I need to find the solution sets for the following inequalities:
$$|3+2x|\leq|4-x|$$$$|2x-1|+|1-x|\geq3$$
After a bit of tinkering with the first one, I think the solution set is $[-7, \frac13]$, but I'm not sure, I've only been taught to solve inequalities with abs. values on either side of the sign, not on both, and I couldn't find any online resource I understood. On the first one, I tried finding the values for $x$ in $4-x=3+2x$ and $4-x=-3-2x$, then dividing the real line into 3 intervals with these numbers and see in which of them the inequality held true. I have no idea what to do with the second one. Is my solution alright, and how are these kinds of inequalities solved?
Best Answer
For the first inequality, since we have modulus on both sides, we can safely square the expression.
$ |3+2x|^2 \le |4-x|^2 \\ \implies 9 + 12x + 4x^2 \le 16 - 8x + x^2 \\ \implies 3x^2 + 20x - 7 \le 0. $
Since the leading term in the quadratic expression is positive, the inequality only holds between the two roots (both included) of the expression. Hence, the inequality holds in the interval
$$ x\in \left[-4,\frac{10}{3}\right]. $$