Intermediate Algebra/Solving Absolute Value Equations

Absolute Values

Absolute Values represented using two vertical bars ($\vert$) are common in Algebra. They are meant to signify the number's distance from 0 on a number line. If the number is negative, it becomes positive. And if the number was positive, it remains positive:

$\left\vert 4\right\vert =4\,$
$\left\vert -4\right\vert =4\,$

For a formal definition:

If $x\geq 0$, then $\left\vert x\right\vert =x$
If $x<0$, then $\left\vert x\right\vert =-x$

The formal definition is simply a declaration of what the function represents at certain restrictions of the $x$-value. For any $x<0$, the output of the graph of the function on the $xy$ plane is that of $y=-x$.

Please note that the opposite (the negative, -) of a negative number is a positive. For example, the opposite of $-1$ is $1$.

Practice Problems

For all of these problems, a = -2 and b = 3. Evaluate the following expressions.

1

$|a|=$

2

$|b|=$

3

$|b+a|=$

Absolute Value Equations

Now, let's say that we're given the equation $\left\vert k\right\vert =8$ and we are asked to solve for $k$. What number would satisfy the equation of $\left\vert k\right\vert =8$? 8 would work, but -8 would also work. That's why there can be two solutions to one equation (and later, even more solutions). (Answer this: why?)

Example 1: Solve for $k$:$|2k+6|=8$ Recall what the absolute value represents: it is the distance of that number to the left or right of the starting point, zero. This means that whatever the inside value represents, it must be either $8$ or $-8$. As such, $2k+6=8\quad {\text{OR}}\quad 2k+6=-8$. All that is left to do is to solve the two equations for $k$: $2k+6=8\qquad 2k+6=-8$ $\Leftrightarrow 2k=2\qquad 2k=-14$ $\Leftrightarrow k=1\qquad k=-7$

A basic principle of solving these absolute value equations is the need to keep the absolute value by itself. This should be enough for most people to understand, yet this phrasing can be a little ambiguous to some students. As such, a lot of practice problems may be in order here.

Example 2: Solve for $k$:$3|2k+6|=12$ We will show you two ways to solve this equation. The first is the standard way, the second will show you something incredible. Standard way: Multiply the constant multiple by its inverse. We'd have to divide both sides by $3$ to get the absolute value by itself. We would set up the two different equations using similar reasoning as in the first example: $2k+6=4\quad {\text{OR}}\quad 2k+6=-4$. Then, we'd solve, by subtracting the 6 from both sides and dividing both sides by 2 to get the $k$ by itself, resulting in $k=-5,-1$. We will leave the solving part as an exercise to the reader. Other way: "Distribute" the three into the absolute value. Play close attention to the steps and reasoning laid out herein, for the reasoning for why this works is just as important as the person using the trick, if not moreso. Let us first generalize the problem. Let there be a positive, non-zero constant multiple $c$ multiplied to the absolute value equation $|2k+6|$: $c\cdot |2k+6|=|c|\cdot |2k+6|\quad {\text{OR}}\quad c\cdot |2k+6|=|-c|\cdot |2k+6|$. Let us assume both are true. If both statements are true, then you are allowed to distribute the positive constant $c$ inside the absolute value. Otherwise, this method is invalid! ${\begin{aligned}|c|\cdot |2k+6|&=|c(2k+6)|&\qquad |-c|\cdot |2k+6|&=|-c(2k+6)|\\&=|2ck+6c|&\qquad &=|-2ck-6c|=|-(2ck+6c)|\\&=|1|\cdot |2ck+6c|={\color {red}1\cdot |2ck+6c|}&\qquad &=|-1|\cdot |2ck+6c|={\color {red}1\cdot |2ck+6c|}\end{aligned}}$ Notice the two equations have the same highlighted answer in red, meaning so long as the value of the constant multiple $c$ is positive, you are allowed to distribute the $c$ inside the absolute value bars. However, this "distributive property" needed the property that multiplying two absolute values is the same as the absolute value of the product. We need to prove this is true first before one can use this in their proof. For the student that spotted this mistake, you may have a good logical mind on one's shoulder, or a good eye for detail. Prove:$|b|\cdot |c|=|bc|$ We will first look at the simplest case: the two values $b{\text{ and }}c$ are constants. We know the following properties are true: $|-b|=|b|=b$ $|-c|=|c|=c$ From this, we can conclude that $|b|\cdot |c|=bc$ for any $b,c\in \mathbb {R}$. We also know the following is true: $bc=d<0\Leftrightarrow |d|=-d>0$. This simply means that for some product $bc$ that equals a negative number $d$, the absolute value of that is $-d$, or the distance from zero. Because $d<0$, multiplying the two sides by $-1$ will change the less than to a greater than, or $d<0\Leftrightarrow -d>0$. $bc=d=0\Leftrightarrow |d|=d=0$. For some product $bc$ that equals a number $d=0$, the absolute value of that is $0$. $bc=d>0\Leftrightarrow |d|=d>0$. For some product $bc$ that equals a positive number $d$, the absolute of the product is $d$. Because we multiplied two absolute values, the product is either positive or zero. We can therefore use the second and third bullet point to conclude that $|b|\cdot |c|=bc\geq 0\Leftrightarrow |d|=d=bc=|bc|\geq 0$. We can use all five bullet points to show that for any constants $b,c\in \mathbb {R}$, $|b|\cdot |c|=|bc|\blacksquare$. This was the simplest case. However, we have already proven the hardest case, where both $b{\text{ and }}c$ are variables. The five bullet points we have is enough to demonstrate this fact. As a result, this proof will be left as a trivial exercise for the reader. By confirming the general case, we may be employ this trick when we see it again. Let us apply this property to the original problem (this gives us the green result below): $3|2k+6|={\color {green}|6k+18|=12}$ This all implies that $6k+18=12\quad {\text{OR}}\quad 6k+18=-12$. From there, a simple use of algebra will show that the answer to the original problem is again $k=-5,-1$.

Let us change the previous problem a little so that the constant multiple is now negative. Without changing much else, what will be true as a result? Let us find out.

Example 3: Solve for $k$:$-4|2k+6|=8$ We will attempt to the problem in two different ways: the standard way and the other way, which we will explain later. Standard way: Multiply the constant multiple by its inverse. Divide like the previous problem, so the equation would look like this: $|2k+6|=-2$. Recall what the absolute value represents: it is the distance of that number to the left or right of the starting point, zero. With this, do you notice anything strange? When you evaluate an absolute value, you will always get a positive number because the distance must always be positive. Because this is means a logically impossible situation, there are no real solutions. Notice how we specifically mentioned "real" solutions. This is because we are certain that the solutions in the real set, $\mathbb {R}$, do not exist. However, there might be some set out there which would have solutions for this type of equation. Because of this posibility, we need to be mathematically rigorous and specifically state "no real solutions." Other way: "Distribute" the constant multiple into the absolute value. Here, we notice that the constant multiple $c<0$. The problem with that is there is no $g$ such that $|g|<0$. The only way this would be true is for $-|g|<0$ because $-|g|<0\qquad {\text{Divide both sides by }}-1$ $|g|>0$ With this property, we may therefore only distribute the constant multiple as $|c|$ with a negative $-1$ as a factor outside the absolute value. As such, $-4|2k+6|=-|8k+24|=8\qquad {\text{Divide both sides by }}-1$ $|8k+24|=-8$ It seems the other way has us multiply a constant by its inverse to both sides. Either way, this "other method" still gave us the same answer: there is no real solution.

The problem this time will be a little different. Keep in mind the principle we had in mind throughout all the examples so far, and be careful because a trap is set in this problem.

Example 4: Solve for $x$:$|3x-3|-3=2x-10$ There are many we ways can attempt to find solutions to this problem. We will do this the standard and allow any student to do it however they so desire. $|3x-3|-3=2x-10\qquad {\text{Add the }}3{\text{ to both sides.}}$ $|3x-3|=2x-7$ Because the absolute value is isolated, we can begin with our generalized procedure. Assuming $2x-7>0$, we may begin by denoting these two equations: (1) $3x-3=2x-7$ (2) $3x-3=-(2x-7)$ These are only true if $2x-7>0$. For now, assume this condition is true. Let us solve for $x$ with each respective equation: Equation (1) $3x-3=2x-7\qquad {\text{Add }}3{\text{ and subtract }}2x{\text{ on both sides.}}$ $x=-4$ Equation (2) $3x-3=-(2x-7)\qquad {\text{Distribute }}-1{\text{.}}$ $3x-3=-2x+7\qquad \quad {\text{Add }}3{\text{ and add }}2x{\text{ on both sides.}}$ $5x=10\qquad \qquad \qquad \quad {\text{Divide }}5{\text{ on both sides.}}$ $x=2$ We have two potential solutions to the equation. Try to answer why we said potential here based on what you know so far about this problem. Why did we state we had two potential solutions? Because we had to assume that $2x-7>0$ and $|3x-3|=2x-7$ is true for the provided $x$. Because of this, we have to verify the solutions to this equation exist. Therefore, let us substitute those values into the equation: $|3(-4)-3|=2(-4)-7$. Notice that the right-hand side is negative. Also, the left-hand side and the right-hand side are not equivalent. Therefore, this is not a solution. $|3(2)-3|=2(2)-7$. Notice the right-hand side is negative, again. Also, the left-hand side and the right-hand side are not equivalent. Therefore, this cannot be a solution. This equations has no real solutions. More specifically, it has two extraneous solutions (i.e. the solutions we found do not satisfy the equality property when we substitute them back in).

Despite doing the procedure outlined since the first problem, you obtain two extraneous solutions. This is not the fault of the procedure but a simple result of the equation itself. Because the left-hand side must always be positive, it means the right-hand side must be positive as well. Along with that restriction is the fact that the two sides may not equal the other for the values whereby only positive values are given. This is all a matter of properties of functions.

Example 5: Solve for $a$:$6\left\vert 5{\frac {a}{6}}+{\frac {1}{12}}\right\vert ={\frac {3}{5}}|15a+15|$ All the properties learned will be needed here, so let us hope you did not skip anything here. It will certainly make our lives easier if we know the properties we are about to employ in this problem. $6\left\vert 5{\frac {a}{6}}+{\frac {1}{12}}\right\vert ={\frac {3}{5}}|15a+15|\qquad {\text{Distribute, so to speak, the constant terms.}}$ $\left\vert 5a+{\frac {1}{2}}\right\vert =|9a+9|$ Looking at the second equation might be the first declaration of absurdity. However, an application of the fundamental properties of absolute values is enough to do this problem. (3) $5a+{\frac {1}{2}}=|9a+9|$ (4) $5a+{\frac {1}{2}}=-|9a+9|$ Peel the problem one layer at a time. For this one, we will categorize equations based on where they come from; this should hopefully explain the dashes: 3-1 is first equation formulated from (3), for example. (3-1) $9a+9=5a+{\frac {1}{2}}$ (3-2) $9a+9=-\left(5a+{\frac {1}{2}}\right)$ (4-1) $-(9a+9)=5a+{\frac {1}{2}}$ (4-2) $-(9a+9)=-\left(5a+{\frac {1}{2}}\right)$ We can demonstrate that some equations are equivalents of the other. For example, (3-1) and (4-2) are equivalent, since dividing both sides of (4-2) by $-1$ gives (3-1). After determining all the equations that are equivalent, distribute $-1$ to the corresponding parentheses. (5) $9a+9=5a+{\frac {1}{2}}$ (6) $9a+9=-5a-{\frac {1}{2}}$ (7) $-9a-9=5a+{\frac {1}{2}}$ Now all that is left to do is solve the equations. We will leave this step as an exercise for the reader. You will find that two of three potential solutions are identical, so that means there are two potential solutions: $a=-{\frac {19}{28}},-{\frac {17}{8}}$. All that is left to do is verify that the equation in the question is true when looking at these specific values of $a$: $a=-{\frac {19}{28}}$ $\left\vert 5\left(-{\frac {19}{28}}\right)+{\frac {1}{2}}\right\vert =\left\vert 9\left(-{\frac {19}{28}}\right)+9\right\vert$ is true. The two sides give the same value: ${\frac {81}{8}}=10.125$. $a=-{\frac {17}{8}}$ $\left\vert 5\left(-{\frac {17}{8}}\right)+{\frac {1}{2}}\right\vert =\left\vert 9\left(-{\frac {17}{8}}\right)+9\right\vert$ is true. The two sides give the same value: ${\frac {81}{28}}\approx 2.893$. Because both solutions are true, the two solutions are $a=-{\frac {19}{28}},-{\frac {17}{8}}\blacksquare$.

Practice Problems

1 $|k+6|=2k$

$k=$

2 $|7+3a|=11-a$

$a\in \{$ , $\}$

3 $|2k+6|+6=0$

How many solutions?

Lesson Review

An absolute value (represented with |'s) stands for the number's distance from 0 on the number line. This essentially makes a negative number positive although a positive number remains the same. To solve an equation involving absolute values, you must get the absolute value by itself on one side and set it equal to the positive and negative version of the other side, because those are the two solutions the absolute value can output. However, check the solutions you get in the end; some might produce negative numbers on the right side, which are impossible because all outputs of an absolute value symbol are positive!

Lesson Quiz

Evaluate each expression.

1

$|-4|=$

2

$|6-8|=$

Solve for $a$. Type NS (with capitalization) into either both fields or the right field for equations with no solutions. Any solutions that are extraneous (don't work when substituted into the equation) should be typed with XS on either the right field or both. Order the solutions from least to greatest.

3 $|3a-4|=5$

$a\in \{$ $,$ $\}$

4 $5|2a+3|=15$

$a\in \{$ $,$ $\}$

5 $3|4a-2|-12=-3$

$a\in \{$ $,$ $\}$

6 $|a+1|-18=a-15$

$a\in \{$ $,$ $\}$

7 $2\left\vert {\frac {a}{2}}-1\right\vert -2a=-4a$

$a\in \{$ $,$ $\}$

Read the situations provided below. Then, answer the prompt or question given. Type NS (with capitalization) into either both fields or the right field for equations that have no solutions. Any solutions that are extraneous should be typed with XS on either the right field or both. Order the solutions from least to greatest.

8 The speed of the current of a nearby river deviates $1.5{\tfrac {\text{m}}{\text{s}}}$ from the average speed $20{\tfrac {\text{m}}{\text{s}}}$. Let $s$ represent the speed of the river. Select all possible equations that could describe the situation.

	$|s-1.5|=20$
	$|s+1.5|=20$
	$|20-s|=1.5$
	$|s-20|=1.5$
	$|s+20|=1.5$
	$|1.5-s|=20$

9 A horizontal artificial river has an average velocity of $-4{\tfrac {\text{m}}{\text{s}}}$. The velocity increases proportionally to the mass of the rocks, $r$, in kilograms, blocking the path of the current. Assume the river's velocity for the day deviates a maximum of $6{\tfrac {\text{m}}{\text{s}}}$. If the proportionality constant is $k={\frac {2}{5}}$ meters per kilograms-seconds, what is the maximum mass of the rocks in the river for that day?

	$5$ kilograms.
	$15$ kilograms.
	$25$ kilograms.
	$35$ kilograms.