I need to find the solution sets for the following inequalities:


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?


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

$$ x\in \left[-4,\frac{10}{3}\right]. $$


Your answer $[-7,\frac 13]$ for the first inequality is correct.

Let's solve it graphically.

Draw the graphs of $y_1=|3+2x|$ and $y_2=|4-x|$ and find the interval where $y_2\ge y_1$.

The graphs intersect at:

$$3+2x=4-x\Rightarrow x=\frac 13$$ and $$3+2x=x-4 \Rightarrow x=-7$$

Referring to the graph below, we find the solution:

enter image description here


In both cases, you split each absolute value into two inequalities without it: $$|3+2x| \le |4-x|$$ becomes $$3+2x \le |4-x| \text{ and } (3+2x) \le -|4-x| \Leftrightarrow |4-x| \le -(3+2x)$$ (please figure out what intervals does each one correspond to).

Now split the second one same way, you end up with 4 inequalities which are easily solvable. EDIT Let's make one more step. Note that the break of the absolute value occurs at the point $3+2x=0$, i.e. at $x = -1.5$. Over $(-\infty,-1.5]$, the right-hand version will apply, and over $[1.5,\infty)$, the left-hand version will apply.

Let's split the left one. Just as before, the split occurs around the point $4-x=0$, i.e. $x=4$, and over $(-\infty,4]$ we get $4-x>0$, so $|4-x| = 4-x$ and the inequality becomes $$3+2x \le 4-x \Leftrightarrow x \le 1/3,$$ so this results in the solution $$(-\infty, 1/3] \cap (-\infty, 4] \cap [1.5, \infty) = \emptyset.$$

Now let's examine the other side of the second break. Over $[4, +\infty)$, we have $4-x < 0$ so $|4-x| = -(4-x)$ and the inequality becomes $$3+2x \le -(4-x) \Leftrightarrow x \le -7,$$ which results in the solution $$(-\infty, -7] \cap [4, \infty) \cap [1.5, \infty) = \emptyset.$$

Therefore, the left-hand version yields no solutions. Now examine the right-hand version $$-(3+2x) \leq |4-x|$$ in a similar way.

  • $\begingroup$ So, the solution would be the intersection of both intervals? Also, how would I split $|2x-1|+|1-x|\geq3$? $\endgroup$ – Juan José Castro Apr 3 '14 at 22:10
  • $\begingroup$ @Joseph yes, they go by intersection. The second inequality splits, e.g. into $(2x-1) + |1-x| \ge 3$ and $-(2x-1) + |1-x| \ge 3$. $\endgroup$ – gt6989b Apr 3 '14 at 22:11
  • $\begingroup$ by solving both equations (the ones in your first answer), I get $(-\infty,-7]$ and $[\frac13, \infty)$. The intersection between those is zero. What am I doing wrong? I don't think I understood what you meant. $\endgroup$ – Juan José Castro Apr 3 '14 at 22:17
  • $\begingroup$ @Joseph please see the edit $\endgroup$ – gt6989b Apr 3 '14 at 22:26
  • $\begingroup$ @Joseph By the way, formally, the intersection between sets is never zero, but rather is said to be empty. $\endgroup$ – gt6989b Apr 3 '14 at 22:27

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