Whether it's a washer, nut, bolt, or gear, when a machine part is made, it must be made to fit with all of the other parts of the system. Since no manufacturing process is perfect, there are small deviations from the norm when each piece is made. In fact, manufacturers have a range of acceptable values for each measurement of every screw, bolt, etc.
Figure11.3.1 Alternative Video Lesson
Let's say we were examining some new bolts just out of the factory. The manufacturer specifies that each bolt should be within a tolerance of 0.04 mm to 10 mm in diameter. So the lowest diameter that the bolt could be to make it through quality assurance is 0.04 mm smaller than 10 mm, which is 9.96 mm. Similarly, the largest diameter that the bolt could be is 0.04 mm larger than 10 mm, which is 10.04 mm.
Summarizing, we want the difference between the actual diameter and the specification to be less than or equal to 0.04 mm. Since absolute values are used to describe distances, we can summarize our thoughts mathematically as \(\abs{x-10}\le 0.04\text{,}\) where \(x\) represents the diameter of an acceptably sized bolt, in millimeters. Since the minimum value is 9.96 mm and the maximum value is 10.04 mm, our range of acceptable values should be \(9.96 \le x \le 10.04\text{.}\)
In this section we will examine a variety of problems and applications that relate to this sort of math with absolute values.
Subsection11.3.1Solving Absolute Value Equations
Recall in Section 11.1 that we learned that graphs of absolute value function are in general shaped like “V”s. We can now solve some absolute value equations graphically.
Example11.3.2
Solve the equations graphically using the graphs provided.
Since the graph of \(y=\abs{x}\) crosses \(y=3\) at the \(x\)-values \(-3\) and \(3\text{,}\) the solution set to the equation \(\abs{x}=3\) must be \(\{-3,3\}\text{.}\)
Since the graph of \(y=\abs{2x+3}\) crosses \(y=5\) at the \(x\)-values \(-4\) and \(1\text{,}\) the solution set to the equation \(\abs{2x+3}=5\) must be \(\{-4,1\}\text{.}\)
Remark11.3.3
At this point, please note that there is a big difference between the expression \(\abs{3}\) and the equation \(\abs{x}=3\text{.}\)
The expression \(\abs{3}\) is describing the distance from \(0\) to the number \(3\text{.}\) The distance is just \(3\text{.}\) So \(\abs{3}=3\text{.}\)
The equation \(\abs{x}=3\) is asking you to find the numbers that are a distance of \(3\) from \(0\text{.}\) We saw in Explanation 11.3.2.1 that these two numbers are \(3\) and \(-3\text{.}\)
Example11.3.4
Verify that the value \(4\) is a solution to the absolute value equation \(\abs{2x-3}=5\text{.}\)
Verify that the value \(\frac{3}{2}\) is a solution to the absolute value equation \(\abs{\frac{1}{6}x-\frac{1}{2}}=\frac{1}{4}\text{.}\)
We will substitute the value \(\substitute{\frac{3}{2}}\) into the absolute value equation \(\abs{\frac{1}{6}x-\frac{1}{2}}=\frac{1}{4}\text{.}\) We get:
Now we will learn to solve absolute value equations algebraically. To motivate this, we will think about what an absolute value equation means in terms of the “distance from zero” definition of absolute value. If
where \(n\ge0\text{,}\) then this means that we want all of the numbers, \(X\text{,}\) that are a distance \(n\) from \(0\text{.}\) Since we can only go left or right along the number line, this is describing both \(X=n\) as well as \(X=-n\text{.}\)
Fact 11.3.6 says that the equation \(\abs{x}=6\) is the same as
\begin{equation*}
x=6\text{ or } x=-6\text{.}
\end{equation*}
Thus the solution set is \(\{6,-6\}\text{.}\)
Fact 11.3.6 doesn't actually apply to the equation \(\abs{x}=-4\) because the value on the right side is negative. How often is an absolute value of a number negative? Never! Thus, there are no solutions and the solution set is the empty set, denoted \(\emptyset\text{.}\)
The equation \(\abs{5x-7}=23\) breaks into two pieces, each of which needs to be solved independently.
Thus, the equation \(\abs{3-4x}=0\) only has one solution, and the solution set is \(\left\{\frac{3}{4}\right\}\text{.}\)
Now we will look at an equation with an absolute value expression on each side, such as \(\abs{x}=\abs{2x+6}\text{.}\) Since \(\abs{x}=5\) has two solutions, you might be wondering how many solutions \(\abs{x}=\abs{2x+6}\) will have. Let's look at a graph to find out.
Figure 11.3.8 shows that there are also two points of intersection between the graphs of \(y=\abs{x}\) and \(y=\abs{2x+6}\text{.}\) The solutions to the equation \(\abs{x}=\abs{2x+6}\) are the \(x\)-values where the graphs cross. So, the solution set is \(\{-6,-2\}\text{.}\)
Example11.3.9
Solve the equation \(\abs{x+1}=\abs{2x-4}\) graphically.
First break up the equation into the left side and the right side and graph each separately, as in \(y=\abs{x+1}\) and \(y=\abs{2x-4}\text{.}\) We can see in the graph that the graphs intersect twice. The \(x\)-values of those intersections are \(1\) and \(5\) so the solution set to the equation \(\abs{x+1}=\abs{2x-4}\) is \(\{1,5\}\text{.}\)
Figure11.3.10 \(y=\abs{x+1}\) and \(y=\abs{2x-4}\)
Fact11.3.11Equations with Two Absolute Value Expressions
Let \(X\) and \(Y\) be linear algebraic expressions. Then, the equation
\begin{equation*}
\abs{X}=\abs{Y}
\end{equation*}
has the same solutions as
\begin{equation*}
X=Y\text{ or } X=-Y\text{.}
\end{equation*}
Remark11.3.12
You might be confused as to why the negative sign has to go on the right side of the equation in \(X=-Y\text{.}\) Well, it doesn't: it can go on either side of the equation. The equations \(X=-Y\) and \(-X=Y\) are equivalent. Similarly, \(-X=-Y\) is equivalent to \(X=Y\text{.}\) That's why we only need to solve two of the four possible equations.
Note that one of the two pieces gives us an equation with no solutions. Since \(0\ne3\text{,}\) we can safely ignore this piece. Thus the only solution is \(\frac{1}{2}\text{.}\)
We should visualize this equation graphically because our previous assumption was that two absolute value graphs would cross twice. The graph shows why there is only one crossing: the left and right sides of each “V” are parallel.
Note that our second equation is an identity so recall from Section 3.6 that the solution set is “all real numbers.”
So, our two pieces have solutions \(1\) and “all real numbers.” Since \(1\) is a real number and we have an or statement, our overall solution set is \((-\infty,\infty)\text{.}\) The graph confirms our answer since the two “V” graphs are coinciding.
Subsection11.3.2Solving Absolute Value Inequalities
Now we turn our attention away from equations and onto absolute value inequalities. Don't dismiss this topic as it will actually be used in some capacity in many subsequent math courses. So let's give these the full treatment. We start with a graphical interpretation of what \(\abs{2x-1} \le 5\) means.
Graphically solving the inequality \(\abs{2x-1} \le 5\) means looking for the \(x\)-values where the graph of \(y=\abs{2x-1}\) is below (or touching) the line \(y=5\text{.}\) On the graph the highlighted region of \(y=\abs{2x-1}\) is the portion that is below the line \(y=5\text{,}\) and the \(x\)-values in that region are \([-2,5]\text{.}\)
To solve the inequality \(\abs{\frac{2}{3}x+1} \lt 3\text{,}\) we will start by making a graph with both \(y=\abs{\frac{2}{3}x+1}\) and \(y=3\text{.}\)
The portion of the graph of \(y=\abs{\frac{2}{3}x+1}\) that is below \(y=3\) is highlighted and the \(x\)-values of that highlighted region are trapped between \(-6\) and \(3\text{:}\) \(-6 \lt x \lt 3\text{.}\) That means that the solution set is \((-6,3)\text{.}\) Note that we shouldn't include the endpoints of the interval because at those values, the two graphs are equal whereas the original inequality was only less than and not equal.
Figure11.3.17 \(y=\abs{\frac{2}{3}x+1}\) and \(y=3\)
For a more verbal approach to understanding the concept, let's try to describe “values that are less than \(4\) units from \(0\text{.}\)” We would say that those are “numbers between \(-4\) and \(4\text{.}\)” Let's translate each sentence into math. “Values that are less than \(4\) units from \(0\)” translates to “\(\abs{x}\lt 4\text{,}\)” and the piece “numbers between \(-4\) and \(4\)” translates to be “\(-4 \lt x \lt 4\text{.}\)”
For a graphical interpretation, let's think in terms of the “distance from zero” definition of absolute value. If
where \(n\ge0\text{,}\) then we want all of the numbers, \(X\text{,}\) that are a distance \(n\) or less from \(0\text{.}\) Since we can only go left or right along the number line, this is describing all numbers from \(-n\) to \(n\text{.}\)
The inequality \(\abs{x} \lt 9\) breaks down into a triple inequality:
\begin{equation*}
-9 \le x \le 9
\end{equation*}
This inequality is already written in simplest form and all that remains for us to do is to write the solution set in interval notation: \([-9,9]\text{.}\)
Fact 11.3.19 doesn't apply to the inequality \(\abs{x} \lt -6\) because the right side is a negative number. Let's translate the meaning of the inequality into English. It says, “The distance from \(0\) to what numbers is less than \(-6\text{?}\)” Since we define distance to be non-negative, there are no possible numbers that are less than \(-6\) units distance from \(0\text{.}\) Thus, the solution set is the empty set, denoted \(\emptyset\text{.}\)
The inequality \(\abs{4x+3} \lt 9\) breaks down into a triple inequality that we can then solve:
So, the solution set to the inequality is \(\left(-3,\frac{3}{2}\right)\text{.}\)
The inequality \(3\cdot\abs{3-x}+1\le 13\) must be simplified into the form that matches Fact 11.3.19, so we will first isolate the absolute value expression on the left side of the inequality:
So, the solution set to the inequality is \([-1,7]\text{.}\)
Example11.3.21
If a machined circular washer must have a circumference that is within 0.2 mm of 36 mm, then what is the acceptable range for the radius of the washer? Round your answers to the nearest hundredth of a millimeter.
We will first define the radius of the washer to be \(r\text{,}\) measured in millimeters. The formula \(C=2\pi r\) gives us the circumference, \(C\text{,}\) of a circle with radius \(r\text{.}\) Now we know that “distance” between the circumference and our preferred circumference of 36 mm must be less than or equal to 0.2 mm. In math, this translates to
To solve this we will use Fact 11.3.19 to break the absolute value inequality into a triple inequality:
\begin{align*}
-0.2 \amp\le 2\pi r-36 \le 0.2\\
-0.2\addright{36} \amp\le 2\pi r-36\addright{36} \le 0.2\addright{36}\\
35.8 \amp\le 2\pi r \le 36.2\\
\divideunder{35.8}{2\pi} \amp\le \divideunder{2\pi r}{2\pi} \le \divideunder{36.2}{2\pi}\\
5.70 \amp\le r \le 5.76\amp\text{(note: these values are rounded)}
\end{align*}
This shows that the radius must be somewhere between 5.70 mm and 5.76 mm, inclusive.
The last few examples have all revolved around absolute values being less than some value. We now need to investigate what happens when we have an absolute value that is greater than a value. We will again start with a graphical interpretation.
Example11.3.22
To graphically solve the inequality \(\abs{x-1} \gt 3\) would mean looking for the \(x\)-values where the graph of \(y=\abs{x-1}\) is above the line \(y=3\text{.}\)
On the graph the highlighted region of \(y=\abs{x-1}\) is the portion that is above the line \(y=3\) and the \(x\)-values in that region can be represented by \((-\infty,-2)\cup(4,\infty)\text{.}\)
To solve the inequality \(\abs{\frac{1}{3}x+2} \ge 6\text{,}\) we will start by making a graph with both \(y=\abs{\frac{1}{3}x+2}\) and \(y=6\text{.}\)
The portion of the graph of \(y=\abs{\frac{1}{3}x+2}\) that is above \(y=6\) is highlighted and the \(x\)-values of that highlighted region are those below (or equal to) \(-24\) and those above (or equal to) \(12\text{:}\) \(x \le -24 \text{ or } x \ge 12\text{.}\) That means that the solution set is \((-\infty,-24)\cup(12,\infty)\text{.}\)
Figure11.3.25 \(y=\abs{\frac{1}{3}x+2}\) and \(y=3\)
Again, for a more verbal approach to understanding the concept, lets try to describe “values that are more than \(4\) units from \(0\text{.}\)” We would say that those are “numbers below \(-4\) as well as numbers above \(4\text{.}\)” We will again translate each sentence into math. “Values that are more than \(4\) units from \(0\)” translates to “\(\abs{x}\gt4\text{,}\)” and the piece “numbers below \(-4\) as well as numbers above \(4\)” translates to be “\(x \lt -4 \text{ or } x \gt 4\text{.}\)”
For a graphical interpretation, let's think in terms of the “distance from zero” definition of absolute value. If
where \(n\ge0\text{,}\) then we want all of the numbers, \(X\text{,}\) that are a distance \(n\) or more from \(0\text{.}\) Since we can only go left or right along the number line, this is describing all numbers below \(-n\) as well as those above \(n\text{.}\)
Figure11.3.26 A Numberline with Points a Distance \(n\) or less from \(0\)
Fact11.3.27An Absolute Value Expression Greater Than a Value
Let \(n\) be a non-negative number and \(X\) be a linear algebraic expression.
Then, the inequality \(\abs{X} \gt n\) has the same solutions as the compound inequality \(X \lt -n \text{ or } X \gt n\text{.}\)
Likewise, the inequality \(\abs{X} \ge n\) has the same solutions as the compound inequality \(X \le -n \text{ or } X \ge n\text{.}\)
Remark11.3.28
Since Fact 11.3.27 specifies that an “absolute value greater than a number”-type inequality breaks down into an or statement, we will therefore need to find the union of the solution sets of the pieces.
Example11.3.29
Solve the absolute value inequalities using Fact 11.3.27.
The inequality \(\abs{x} \ge 4\) breaks down into a compound inequality:
\begin{align*}
x \amp \le -4 \amp\amp\text{or}\amp x \amp \ge 4
\end{align*}
So, the solution set is \((-\infty,-4]\cup[4,\infty)\text{.}\)
Fact 11.3.27 doesn't apply to the inequality \(\abs{x} \gt -2\) because the right side is negative. Instead, we will make sense of it logically. This is asking, “When is an absolute value greater than a negative number?” The answer is that absolute values are always bigger than negative numbers! So, our solution set is \((-\infty,\infty)\text{.}\)
The inequality \(\abs{5x-7} \gt 7\) breaks down into a compound inequality:
Our final simplified solution set is \((-\infty,3]\cup[6,\infty)\text{.}\)
Example11.3.30
Phuong is taking the standard climbing route on Mount Hood from Timberline Lodge up the Southside Hogsback and back down. Her altitude can be very closely modeled by an absolute value function since the angle of ascent is nearly constant. Let \(x\) represent the number of miles walked from Timberline Lodge, and let \(f(x)\) represent the altitude, in miles, after walking for a distance \(x\text{.}\) The altitude can be modeled by \(f\left(x\right)=2.1-0.3077\cdot\abs{x-3.25}\text{.}\) Note that below Timberline Lodge this model fails to be accurate.
Solve the equation \(f(x)=1.1\) and interpret the results in the context of the problem.
Altitude sickness can occur at altitudes above \(1.5\) miles. Set up and solve an inequality to find out how far Phuong can walk the trail and still be under \(1.5\) miles of elevation.
At this point, we can use Fact 11.3.6 to split apart the equation:
\begin{align*}
x-3.25 \amp\approx 3.25 \amp \text{or} \amp\amp x-3.25 \amp\approx -3.25\\
x \amp\approx 6.5 \amp \text{or} \amp\amp x \amp\approx 0
\end{align*}
According to the model, Phuong will be at \(1.1\) miles of elevation after walking about \(0\) miles and about \(6.5\) miles along the trail. This seems to imply that Timberline Lodge is very close to \(1.1\) miles of elevation. In addition, it implies that the entire hike is \(6.5\) miles round trip, ending at Timberline Lodge again.
The inequality we are looking for will describe when the altitude is below \(1.5\) miles. Since \(f(x)\) is the altitude, the inequality we need is:
\begin{equation*}
f(x)\lt 1.5
\end{equation*}
To solve this, we need to input the formula and simplify before using one of the absolute value inequality rules.
\begin{align*}
f(x)\amp\lt 1.5\\
2.1-0.3077\cdot\abs{x-3.25} \amp\lt 1.5\\
-0.3077\cdot\abs{x-3.25} \amp\lt -0.6\\
\divideunder{-0.3077\cdot\abs{x-3.25}}{-0.3077} \amp\mathbin{\highlight{\gt}} \divideunder{-0.6}{-0.3077}\\
\abs{x-3.25} \amp\gt 1.95\amp\text{(note: this value is rounded)}
\end{align*}
At this point, we can use Fact 11.3.27 to split apart the inequality:
\begin{align*}
x-3.25 \amp\lt -1.95 \amp \text{or} \amp\amp x-3.25 \amp\gt 1.95\\
x \amp\lt 1.3 \amp \text{or} \amp\amp x \amp\gt 5.2
\end{align*}
The image only shows the portion of the graph that is above Timberline Lodge, which we learned was at \(1.1\) miles in elevation in the previous part. The highlighted portions of the graph are those indicated by \(x\gt 5.2 \text{ or } x \lt 1.3\text{.}\)
Figure11.3.31 \(y=f(x)\text{,}\) the Graph of the Mt Hood Ascent and Descent
In conclusion, based both on our math and the reality of the situation, regions of the trail that are below \(1.5\) miles are those that are from Timberline Lodge (at \(0\) miles on the trail), to \(1.3\) miles along the trail and then also from \(5.2\) miles along the trail (and by now we are on our way back down) to \(6.5\) miles along the trail (back at Timberline Lodge). If we wanted to write this in interval notation, we might write \([0,1.3)\cup(5.2,6.5]\text{.}\) There is a big portion along the trail (from \(1.3\) miles to \(5.2\) miles) that Phuong will be above the \(1.5\) mile altitude and should watch for signs of altitude sickness.
Write the equation \(6 = |3 x| - 5\) as two separate equations. Neither of your equations should use absolute value.
Solve both equations above.
14
Write the equation \(7 = |5 x| - 3\) as two separate equations. Neither of your equations should use absolute value.
Solve both equations above.
15
Write the equation \(\displaystyle \left| 8 - \frac{r}{7} \right| = 3\) as two separate equations. Neither of your equations should use absolute value.
Solve both equations above.
16
Write the equation \(\displaystyle \left| 2 - \frac{r}{5} \right| = 7\) as two separate equations. Neither of your equations should use absolute value.
Solve both equations above.
17
Verify that the value \(-1\) is a solution to the absolute value equation \(\abs{\frac{x-3}{2}}=2\text{.}\)
Verify that the value \(\frac{2}{3}\) is a solution to the absolute value equation \(\abs{6x-5}\lt 4\text{.}\)
18
Verify that the value \(8\) is a solution to the absolute value equation \(\abs{\frac{1}{2}x-2}=2\text{.}\)
Verify that the value \(6\) is a solution to the absolute value equation \(\abs{7-2x}\ge 5\text{.}\)
19
Solve the following equation.
\(\displaystyle{ \left\lvert 3 x - 9 \right\rvert = 9 }\)
