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What can ensure a quality approach when analyzing elements such as water or soil? How about when making chocolate? Predetermined ranges for specific characteristics, that is. For instance, it is possible to use inequalities that involve absolute values to ensure the quality of a chocolate bar or in understanding elements of nature. Here, one-variable absolute value inequalities will be solved, and their solution sets will be drawn.
### Catch-Up and Review

**Here are a few recommended readings before getting started with this lesson.**

Challenge

From the school yard to the kitchen, three states of water are observable at a moments notice: solid, liquid, and gas.

It is known that, at atmospheric pressure, water freezes at $32_{∘}F$ and vaporizes at $212_{∘}F.$

a Write the range of temperatures in which water is not liquid.

b Graph the given range of when water freezes and vaporizes.

c Write the absolute value inequality that describes this situation.

Explore

An inequality that results from combining two inequalities by using the the word

andor the word

or, is called a compound inequality. The applet shows compound inequalities and their solution sets. Examine the solution set for the indicated inequality and explore how it changes for different compound inequalities.

Can these compound inequalities be represented in a different form? Is it possible to use an absolute value expression?

Discussion

An absolute value inequality is an inequality that involves the absolute value of an expression containing a variable. As with other inequalities, absolute value inequalities can be strict or non-strict.

Strict Absolute Value Inequalities | Non-Strict Absolute Value Inequalities | ||
---|---|---|---|

$∣x+2∣>5$ | $∣x+7∣<5$ | $∣2x∣≥10$ | $∣x−2∣≤4$ |

$∣x∣<a⇕-a<x<a ∣x∣≤a⇕-a≤x≤a $

Likewise, if $a≥0,$ an absolute value inequality of the form $∣x∣>a$ can be seen as the set of all numbers that are less than $-a$ $∣x∣>a⇕x<-aorx>a ∣x∣≥a⇕x≤-aorx≥a $

As with other inequalities, absolute value inequalities can be represented by an interval on a number line. Open points at the ends of the interval represent strict inequalities where the corresponding values are not included in the interval. Conversely, closed points represent non-strict inequalities and the corresponding values are included in the interval.
If the absolute value inequality involves an expression of the form $∣x−b∣$ with $b>0$ rather than just $∣x∣,$ the interval will be translated $b$ units to the right. Likewise, if the expression is of the form $∣x+b∣,$ the interval will be translated $b$ units to the left.

Method

Using the definition of absolute value, an absolute value inequality can be written as a compound inequality. Therefore, solving absolute value inequalities is quite similar to solving compound inequalities. To detail the reasoning, the following inequality will be solved. *expand_more*
*expand_more*
*expand_more*
*expand_more*

$3∣x−7∣+2>20 $

There are four steps to follow.
1

Isolate the Absolute Value Expression

To be able to rewrite the given inequality as a compound inequality, the absolute value expression should be isolated first. To do so, the Properties of Inequalities and inverse operations will be used.
Here, the number on the other side of the absolute value is positive. However, it should be noted that if that number were negative, then the inequality either has *no solution* or *all real numbers* as a solution depending on the inequality symbol.

$Example Inequality∣x−7∣<-5∣x−7∣>-5 Solution SetNo SolutionAll Real Numbers $

Since the absolute value of an expression is always non-negative, $∣x−7∣$ should be greater than or equal to $0.$ Therefore, all real numbers satisfy $∣x−7∣>-5,$ and there is no $x-$value which satisfies $∣x−7∣<-5.$ 2

Rewrite It as a Compound Inequality

Now, the absolute value inequality $∣x−7∣>6$ will be rewritten. It represents the set of all numbers that are *less than* $-6$ or *greater than* $6.$

$∣x−7∣>6⇓x−7<-6orx−7>6 $

3

Solve the Individual Inequalities

Next, the individual inequalities will be solved by adding $7$ to both sides of the inequalities.

$Inequality Ix−7<-6⇓x<1 Inequality IIx−7>6⇓x>13 $

The solution set of Inequality I contains all real numbers less than $1.$ The solutions to Inequality II are all real numbers greater than $13.$ 4

Combine the Solution Sets

The combination of the solution sets for the individual inequalities is the solution set of the given absolute value inequality. Since the derived compound inequality was combined with the word

or,the combination of the solution sets is also written with that word.

$x<1orx>13 $

Example

Izabella goes on a tour of a chocolate factory. They aim to produce chocolate bars weighing $93$ grams. A machine at the factory weighs chocolates chosen at random. The bars must not deviate from the predetermined weight by more than $5$ grams. Otherwise, the machine has to send them back.

a Write a range of allowable weights $w$ for a chocolate bar using a compound inequality.

b Graph the range on a number line.

c Write an absolute value inequality that describes this situation.

a $88≤w≤98$

b **Graph:**

c $∣w−93∣≤5$

a Determine the maximum and minimum acceptable weights.

b Should the solution set contain the maximum and minimum $w-$values?

c Which inequality symbol should be used?

a For a chocolate bar to be sent back, the difference between the weight of the chocolate bar and the predetermined weight is more than $5$ grams. Therefore, an acceptable weight $w$ for a chocolate bar, not to be sent back, must be *greater than or equal to* $93−5=88$ grams.

$88≤w $

At the same time, it must weigh $w≤98 $

Consequently, the combination of these individual inequalities will result in a compound inequality describing the range of acceptable weights.
$88≤wandw≤98⇕88≤w≤98 $

b The inequality $88≤w$ is the set of values which are *greater than or equal to* $88.$ Since the inequality sign is non-strict, $88$ itself is included.

The other inequality represents all the points to the left of $98.$ Again, the inequality sign is non-strict, so $98$ is included.

Because the inequalities are combined using word

the solution set of the resulting compound inequality is equal to the union of the solution sets of the individual inequalities. Therefore, the graph of the acceptable weight range is the combination of the graphs.
*and*,

c Any chocolate bar that varies from the predetermined weight by more than $5$ grams is sent back. Therefore, the absolute value of the difference between the weight of a bar $w$ and the predetermined weight should be *less than or equal to* $5.$ This can be expressed as follows.

$∣w−93∣≤5 $

Since this absolute value inequality can be rewritten as the compound inequality obtained in the previous steps, both have the same set of solutions.
Example

Izabella, looking around the chocolate factory, discovers a room full of chocolate fondue fountains on sale for special events! The prices are high and they vary significantly. She decides to make a list of the prices to see which is a fair price.

a To define a fair price, Izabella decides a fondue with a price within $$100$ of the average price of all the fondues is fair. Write an absolute value inequality describing the situation. Which prices meet this condition?

b Izabella decides to narrow the range. She thinks that a fondue that is no more than $20$ above the average price is what is really fair. Write an absolute value inequality for this case and determine the prices that meet this condition.

a **Absolute Value Inequality:** $∣x−447∣<100$

**Prices:** $$358,$ $$450,$ $$476,$ and $$480$

b **Absolute Value Inequality:** $∣x−447∣≤20$

**Price:** $$450$

a The average price of the chocolate fondues is the sum of the prices divided by the number of prices. The absolute value of the difference between the amount of money Izabella thinks is fair and the average price should be *less than* $100.$

b The absolute value of the difference between the amount of money Izaballa thinks is fair and the average price should be *less than or equal to* $20.$

a To find the average price $p$ of the chocolate fondues, the sum of the prices will be divide by the number of prices, $8.$

$p=Number of ValuesSum of Values $

SubstituteValues

Substitute values

$p=8450+476+560+327+358+325+600+480 $

$p=447$

$Absolute Value Inequality ∣x−447∣<100 $

By solving this absolute value inequality, the chocolate fondues satisfying this condition can be determined. To do so, first rewrite it as a compound inequality and then solve for $x.$
$∣x−447∣<100⇓-100<x−447<100⇓347<x<547 $

The chocolate fondues whose prices are within this range are $$450,$476,$358,and$480 $

b In the previous part, the average price of the fondue's was found to be $$447.$ Now, the range is narrowed to $$20.$ The absolute value of the difference between $x$ and $447$ should be *less than or equal to* $20.$

$∣x−447∣≤20 $

By solving the inequality, the fondue prices satisfying this condition can be determined.
$∣x−447∣≤20⇓-20≤x−447≤20⇓427≤x≤467 $

The fondue prices are within this range are those which Izabella would consider to be priced fairly.
There is only one price that meets Izabella's condition.
$$450 $

Example

Izabella learns that the factory has future plans to develop packaging that can withstand the harshest conditions, including Mars! The temperature of the surface of Mars reaches its highest value at the equator where it is less than $36_{∘}C.$ Mars reaches its lowest temperature at the poles where it is always greater than $-144_{∘}C.$

External credits: ESA & MPS for OSIRIS Team MPS/UPD/LAM/IAA/RSSD/INTA/UPM/DASP/IDA

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Start by writing a compound inequality for the possible temperature values $t$ on Mars.

From the given information, the temperature on Mars is always *less than* $36_{∘}C$ but *greater than* $-144_{∘}C.$
The distance $d$ to the midpoint from the endpoints is $90.$ Therefore, the midpoint is the number $-144+90=-54.$
The points that are within $90$ units of the midpoint represent the range can be written as the following absolute inequality. Note that since the endpoints are not included, the inequality should be strict.

$At the equator:At the poles: t<36t>-144 $

These individual inequalities can be combined to form an equivalent compound inequality that describe the possible temperatures on Mars.
$-144<tandt<36⇕-144<t<36 $

To write this compound inequality as an absolute value inequality, the midpoint between $-144$ and $36$ on the number line should be found. Let $d$ be the distance to the midpoint.
To find $d,$ half of the difference between the endpoints will be calculated.
$d=236−(-144) $

$d=90$

$∣t−(-54)∣<90⇕∣t+54∣<90 $

Now that Izabella has this information, she is even more impressed with the chocolate factory!
Example

At this special chocolate factory the average salary for a Chocolatier is a whopping $$45700.$

As a company policy, a new Chocolatier's actual salary can only differ from the company average by less than $$1250.$

a Write and solve an absolute value inequality to find a range for the possible salaries $s$ for a new Chocolatier. Draw the range on a number line.

b Write an absolute value inequality for the salaries that are **not** offered to new Chocolatiers.

a **Absolute Value Inequality:** $∣s−45700∣<1250$

**Graph:**

**Range:** $44450<s<49950$

b **Absolute Value Inequality:** $∣s−45700∣≥1250$

a The absolute value of the difference between the salary for a new Chocolatier and the average salary of any Chocolatier is less than $1250.$

b Use the graph drawn in the part A.

a This special chocolate factory pays a new Chocolatier's salary within $$1250$ of the company's average salary of $$45700.$ This means that the absolute value of the difference between the salary $s$ for a new employee and the average salary of any Chocolatier is *less than* $1250.$

$Absolute Value Inequality ∣s−45700∣<1250 $

By solving this absolute value inequality, the range for the possible salaries can be found. To do so, the inequality will be rewritten. $∣s−45700∣<1250⇓-1250<s−45700<1250⇓44450<s<46950 $

The solution set of the inequality is the set of values between $44450$ and $46950.$ Since the inequality sign is strict, those numbers are not included.
b Consider the graph of the absolute value inequality found in Part A.

The values that are more than $1250$ units away from the average salary in the number line, need to be represented by an absolute value inequality.

These values can be written as the following absolute value inequality.$∣s−45700∣≥1250 $

Pop Quiz

Analyze the given graph and determine the corresponding absolute value inequality.

Closure

In this lesson, the relationship between absolute value inequalities and compound inequalities has been explained using real-world examples. Considering those examples, the challenge presented at the beginning can now be solved seamlessly.

It is known that, at atmospheric pressure, water freezes at $32_{∘}F$ and vaporizes at $212_{∘}F.$

a Write the range of temperatures in which water is not liquid.

b Graph this range.

c Write the absolute value inequality that describes this situation.

a $t<32$ or $t>212$

b **Graph:**

c $∣t−122∣>90$

a Determine the values of temperature that water freezes or vaporizes.

b Does the solution set contain the numbers $32$ and $212?$

c Find the middle point between $32$ and $212$ on the number line.

a Water is not liquid when it is in frozen form nor when it is vapor form. Noting the given information, it is known that, at atmospheric pressure, water is not liquid for temperatures *less than* $32_{∘}F$ or *greater than* $212_{∘}F$

$Solid:Gas: t<32t>212 $

The values $32_{∘}F$ and $212_{∘}F$ are not included because at these temperatures water starts to change its physical state and the liquid form of water can be present in both states of the transition. Under these conditions, the following compound inequality shows the temperatures in which water is not a liquid. $t<32ort>212 $

b The compound inequality is the combination of two inequalities. The inequality $t<32$ shows the set of values *less than* $32.$ Since the inequality sign is strict, the value $32$ is not included.

The other inequality represents all of the points to the right of $212.$ Again, the inequality sign is strict, so $212$ is not included.

Since the inequalities were combined with the word

the solution set of the resulting compound inequality is the union of the solutions sets of the two individual inequalities. Therefore, the graph of the range is the combination of the above graphs.
*or*,

c To write the absolute value equation, the midpoint between $32$ and $212$ on the number line should be found. To do so, half the difference between these numbers will be calculated.

$2212−32 =2180 ⇒2180 =90 $

The midpoint is $90$ units away from the endpoints. Therefore, the midpoint is $32+90=122.$
The points that are further away in units than the calculated distance of $90$ represent the range, which can be written as the following absolute inequality.
$∣t−122∣>90 $

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