How do I solve an inequality that involves two absolute values?

Question

I have to solve the following inequality: $$|6x−2|≤|3x−5|$$ I do know that I have to do this first step: $$|6x−2|-|3x−5|≤ 0$$ From here I got confused what I should do with the absolute values. Could someone give me a push in the right direction?

Answer 1

Square, square, you just square! Then you get $$(6x-2)^2\le (3x-5)^2.$$ Now you may perform a transposition of one term, and factorise, then the rest is trivial -- hopefully. :)

Answer 2

Hint: Start by making a sketch. Find the points of intersection of $y=|3x-5|$ and $y=|6x-2|$ by solving $|3x-5|=|6x-2|$. Then determine the range of values of $x$ for which $|3x-5|\geq|6x-2|$ (i.e. when $y=|3x-5|$ is above $y=|6x-2|$).

Answer 3

Consider three intervals:

$x<\dfrac13\tag1$

$\dfrac13\le x\le\dfrac53\tag2$

$x>\dfrac53.\tag3$

In interval ($1$), $3x-5, 6x-2<0$.

In interval ($2$), $6x-2\ge0 $ but $3x-5\le0$.

In interval $(3)$, $3x-5, 6x-2>0$.

Can you take it from here?

Answer 4

You can multiply inequality by $ |6x−2|+|3x−5| $. It is always greater than $ 0 $ so solutions remain the same. Then you have a simple quadratic inequality.