# How to arrive at the particular form of inequality as answer?

The stratosphere is the layer of earth's atmosphere that is more than $$10$$ km and less than $$50$$ km above the earth's surface. Write an inequality which describes all possible heights $$x$$, in km, above the earth's surface that are in stratosphere.

Answer: $$|x-30|< 20$$

I can't seem to understand how we got to the answer. Why did we subtract $$30$$ from $$x$$ and where did the $$20$$ come from?

$$10 < x < 50\\ 10-30 < x - 30 < 50-30\\ -20 < x - 30 < 20\\ |x - 30| < 20\\$$

From your question, we could plainly get

$$10 < x < 50$$

It is equivalent to $$-20 < x - 30 < 20$$ Thus $$|x-30| < 20$$

But to be honest, I see no reason the answer wants to write it this way.

• Why it's been given as the answer? Probably to keep the students from cheating. It's an answer a student is unlikely to come up with. Commented Jul 28, 2017 at 19:28
• @A.Ellett Could be.. Commented Jul 28, 2017 at 19:29
• @A.Ellett I do not understand what you mean. // Generally, while I agree this might be a bit of an odd type of problem, the reason why it should be written like this is "an inequality" that is one. Not two as in $10 \lt x \lt 50$.
– quid
Commented Jul 28, 2017 at 19:31
• @quid Right, that could be a stronger reason - the save one side of inequality Commented Jul 28, 2017 at 19:36
• @quid It depends on the approach of the instructor or textbook. Inequalities such as $10<x<50$ are some times construed as a compound inequality, but still since as a unit. But since I don't know the greater context, I'm just guessing. Commented Jul 28, 2017 at 19:38

It's possible that the answer you would come up with is $$10 < x < 50$$ As some have noted, this isn't technically one inequality, but a compound inequality.

Nevertheless, there are two questions you can ask.

• Is this saying the same things as $|x-30|<20$?
• How do I write my answer using absolute value?

For the first question, using the definition of absolute value, you get $$-20 < x - 30 < 20 \\ -20 + 30 < x - 30 + 30 < 20 + 30 \\ 10 < x < 50$$ Thus you can see that the absolute value inequality is saying the same thing.

For the second question, you need to find a value $k$ such that $$10 + k < x + k < 50 + k \\ \text{and}\\ -(10+k) = 50 +k$$ Solving for $k$ gives you $k=-30$.

More generally:

$$a \lt x \lt b \;\;\iff\;\; \left|x - \frac{a+b}{2}\right| \lt \frac{b-a}{2}$$

That's saying that $x$ is in the interval $\,(a,b)\,$ iff the distance between $\,x\,$ and the midpoint $\,\frac{a+b}{2}\,$ is smaller than half the length of the interval $\,\frac{b-a}{2}\,$.

The given problem follows from the above with $\,a=10\,$ and $\,b=50\,$.

Further information :

Use this propertie $|y| = \left \lbrace \begin{array}l y\quad &\text{if} \quad y\geq 0\\ -y\quad &\text{if} \quad y< 0 \end{array}\right.$

$|x-30|< 20 \iff \left \lbrace \begin{array}l x-30<20 \quad (x\geq 30)\quad &(1)\\\\-x+30<20 \quad (x< 30)&(2)\end{array}\right.$

$\left .\begin{array}l (1)\quad x-30<20\iff &x<50\\(2)\quad -x+30<20\iff &x>10 \end{array}\right\rbrace \implies \boxed{10<x<50}$

OR :

$|x-30|< 20\iff (x-30)^2<400 \iff x^2-60x+500<0$

$\Delta= (-60)^2-4\times(500)=1600\iff \sqrt{\Delta}=40$

$x_1=\dfrac{60-40}{2}=10 \;\text{ and} \;\quad x_2=\dfrac{60+40}{2}=50 \implies \boxed{10<x<50}$