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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?

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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

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

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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

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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.

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  • $\begingroup$ 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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