Lane ORCCA (2020–2021): Open Resources for Community College Algebra

To solve the equations graphically, first we need to graph the right sides of the equations also.

1. \(\abs{x}=3\)

   

   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 \(\{-3,3\}\text{.}\)

2. \(\abs{2x+3}=5\)

   

   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 \(\{-4,1\}\text{.}\)

1. We will substitute the value \(\substitute{4}\) into the absolute value equation \(\abs{2x-3}=5\text{.}\) We get:

   \begin{align*} \abs{2x-3}\amp=5\\ \abs{2\cdot\substitute{4}-3}\amp\stackrel{?}{=}5\\ \abs{8-3}\amp\stackrel{?}{=}5\\ \abs{5}\amp\stackrel{\checkmark}{=}5 \end{align*}

2. 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:

   \begin{align*} \abs{\frac{1}{6}x-\frac{1}{2}}\amp=\frac{1}{4}\\ \abs{\frac{1}{6}\cdot\substitute{\frac{3}{2}}-\frac{1}{2}}\amp\stackrel{?}{=}\frac{1}{4}\\ \abs{\frac{1}{4}-\frac{1}{2}}\amp\stackrel{?}{=}\frac{1}{4}\\ \abs{-\frac{1}{4}}\amp\stackrel{\checkmark}{=}\frac{1}{4} \end{align*}

1. Fact 12.3.5 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 \(\{6,-6\}\text{.}\)

2. Fact 12.3.5 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 \(\emptyset\text{.}\)

3. The equation \(\abs{5x-7}=23\) breaks into two pieces, each of which needs to be solved independently.

   \begin{align*} 5x-7\amp=23\amp\amp\text{or}\amp 5x-7\amp=-23\\ 5x\amp=30\amp\amp\text{or}\amp 5x\amp=-16\\ x\amp=6\amp\amp\text{or}\amp x\amp=-\frac{16}{5} \end{align*}

   Thus, the solution set is \(\left\{5,-\frac{16}{5}\right\}\)

4. The equation \(\abs{14-3x}=8\) breaks into two pieces, each of which needs to be solved independently.

   \begin{align*} 14-3x\amp=8\amp\amp\text{or}\amp 14-3x\amp=-8\\ -3x\amp=-6\amp\amp\text{or}\amp -3x\amp=-22\\ x\amp=2\amp\amp\text{or}\amp x\amp=\frac{22}{3} \end{align*}

   Thus, the solution set is \(\left\{2,\frac{22}{3}\right\}\)

5. The equation \(\abs{3-4x}=0\) breaks into two pieces, each of which needs to be solved independently.

   \begin{align*} 3-4x\amp=0\amp\amp\text{or}\amp 3-4x\amp=-0\\ \end{align*}

   Since these are identical equations, all we have to do is solve one equation.

   \begin{align*} 3-4x\amp=0\\ -4x\amp=-3\\ x\amp=\frac{3}{4} \end{align*}

   Thus, the equation \(\abs{3-4x}=0\) only has one solution and the solution set is \(\left\{\frac{3}{4}\right\}\text{.}\)

1. The inequality \(\abs{x} \lt 9\) breaks down into a double 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: \([-9,9]\text{.}\)

2. Fact 12.3.8 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 \(\emptyset\text{.}\)

3. The inequality \(\abs{4x+3} \lt 9\) breaks down into a triple inequality that we can then solve:

   \begin{gather*} -9 \lt 4x+3 \lt 9\\ -9\subtractright{3} \lt 4x+3\subtractright{3} \lt 9\subtractright{3}\\ -12 \lt 4x \lt 6\\ \divideunder{-12}{4} \lt \divideunder{4x}{4} \lt \divideunder{6}{4}\\ -3 \lt x \lt \frac{3}{2} \end{gather*}

   So, the solution set to the inequality is \(\left(-3,\frac{3}{2}\right)\text{.}\)

4. The inequality \(3\cdot\abs{3-x}+1\le 13\) must be simplified into the form that matches Fact 12.3.8, so we will first isolate the absolute value expression on the left side of the inequality:

   \begin{align*} 3\cdot\abs{3-x}+1 \amp \le 13\\ 3\cdot\abs{3-x} \amp\le 12\\ \abs{3-x} \amp\le 4 \end{align*}

   Now that we have the absolute value isolated, we can rewrite it into a double inequality that we can finish solving:

   \begin{align*} -4 \amp \le 3-x \le 4\\ -4\subtractright{3} \amp \le 3-x \subtractright{3} \le 4 \subtractright{3}\\ -7 \amp\le -x \le 1\\ \divideunder{-7}{-1} \amp \mathbin{\highlight{\ge}} \divideunder{-x}{-1} \mathbin{\highlight{\ge}} \divideunder{1}{-1}\\ 7 \amp \ge x \ge -1 \end{align*}

   So, the solution set to the inequality is \([-1,7]\text{.}\)

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

Now we can substitute our formula for circumference and solve for \(r\text{.}\)

To solve this we will use Fact 12.3.8 to break the absolute value inequality into a double inequality:

This shows that the radius must be somewhere between 5.70 mm and 5.76 mm, inclusive.

1. 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 \((-\infty,-4]\cup[4,\infty)\text{.}\)

2. Fact 12.3.12 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 \((-\infty,\infty)\text{.}\)

3. The inequality \(\abs{5x-7} \gt 7\) breaks down into a compound inequality:

   \begin{align*} 5x-7 \amp \lt -7 \amp\amp\text{or}\amp 5x-7 \amp \gt 7\\ 5x \amp \lt 0 \amp\amp\text{or}\amp 5x \amp \gt 14\\ x \amp \lt 0 \amp\amp\text{or}\amp x \amp \gt \frac{14}{5} \end{align*}

   We will write the solution set as \((-\infty,0)\cup\left(\frac{14}{5},\infty\right)\text{.}\)

4. Before we break up the inequality \(2\cdot\abs{3-2x}-5\ge 13\) into an “or” statement, we must isolate the absolute value expression:

   \begin{align*} 2\cdot\abs{3-2x}-5\amp\ge 13\\ 2\cdot\abs{3-2x}\amp\ge 18\\ \abs{3-2x}\amp\ge 9 \end{align*}

   Now that the absolute value expression has been isolated on the left side, we can use Fact 12.3.12 to break it into an “or” statement:

   \begin{align*} 3-2x \amp\le -9\amp\amp\text{or}\amp 3-2x \amp\ge 9\\ -2x \amp\le -12 \amp\amp\text{or}\amp -2x \amp\ge 6\\ x \amp\ge 6 \amp\amp\text{or}\amp x \amp\le -3 \end{align*}

   Our final simplified solution set is \((-\infty,-3]\cup[6,\infty)\text{.}\)

1. First, we substitute the formula for \(f(x)\) and simplify the equation.

   \begin{align*} f(x)\amp=1.1\\ 2.1-0.3077\cdot\abs{x-3.25} \amp=1.1\\ -0.3077\cdot\abs{x-3.25} \amp= -1\\ \divideunder{-0.3077\cdot\abs{x-3.25}}{-0.3077} \amp= \divideunder{-1}{-0.3077}\\ \abs{x-3.25} \amp\approx 3.25 \end{align*}

   At this point, we can use Fact 12.3.5 to rewrite 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.

2. 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 that this value is rounded\}}\amp \end{align*}

   At this point, we can use Fact 12.3.12 to rewrite 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 \(x\gt 5.2 \text{ or } x \lt 1.3\text{.}\)

   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.