What is my strategy for solving inequalities like these?

Question

I am trying to determine the solution to:

$$3\lvert2-x\rvert \ge\lvert x+5\rvert-3$$

I looked at some other similar questions here and I tried to follow their strategy by splitting the inequality up into cases:

$$\lvert 2-x\rvert = \begin{cases} 2-x, & \text{if $x \le 2$} \\ -2+x, & \text{if $x >2$} \end{cases}$$

$$\lvert x+5\rvert = \begin{cases} x+5, & \text{if $x \ge -5$} \\ -x-5, & \text{if $x <-5$} \end{cases}$$

As far as I understand I have three intervals to check:

$$1) \space x<-5 \\2) \space -5 \le x\le2 \\3) \space x>2$$

Case 1: $3(2-x) \ge -x-5-3 \implies x\le 7$

Case 2: $-3(2-x) \ge (x+5)-3 \implies x\le 1$

Case 3: $-3(2-x) \ge x+5-3 \implies x\ge 4$

Final answer: $$ x \le 1;\space \space 4 \le x\le7$$

My questions:

• I am currently doing this like a robot without really knowing whats going on. What am I trying to figure out when looking at the different cases 1,2,3? In other words, what do the solutions of the cases tell me about my inequality.
• Is there any other strategy I can use to solve these inequalities?
• The solution wolframalpha gives me is $(4, \infty)$ and $(-\infty,1]$. What am I doing wrong?

Answer 1

You start of with the expression $$3\lvert2-x\rvert \ge\lvert x+5\rvert-3$$ and you are looking for all the possible $x \in \mathbb{R}$ that make this inequality true. Now, why are we breaking it up into those intervals? Well that is because of how absolute value is defined:

$|a| = \begin{cases} a, \quad \ \ \ \text{if}\ \ a\geq 0 \\ -a, \quad \text{if}\ \ a < 0. \end{cases}$

Notice that:

• If $x<-5$ then $(x+5)$ is negative and $(2-x)$ is positive, hence by definition $|2-x|= 2-x$, and $|x+5| = -(x+5) = -x-5.$
• If $-5 \leq x<2 $ then $(2-x)$ is still positive, and $x+5$ is now also positive. Hence by definition $|2-x|= 2-x $ but now $|x+5| = x+5.$
• If $x\geq2$ then $(2-x)$ turns negative and $(x+5)$ is positive, hence $|2-x| = -(2-x)=x-2$ and $|x+5| = x+5.$

Now, we sum up the possible cases:

• Case 1: $x < -5.$ Then the expression $\ \ $ $3\lvert2-x\rvert \ge\lvert x+5\rvert-3$ is equivalent to $$3(2-x) \geq -x-5-3$$ which reduces to $x \leq 7$. Then the possible solutions in this case are those such that $x < -5$ AND $x \leq 7$, which is simply $x <-5$.
• Case 2: $-5\leq x < 2.$ Then the expression $\ \ $ $3\lvert2-x\rvert \ge\lvert x+5\rvert-3$ is equivalent to $$3(2-x) \geq x+5-3$$ which reduces to $x \leq 1$. Thus now the possible solutions are $x$ such that $-5 \leq x <2$ AND $x \leq 1$, which is the interval$ -5 \leq x \leq 1.$
• Case 3: $x \geq 2.$ Then the expression $\ \ $ $3\lvert2-x\rvert \ge\lvert x+5\rvert-3$ is equivalent to $$3(x-2) \geq x+5-3$$ which reduces to $x \geq 4$. Thus now the possible solutions are $x$ such that $x \geq 2$ AND $x \geq 4$, which is simply $x \geq 4$.

To this we can conclude that the following intervals fit the criteria: $x<-5,\ -5 \leq x \leq 1$ and $x \geq 4$. We can type it more neatly so the solution is given by $ x \in (-\infty,1] \cup [4, \infty).$

Answer 2

If you have $a \ge b$, where $a$ and $b$ are relatively simple absolute value expressions, one strategy might be to graph $y = a$ and $y = b$ on the same coordinate axes.

So in this particular case, you would graph $y = 3|2−x|$ and $y = |x+5|−3$.

That may make it easier to determine where the graph of $y = 3|2−x|$ is "above" the graph of $y = |x+5|−3$, in other words where $ 3|2−x| \ge |x+5|−3$.

You can then use your systematic, algebraic approach to determine the exact boundaries for your solution.

Answer 3

You have made at least three mistakes.

1) Solution for case 2 is just plain wrong, please check it.

2) You have to combine intervals with solutions to draw the right conclusion. In case 1) the interval is $x\lt -5$ and the solution for that interval is $x\le7$. You have to combine (intersect) the solution with the interval. So in case 1) you have only $x\lt -5$ as part of your final solution. The final solution is a union of three different cases.

3) Equation for case 3 looks exactly the same as case 2?!

It might look mechanical but this approach is, in my opinion, the best one. Even if it is not, you still have to learn to use it properly.