ORCCAOpen 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{?}{=}5\\ 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{?}{=}\frac{1}{4}\\ \frac{1}{4}\amp\stackrel{\checkmark}{=}\frac{1}{4} \end{align*}

1. Fact 11.3.10 says that the equation \(\abs{x}=6\) is the same as

   \begin{equation*} x=6\quad\text{or}\quad x=-6\text{.} \end{equation*}

   Thus the solution set is \(\{6,-6\}\text{.}\)

2. Fact 11.3.10 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-7\addright{7}\amp=23\addright{7}\amp\amp\text{or}\amp 5x-7\addright{7}\amp=-23\addright{7}\\ 5x\amp=30\amp\amp\text{or}\amp 5x\amp=-16\\ \divideunder{5x}{5}\amp=\divideunder{30}{5}\amp\amp\text{or}\amp \divideunder{5x}{5}\amp=-\divideunder{16}{5}\\ x\amp=5\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\\ 14-3x\subtractright{14}\amp=8\subtractright{14}\amp\amp\text{or}\amp 14-3x\subtractright{14}\amp=-8\subtractright{14}\\ -3x\amp=-6\amp\amp\text{or}\amp -3x\amp=-22\\ \divideunder{-3x}{-3}\amp=\divideunder{-6}{-3}\amp\amp\text{or}\amp \divideunder{-3x}{-3}\amp=\divideunder{-22}{-3}\\ 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\\ 3-4x\subtractright{3}\amp=0\subtractright{3}\\ -4x\amp=-3\\ \divideunder{-4x}{-4}\amp=\divideunder{-3}{-4}\\ 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 equation \(\abs{x-4}=\abs{3x-2}\) breaks down into two pieces:

   \begin{align*} x-4\amp=3x-2\amp\amp\text{or}\amp x-4\amp=-(3x-2)\\ x-4\amp=3x-2\amp\amp\text{or}\amp x-4\amp=\highlight{-3x+2}\\ -2\amp=2x\amp\amp\text{or}\amp 4x\amp=6\\ \divideunder{-2}{2}\amp=\divideunder{2x}{2}\amp\amp\text{or}\amp \divideunder{4x}{4}\amp=\divideunder{6}{4}\\ -1\amp=x\amp\amp\text{or}\amp x\amp=\frac{3}{2} \end{align*}

   So, the solution set is \(\left\{-1,\frac{3}{2}\right\}\text{.}\)

2. The equation \(\abs{\frac{1}{2}x+1}=\abs{\frac{1}{3}x+2}\) breaks down into two pieces:

   \begin{align*} \frac{1}{2}x+1\amp=\frac{1}{3}x+2\amp\amp\text{or}\amp \frac{1}{2}x+1\amp=-\left(\frac{1}{3}x+2\right)\\ \frac{1}{2}x+1\amp=\frac{1}{3}x+2\amp\amp\text{or}\amp \frac{1}{2}x+1\amp=\highlight{-\frac{1}{3}x-2}\\ \multiplyleft{6}\left(\frac{1}{2}x+1\right)\amp=\multiplyleft{6}\left(\frac{1}{3}x+2\right)\amp\amp\text{or}\amp \multiplyleft{6}\left(\frac{1}{2}x+1\right)\amp=\multiplyleft{6}\left(-\frac{1}{3}x-2\right)\\ 3x+6\amp=2x+12\amp\amp\text{or}\amp 3x+6\amp=-2x-12\\ x\amp=6 \amp\amp\text{or}\amp 5x\amp=-18\\ x\amp=6 \amp\amp\text{or}\amp \divideunder{5x}{5}\amp=-\divideunder{18}{5}\\ x\amp=6 \amp\amp\text{or}\amp x\amp=-\frac{18}{5} \end{align*}

   So, the solution set is \(\left\{6,-\frac{18}{5}\right\}\text{.}\)

3. The equation \(\abs{x-2}=\abs{x+1}\) breaks down into two pieces:

   \begin{align*} x-2\amp=x+1\amp\amp\text{or}\amp x-2\amp=-(x+1)\\ x-2\amp=x+1\amp\amp\text{or}\amp x-2\amp=\highlight{-x-1}\\ x-2\addright{2}\subtractright{x}\amp=x+1\addright{2}\subtractright{x}\amp\amp\text{or}\amp x-2\addright{2}\addright{x}\amp=-x-1\addright{2}\addright{x}\\ 0\amp=3\amp\amp\text{or}\amp 2x\amp=1\\ 0\amp\ne3\amp\amp\text{or}\amp x\amp=\frac{1}{2} \end{align*}

   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.

   

4. The equation \(\abs{x-1}=\abs{1-x}\) breaks down into two pieces:

   \begin{align*} x-1\amp=1-x\amp\amp\text{or}\amp x-1\amp=-(1-x)\\ x-1\amp=1-x\amp\amp\text{or}\amp x-1\amp=\highlight{-1+x}\\ \addleft{x}x-1\addright{1}\amp=\addleft{1}1-x\addright{x}\amp\amp\text{or}\amp x\subtractright{x}-1\addright{1}\amp=-1\addright{1}+x\subtractright{x}\\ 2x\amp=2\amp\amp\text{or}\amp 0\amp=0\\ x\amp=1\amp\amp\text{or}\amp 0\amp=0 \end{align*}

   Note that our second equation is an identity so recall from Section 3.5 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 the two “V” graphs are coinciding.

   

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.

1. 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: \([-9,9]\text{.}\)

2. Fact 11.3.23 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 equation 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 11.3.23, so we will first isolate the absolute value expression on the left side of the equation:

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

   Now that we have the absolute value isolated, we can split it into a triple 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 11.3.23 to break the absolute value inequality into a triple inequality:

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

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) \(12\text{:}\) \(x \le -24 \quad\text{or}\quad x \ge 12\text{.}\) That means that the solution set is \((-\infty,-24)\cup(12,\infty)\text{.}\)

1. The inequality \(\abs{x} \ge 4\) breaks down into a compound inequality:

   \begin{align*} x \amp \ge 4 \amp\amp\text{or}\amp -x \amp \ge 4\\ x \amp \ge 4 \amp\amp\text{or}\amp \divideunder{-x}{-1} \amp \mathbin{\highlight{\le}} \divideunder{4}{-1}\\ x \amp \ge 4 \amp\amp\text{or}\amp x \amp \le -4 \end{align*}

   So, the solution set is \([4,\infty)\cup(-\infty,-4]\text{.}\) At this point we should mention that it is “nicer” to write this set in “order”: \((-\infty,-4]\cup[4,\infty)\text{.}\) They are, however, identical and either answer would be correct.

2. Fact 11.3.30 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 \gt 7 \amp\amp\text{or}\amp -(5x-7) \amp \gt 7\\ 5x-7 \amp \gt 7 \amp\amp\text{or}\amp \highlight{-5x+7} \amp \gt 7\\ 5x-7\addright{7} \amp \gt 7\addright{7} \amp\amp\text{or}\amp -5x+7\subtractright{7} \amp \gt 7\subtractright{7}\\ 5x \amp \gt 14 \amp\amp\text{or}\amp -5x \amp \gt 0\\ \divideunder{5x}{5} \amp \gt \divideunder{14}{5} \amp\amp\text{or}\amp \divideunder{-5x}{-5} \amp \mathbin{\highlight{\lt}} \divideunder{0}{-5}\\ x \amp \gt \frac{14}{5} \amp\amp\text{or}\amp x \amp \lt 0 \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}-5\addright{5}\amp\ge 13\addright{5}\\ 2\cdot\abs{3-2x}\amp\ge 18\\ \divideunder{2\cdot\abs{3-2x}}{2}\amp\ge \divideunder{18}{2}\\ \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 11.3.30 to break it into an “or” statement:

   \begin{align*} 3-2x \amp\ge 9 \amp\amp\text{or}\amp -(3-2x) \amp\ge 9\\ 3-2x \amp\ge 9 \amp\amp\text{or}\amp \highlight{-3+2x}\amp\ge 9\\ 3-2x\subtractright{3} \amp\ge 9\subtractright{3} \amp\amp\text{or}\amp -3+2x\addright{3} \amp\ge 9\addright{3}\\ -2x \amp\ge 6 \amp\amp\text{or}\amp 2x \amp\ge 12\\ \divideunder{-2x}{-2} \amp\mathbin{\highlight{\le}} \divideunder{6}{-2} \amp\amp\text{or}\amp \divideunder{2x}{2} \amp\ge \divideunder{12}{2}\\ x \amp\le -3 \amp\amp\text{or}\amp x \amp\ge 6 \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\\ 2.1-0.3077\cdot\abs{x-3.25}\subtractright{2.1} \amp= 1.1\subtractright{2.1}\\ -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 11.3.10 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, a hiker will be at \(1.1\) miles of elevation after walking for both about \(6.5\) miles along the trail or after hiking about \(0\) miles. 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\\ 2.1-0.3077\cdot\abs{x-3.25}\subtractright{2.1} \amp\lt 1.5\subtractright{2.1}\\ -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 approximate\}}\amp \end{align*}

   At this point, we can use Fact 11.3.30 to split apart the inequality:

   \begin{align*} x-3.25 \amp\gt 1.95 \amp \text{or} \amp\amp -(x-3.25) \amp\gt 1.95\\ x-3.25 \amp\gt 1.95 \amp \text{or} \amp\amp -x+3.25 \amp\gt 1.95\\ x \amp\gt 5.2 \amp \text{or} \amp\amp -x \amp\gt -1.3\\ x \amp\gt 5.2 \amp \text{or} \amp\amp x \amp\mathbin{\highlight{\lt}} 1.3 \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 up 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 the hiker will be above the \(1.5\) mile altitude and thus prone to altitude sickness if not in good health.