

Answer 15

What if i wanna use my method? Is there any way to find ou t whether it's less or ,ore?

Answer 16

@wio 's method is the easiest and simplest method.

Answer 17

I,m just asking.

Answer 18

For either to be true: \(-1/2<x\).
For both to be true: \(5/4<x\).
I believe that with modulus, we use the either restriction.

Answer 19

I actually tried proving x<5/4 in the equation and it can't seem To be proven

Answer 20

Here is the problem with your method.

Answer 21

\[
-10<|5|
\]Square both sides: \[
100 < 25
\]This is false.

Answer 22

Is your answer x>-1/2???

Answer 23

????

Answer 24

@Kainui

Answer 25

i think it should be 2-3x < x+3

Answer 26

bc its a absolute value the sign changes

Answer 27

@LXelle

Answer 28

Uh no. Modulus means sqyare both sides

Answer 29

ohh yeah i forgot