Absolute Value Inequalities

Absolute Value Inequalities exercise

• Evaluate the absolute value inequality.

  Hints

  You can visualize how to solve this problem with a number line, like on a thermometer.

  If you subtract the normal body temperature from the highest, or the lowest, healthy body temperature you will always get $1~^\circ F$, or $-1~^\circ F$.

  Solution

  Daisy knows that a healthy body temperature is $98.6~^\circ F$, plus or minus $1~^\circ F$.

  To say this fact using inequalities, we let $x$ stand for a healthy body temperature, and we have that $|x-98.6|\le 1$.

  $~$

  Solving any absolute value inequality $|x|\le v$ is the same as solving two different inequalities, $-v\le x$ and $x\le v$, or when written together, $-v\le x\le v$.

  $~$

  So for the inequality $|x-98.6|\le 1$, we get

  $-1\le x-98.6$ and $x-98.6\ge 1$, or $-1\le x-98.6\le 1$.

  We can then solve both inequalities by using opposite operations:

  $\begin{array}{rcr} x-98.6&\ge&-1~~~\\ \color{#669900}{+98.6}&&\color{#669900}{+98.6}\\ x&\ge&97.6 \end{array}$

  and similarly

  $\begin{array}{rcr} x-98.6&\le&1~~~\\ \color{#669900}{+98.6}&&\color{#669900}{+98.6}\\ x&\le&99.6 \end{array}$

  Combining both solutions, we get

  $97.6\le x\le 99.6$.

  $~$

  So a healthy body temperature is at least $97.6~^\circ F$ and at most $99.6~^\circ F$.

• Identify the critical body temperature range.

  Hints

  You can also solve this problem with a number line.

  The inequality $|x|\le v$ can also be written as $-v\le x$ and $x\le v$.

  Solution

  A body temperature $x$ which is $6~^\circ F$ more or less than the normal body temperature of $98.6^{\circ} F$ gives us the absolute value inequality

  $|x-98.6|\ge 6$.

  Looking at a general inequality with absolute values $|x|\ge v$, we see that we get two statements, which we combine with or:

  $x\le -v$ or $x\ge v$.

  $~$

  Applying this observation to the inequality $|x-98.6|\ge 6$, we get

  $x-98.6\le -6$ or $x-98.6\ge 6$.

  We solve each of these equations using opposite operations:

  $\begin{array}{rcr} x-98.6&\le&-6~~~\\ \color{#669900}{+98.6}&&\color{#669900}{+98.6}\\ x&\le&92.6 \end{array}$

  and similarly,

  $\begin{array}{rcr} x-98.6&\ge&6~~~\\ \color{#669900}{+98.6}&&\color{#669900}{+98.6}\\ x&\ge&104.6 \end{array} $

  $~$

  If Daisy choses a temperature less than or equal to $92.6~^\circ F$ or greater than or equal to $104.6~^\circ F$ she risks a visit to the hospital and could wave byebye to her Super Bowl parade plan.

• Assign the absolute value inequality its corresponding number line.

  Hints

  First look at the number in the middle of the boundary values.

  The absolute value inequality for this number line is given by $|x-0|\le3$.

  The right side of the inequality is given by the distance between the boundary values divided by $2$.

  Solution

  We can solve an absolute value inequality with a number line.

  Any absolute value inequality can be described as a distance from a given value: "The distance from $x$ and $95$ should be less than or equal to $4$" can be represented as $|x-95|\le 4$.

  On the number line:

  • First you have to go four steps to the right of $95$, as well as to the left of $95$.
  • Then, because of the $\le$ sign, you have to fill in the circles at $99$ and $91$.
  • The solution lies between those circles, including the boundary values $91$ and $99$.

  This is shown in the number line to the left.

  When given a number line, first you can check what number is in the middle of the boundary values; let's call it $a$. Next we can find the distance between $a$ and either of the boundary values (as the distance between $a$ and either of the boundary values is the same); let's say the distance between $a$ and either of the boundary values is $v$.

  We now we have $|x-a|..?..v$.

  Our last step is deciding which relation sign we need: if the solution lies between the circles we have $<$ or $\le$, otherwise we have $>$ or $\ge$.

  • If the circles are filled in, we have an $\le$ or $\ge$ sign.
  • If the circles are empty, we have an $<$ or $>$ sign.

• Determine the dangerous and critical temperatures.

  Hints

  Pay attention to and and or.

  Remember:

  • $|x|\le v$ gives us $-v\le x \le v$

  • $|x|\ge v$ gives us $x\le -v$ or $x\ge v$.

  Solution

  Let $x$ represent the warmest or coldest temperature.

  $~$

  Less than or equal to gives us two statements combined by and:

  $|x-24|\le 4.5$

  i.e. $-4.5\le x-24\le 4.5$, which is equivalent to $-4.5\le x-24$ and $x-24\le 4.5$

  Solving these equations,

  $\begin{array}{rcr} -4.5&\le&x-24\\ \color{#669900}{+24}&&\color{#669900}{+24}\\ 19.5&\le& x \end{array}$

  and similarly

  $\begin{array}{rcr} x-24&\le&4.5\\ \color{#669900}{+24}&&\color{#669900}{+24}\\ x&\le& 28.5 \end{array}$

  We know the harmless water temperatures: $x\ge 19.5~^\circ C$ and $x\le 28.5~^\circ C$.

  Now let's find the dangerous water temperatures.

  $~$

  Greater than or equal to leads to two statements combined by or:

  $|x-24|\ge 6$

  i.e. $x-24\le -6$ or $x-24\ge 6$.

  Soliving these equations:

  $\begin{array}{rcr} x-24&\le&-6\\ \color{#669900}{+24}&&\color{#669900}{+24}\\ x&\le& 18 \end{array}$

  and similarly

  $\begin{array}{rcr} x-24&\ge&6\\ \color{#669900}{+24}&&\color{#669900}{+24}\\ x&\ge& 28.5 \end{array}$

  We know the dangerous water temperatures: $x\le 18~^\circ C$ or $x\ge 30~^\circ C$.

• Analyze the absolute value inequalities.

  Hints

  You can see the solution to the inequality $|x|\le 3$ using the number line.

  This number line shows the solution to $|x|\ge 3$.

  Solution

  We can break down absolute value inequalities into two cases:
  $~$
  (1) Less than or equal to gives us two statements combined by and:

  First inequality $|x|<a$

  • $-a<x<a$
  • this is equivalent to $-a<x$ and $x<a$

  Second inequality $|x|\le a$

  • $-a\le x\le a$
  • is equivalent to $-a\le x$ and $x\le a$

  $~$

  (2) Greater than or equal to gives us two statements combined by or:
  $~$
  Third inequality $|x|>a$
  

  • $x<-a$ or $a<x$

  Fourth inequality $|x|\ge a$
  

  • $x\le -a$ or $a\le x$

• Solve the following absolute value inequalities.

  Hints

  For $|x|\le v$, solve the inequalities $-v\le x$ as well as $x\le v$.

  You can also solve such an equation using a number line. So you can see if you have to use and or or.

  For $|x|>v$ solve the inequalities $x<-v$ or $x>v$.

  This number line represents $|x-95|\le 4$.

  Solution

  Keep the following in mind:

  • $|x|\le v$ $\rightarrow$ $-v\le x$ and $x\le v$
  • $|x|< v$ $\rightarrow$ $-v< x$ and $x< v$
  • $|x|\ge v$ $\rightarrow$ $x\le -v$ or $v\le x$
  • $|x|> v$ $\rightarrow$ $x< -v$ or $v> x$

  Because the inequalities are a little more complicated, we have to solve the inequalities each time:

  $~$

  $|x-23|\le 12$

  $\begin{array}{rcr} -12&\le&x-23\\ \color{#669900}{+23}&&\color{#669900}{+23}\\ 11&\le&x \end{array}$

  $\begin{array}{rcr} x-23&\le&12\\ \color{#669900}{+23}&&\color{#669900}{+23}\\ x&\le&35 \end{array}$

  Together we have $11\le x$ and $x\le 35$.

  $~$

  $\left|\frac12x+5\right|\le 7$

  $\begin{array}{rcr} -7&\le&\frac12x+5\\ \color{#669900}{-5}&&\color{#669900}{-5}\\ -12&\le&\frac12x\\ \color{#669900}{\cdot 2}&&\color{#669900}{\cdot 2}\\ -24&\le&x \end{array}$

  $\begin{array}{rcr} \frac12x+5&\le&7\\ \color{#669900}{-5}&&\color{#669900}{-5}\\ \frac12x&\le&2\\ \color{#669900}{\cdot 2}&&\color{#669900}{\cdot 2}\\ x&\le&4 \end{array}$

  Together we have $-24\le x$ and $x\le 4$.

  $~$

  $|x+6|>5$

  $\begin{array}{rcr} x+6&<&-5\\ \color{#669900}{-6}&&\color{#669900}{-6}\\ x&<&-11 \end{array}$

  $\begin{array}{rcr} x+6&>&5\\ \color{#669900}{-6}&&\color{#669900}{-6}\\ x&>&-1 \end{array}$

  Together we have $x<-11$ or $-1<x$.

  $~$

  $|3x-12|\ge 9$

  $\begin{array}{rcr} 3x-12&\le&-9\\ \color{#669900}{+12}&&\color{#669900}{+12}\\ 3x&\le&3\\ \color{#669900}{\overline{\div 3}}&&\color{#669900}{\overline{\div 3}}\\ x&\le&1 \end{array}$

  $\begin{array}{rcr} 3x-12&\ge&9\\ \color{#669900}{+12}&&\color{#669900}{+12}\\ 3x&\ge&21\\ \color{#669900}{\overline{\div 3}}&&\color{#669900}{\overline{\div 3}}\\ x&\ge&7 \end{array}$

  Together we have $x\le 1$ or $7\le x$.