## How to solve absolute value equations

|x + 5| = 3.

Worksheet on Abs Val Equations Abs Val Eqn Solver

The General Steps to solve an absolute value equation are:

- Rewrite the absolute value equation as two separate equations, one positive and the other negative
- Solve each equation separately
- After solving, substitute your answers back into original equation to verify that you solutions are valid
- Write out the final solution or graph it as needed

It's always easiest to understand a math concept by looking at some examples so, check outthe many examples and practice problems below.

You can always check your work with our Absolute value equations solver too

## Practice Problems

Example equation.

Solve the equation: | X + 5| = 3

Click here to practice more problems like this one , questions that involve variables on 1 side of the equation.

Some absolute value equations have variables both sides of the equation. However, that will not change the steps we're going to follow to solve the problem as the example below shows:

Solve the equation: |3 X | = X − 21

Solve the following absolute value equation: | 5X +20| = 80

Solve the following absolute value equation: | X | + 3 = 2 X

This first set of problems involves absolute values with x on just 1 side of the equation (like problem 2 ).

Solve the following absolute value equation: |3 X −6 | = 21

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## Solving Simpler Absolute-Value Equations

Simpler Harder Special Case

When we take the absolute value of a number, we always end up with a positive number (or zero). Whether the input was positive or negative (or zero), the output is always positive (or zero). For instance, | 3 | = 3 , and | −3 | = 3 also.

This property — that both the positive and the negative become positive — makes solving absolute-value equations a little tricky. But once you learn the "trick", they're not so bad. Let's start with something simple:

Content Continues Below

## MathHelp.com

Solving Absolute Value Equations

## Solve | x | = 3

I've pretty much already solved this, in my discussion above:

| −3 | = 3

So then x must be equal to 3 or equal to −3 .

But how am I supposed to solve this if I don't already know the answer? I will use the positive / negative property of the absolute value to split the equation into two cases, and I will use the fact that the "minus" sign in the negative case indicates "the opposite sign", not "a negative number".

For example, if I have x = −6 , then " − x " indicates "the opposite of x " or, in this case, −(−6) = +6 , a positive number. The "minus" sign in " − x " just indicates that I am changing the sign on x . It does not indicate a negative number. This distinction is crucial!

Whatever the value of x might be, taking the absolute value of x makes it positive. Since x might originally have been positive and might originally have been negative, I must acknowledge this fact when I remove the absolute-value bars. I do this by splitting the equation into two cases. For this exercise, these cases are as follows:

a. If the value of x was non-negative (that is, if it was positive or zero) to start with, then I can bring that value out of the absolute-value bars without changing its sign, giving me the equation x = 3 .

b. If the value of x was negative to start with, then I can bring that value out of the absolute-value bars by changing the sign on x , giving me the equation − x = 3 , which solves as x = −3 .

Then my solution is

x = ±3

We can, by the way, verify the above solution graphically. When we attempt to solve the absolute-value equation | x | = 3 , we are, in effect, setting two line equations equal to each other and finding where they cross. For instance:

In the above, I've plotted the graph of y 1 = | x | (being the blue line that looks like a "V") and y 2 = 3 (being the green horizontal line). These two graphs cross at x = −3 and at x = +3 (being the two red dots).

If you're wanting to check your answers on a test (before you hand it in), it can be helpful to plug each side of the original absolute-value equation into your calculator as their own functions; then ask the calculator for the intersection points.

Of course, any solution can also be verified by plugging it back into the original exercise, and confirming that the left-hand side (LHS) of the equation simplifies to the same value as does the right-hand side (RHS) of the equation. For the equation above, here's my check:

x = −3

LHS: | x | = | −3 |

LHS: | x | = | +3 |

If you're ever in doubt about your solution to an equation, try graphing or else try plugging your solution back into the original question. Checking your work is always okay!

The step in the above, where the absolute-value equation was restated in two forms, one with a "plus" and one with a "minus", gives us a handy way to simplify things: When we have isolated the absolute value and go to take off the bars, we can split the equation into two cases; we will signify these cases by placing a "minus" on the opposite side of the equation (for one case) and a "plus" on the opposite side (for the other). Here's how this works:

## Solve | x + 2 | = 7 , and check your solution(s).

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The absolute value is isolated on the left-hand side of the equation, so it's already set up for me to split the equation into two cases. To clear the absolute-value bars, I must split the equation into its two possible two cases, one each for if the contents of the absolute-value bars (that is, if the "argument" of the absolute value) is negative and if it's non-negative (that is, if it's positive or zero). To do this, I create two new equations, where the only difference between then is the sign on the right-hand side. First, I'll do the "minus" case:

x + 2 = −7

x = −9

Now I'll do the non-negative case, where I can just drop the bars and solve:

Now I need to check my solutions. I'll do this by plugging them back into the original equation, since the grader can't see me checking plots on my graphing calculator.

x = −9:

LHS: |(−9) + 2|

= |−7| = 7 = RHS

LHS: |(5) + 2|

= |7| = 7 = RHS

Both solutions check, so my answer is:

x = −9, 5

## Solve | 2 x − 3 | − 4 = 3

First, I'll isolate the absolute-value part of the equation; that is, I'll get the absolute-value expression by itself on one side of the "equals" sign, with everything else on the other side:

| 2 x − 3 | − 4 = 3

| 2 x − 3 | = 7

Now I'll clear the absolute-value bars by splitting the equation into its two cases, one for each sign on the argument. First I'll do the negative case:

2 x − 3 = −7

2 x = −4

x = −2

And then I'll do the non-negative case:

2 x − 3 = 7

The exercise doesn't tell me to check, so I won't. (But, if I'd wanted to, I could have plugged "abs(2X−3)−4" and "3" into my calculator (as Y1 and Y2, respectively), and seen that the intersection points were at my x -values.) My answer is:

x = −2, 5

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## 4.3: Absolute Value Equations

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In the previous section, we defined

\[|x|=\left\{\begin{array}{ll}{-x,} & {\text { if } x<0} \\ {x,} & {\text { if } x \geq 0}\end{array}\right. \nonumber \]

and we saw that the graph of the absolute value function defined by f(x) = |x| has the “V-shape” shown in Figure \(\PageIndex{1}\).

It is important to note that the equation of the left-hand branch of the “V” is y = −x. Typical points on this branch are (−1, 1), (−2, 2), (−3, 3), etc. It is equally important to note that the right-hand branch of the “V” has equation y = x. Typical points on this branch are (1, 1), (2, 2), (3, 3), etc.

## Solving |x| = a

We will now discuss the solutions of the equation

\[|x|=a \nonumber \]

There are three distinct cases to discuss, each of which depends upon the value and sign of the number a.

- Case I: a < 0

If a < 0, then the graph of y = a is a horizontal line that lies strictly below the x-axis, as shown in Figure \(\PageIndex{2}\)(a). In this case, the equation |x| = a has no solutions because the graphs of y = a and y = |x| do not intersect.

- Case II: a = 0

If a = 0, then the graph of y = 0 is a horizontal line that coincides with the x-axis, as shown in Figure \(\PageIndex{2}\)(b). In this case, the equation |x| = 0 has the single solution x = 0, because the horizontal line y = 0 intersects the graph of y = |x| at exactly one point, at x = 0.

- Case III: a > 0

If a > 0, then the graph of y = a is a horizontal line that lies strictly above the x-axis, as shown in Figure \(\PageIndex{2}\)(c). In this case, the equation |x| = a has two solutions, because the graphs of y = a and y = |x| have two points of intersection.

Recall that the left-hand branch of y = |x| has equation y = −x, and points on this branch have the form (−1, 1), (−2, 2), etc. Because the point where the graph of y = a intersects the left-hand branch of y = |x| has y-coordinate y = a, the x-coordinate of this point of intersection is x = −a. This is one solution of |x| = a.

Recall that the right-hand branch of y = |x| has equation y = x, and points on this branch have the form (1, 1), (2, 2), etc. Because the point where the graph of y = a intersects the right-hand branch of y = |x| has y-coordinate y = a, the x-coordinate of this point of intersection is x = a. This is the second solution of |x| = a.

This discussion leads to the following key result.

The solution of |x| = a depends upon the value and sign of a.

The equation |x| = a has no solutions.

The equation |x| = 0 has one solution, x = 0.

The equation |x| = a has two solutions, x = −a or x = a.

Let’s look at some examples.

## Example \(\PageIndex{1}\)

Solve |x| = −3 for x.

The graph of the left-hand side of |x| = −3 is the “V” of Figure \(\PageIndex{2}\)(a). The graph of the right-hand side of |x| = −3 is a horizontal line three units below the x-axis. This has the form of the sketch in Figure \(\PageIndex{2}\)(a). The graphs do not intersect. Therefore, the equation |x| = −3 has no solutions.

An alternate approach is to consider the fact that the absolute value of x can never equal −3. The absolute value of a number is always nonnegative (either zero or positive). Hence, the equation |x| = −3 has no solutions.

## Example \(\PageIndex{2}\)

Solve |x| = 0 for x

This is the case shown in Figure \(\PageIndex{2}\)(b). The graph of the left-hand side of |x| = 0 intersects the graph of the right-hand side of |x| = 0 at x = 0. Thus, the only solution of |x| = 0 is x = 0.

Thinking about this algebraically instead of graphically, we know that 0 = 0, but there is no other number with an absolute value of zero. So, intuitively, the only solution of |x| = 0 is x = 0.

## Example \(\PageIndex{3}\)

Solve |x| = 4 for x.

The graph of the left-hand side of |x| = 4 is the “V” of Figure \(\PageIndex{2}\)(c). The graph of the right-hand side is a horizontal line 4 units above the x-axis. This has the form of the sketch in Figure \(\PageIndex{2}\)(c). The graphs intersect at (−4, 4) and (4, 4). Therefore, the solutions of |x| = 4 are x = −4 or x = 4.

Alternatively, | − 4| = 4 and |4| = 4, but no other real numbers have absolute value equal to 4. Hence, the only solutions of |x| = 4 are x = −4 or x = 4.

## Example \(\PageIndex{4}\)

Solve the equation |3 − 2x| = −8 for x.

If the equation were |x| = −8, we would not hesitate. The equation |x| = −8 has no solutions. However, the reasoning applied to the simple case |x| = −8 works equally well with the equation |3 − 2x| = −8. The left-hand side of this equation must be nonnegative, so its graph must lie above or on the x-axis. The right-hand side of |3−2x| = −8 is a horizontal line 8 units below the x-axis. The graphs cannot intersect, so there is no solution.

We can verify this argument with the graphing calculator. Load the left and righthand sides of |3 − 2x| = −8 into Y1 and Y2, respectively, as shown in Figure \(\PageIndex{3}\)(a). Push the MATH button on your calculator, then right-arrow to the NUM menu, as shown in Figure \(\PageIndex{3}\)(b). Use 1:abs( to enter the absolute value shown in Y1 in Figure \(\PageIndex{3}\)(a). From the ZOOM menu, select 6:ZStandard to produce the image shown in Figure \(\PageIndex{3}\)(c).

Note, that as predicted above, the graph of y = |3 − 2x| lies on or above the xaxis and the graph of y = −8 lies strictly below the x-axis. Hence, the graphs cannot intersect and the equation |3 − 2x| = −8 has no solutions.

Alternatively, we can provide a completely intuitive solution of |3 − 2x| = −8 by arguing that the left-hand side of this equation is nonnegative, but the right-hand side is negative. This is an impossible situation. Hence, the equation has no solutions.

## Example \(\PageIndex{5}\)

Solve the equation |3 − 2x| = 0 for x.

We have argued that the only solution of |x| = 0 is x = 0. Similar reasoning points out that |3 − 2x| = 0 only when 3 − 2x = 0. We solve this equation independently.

\[\begin{aligned} 3-2 x &=0 \\-2 x &=-3 \\ x &=\frac{3}{2} \end{aligned} \nonumber \] Thus, the only solution of |3 − 2x| = 0 is x = 3/2.

It is worth pointing out that the “tip” or “vertex” of the “V” in Figure \(\PageIndex{3}\)(c) is located at x = 3/2. This is the only location where the graphs of y = |3 − 2x| and y = 0 intersect.

## Example \(\PageIndex{6}\)

Solve the equation |3 − 2x| = 6 for x.

In this example, the graph of y = 6 is a horizontal line that lies 6 units above the x-axis, and the graph of y = |3 − 2x| intersects the graph of y = 6 in two locations. You can use the intersect utility to find the points of intersection of the graphs, as we have in Figure \(\PageIndex{4}\)(b) and (c).

We need a way of summarizing this graphing calculator approach on our homework paper. First, draw a reasonable facsimile of your calculator’s viewing window on your homework paper. Use a ruler to draw all lines. Complete the following checklist.

- Label each axis, in this case with x and y.
- Scale each axis. To do this, press the WINDOW button on your calculator, then report the values of xmin, xmax, ymin, and ymax on the appropriate axis.
- Label each graph with its equation.
- Drop dashed vertical lines from the points of intersection to the x-axis. Shade and label these solutions of the equation on the x-axis.

Following the guidelines in the above checklist, we obtain the image in Figure \(\PageIndex{5}\).

Algebraic Approach . One can also use an algebraic technique to find the two solutions of |3 − 2x| = 6. Much as |x| = 6 has solutions x = −6 or x = 6, the equation

\[|3-2 x|=6 \nonumber \]

is possible only if the expression inside the absolute values is either equal to −6 or 6. Therefore, write

\[3-2 x=-6 \qquad \text { or } \qquad 3-2 x=6 \nonumber \]

and solve these equations independently

\[\begin{array}{rlrrrl}{3-2 x}&{=}&{-6} & {\text { or }} & {3-2 x}&{=}&{6} \\ {-2 x}&{=}&{-9} && {-2 x}&{=}&{3} \\ {x}&{=}&{\frac{9}{2}} && {x}&{=}&{-\frac{3}{2}}\end{array} \nonumber \]

Because −3/2 = −1.5 and 9/2 = 4.5, these exact solutions agree exactly with the graphical solutions in Figure \(\PageIndex{4}\)(b) and (c).

Let’s summarize the technique involved in solving this important case.

Solving |expression| = a, when a > 0. To solve the equation

\[| \text { expression } |=a, \quad \text { when } a>0 \nonumber \]

\[\text { expression }=-a \qquad \text { or } \qquad \text { expression }=a \nonumber \]

then solve each of these equations independently.

For example:

• To solve |2x + 7| = 5, set \[2x + 7 = −5 \qquad or \qquad 2x + 7 = 5 \nonumber \], then solve each of these equations independently.

• To solve |3 − 5x| = 9, set \[3 − 5x = −9 \qquad or \qquad 3 − 5x = 9 \nonumber \], then solve each of these equations independently.

• Note that this technique should not be applied to the equation |2x + 11| = −10, because the right-hand side of the equation is not a positive number. Indeed, in this case, no values of x will make the left-hand side of this equation equal to −10, so the equation has no solutions.

Sometimes we have to do a little algebra before removing the absolute value bars.

## Example \(\PageIndex{7}\)

Solve the equation \[|x+2|+3=8 \nonumber \] for x.

First, subtract 3 from both sides of the equation. \[\begin{aligned}|x+2|+3 &=8 \\|x+2|+3-3 &=8-3 \end{aligned} \nonumber \]

This simplifies to \[|x+2|=5 \nonumber \]

Now, either \[x+2=-5 \qquad \text { or } \qquad x+2=5 \nonumber \]

each of which can be solved separately.

\[\begin{array}{rrlrrl}{x+2} & {=} & {-5} & {\text { or }} & {x+2} & {=} & {5} \\ {x+2-2} & {=} & {-5-2} && {x+2-2} & {=} & {5-2} \\ {x} & {=} & {-7} && {x} & {=} & {3}\end{array} \nonumber \]

## Example \(\PageIndex{8}\)

Solve the equation \[3|x-5|=6 \nonumber \] for x.

First, divide both sides of the equation by 3

\[\begin{aligned} 3|x-5| &=6 \\ \frac{3|x-5|}{3} &=\frac{6}{3} \end{aligned} \nonumber \]

This simplifies to \[|x-5|=2 \nonumber \]

Now, either \[x-5=-2 \qquad \text { or } \qquad x-5=2 \nonumber \]

\[\begin{array}{rllrrl}{x-5} & {=} & {-2} & {\text { or }} & {x-5} & {=} & {2} \\ {x-5+5} & {=} & {-2+5} && {x-5+5} & {=} & {2+5} \\ {x} & {=} & {3} && {x} & {=} & {7}\end{array} \nonumber \]

## Properties of Absolute Value

An example will motivate the need for some discussion of the properties of absolute value.

## Example \(\PageIndex{9}\)

Solve the equation \[\left|\frac{x}{2}-\frac{1}{3}\right|=\frac{1}{4} \nonumber \] for x.

It is tempting to multiply both sides of this equation by a common denominator as follows.

\[\begin{array}{l}{\left|\dfrac{x}{2}-\dfrac{1}{3}\right|=\dfrac{1}{4}} \\ {12\left|\dfrac{x}{2}-\dfrac{1}{3}\right|=12\left(\dfrac{1}{4}\right)}\end{array} \nonumber \]

If it is permissible to move the 12 inside the absolute values, then we could proceed as follows.

\[\begin{aligned}\left|12\left(\frac{x}{2}-\frac{1}{3}\right)\right| &=3 \\|6 x-4| &=3 \end{aligned} \nonumber \]

Assuming for the moment that this last move is allowable, either

\[6 x-4=-3 \qquad \text { or } \qquad 6 x-4=3 \nonumber \]

Each of these can be solved separately, first by adding 4 to both sides of the equations, then dividing by 6.

\[\begin{array}{rllrrl}{6 x-4} & {=} & {-3} & {\text { or }} & {6 x-4} & {=} & {3} \\ {6 x} & {=} & {1} & &{6 x} & {=} & {7} \\ {x} & {=} & {1 / 6} && {x} & {=} & {7 / 6}\end{array} \nonumber \]

As we’ve used a somewhat questionable move in obtaining these solutions, it would be wise to check our results. First, substitute x = 1/6 into the original equation.

\[\begin{aligned}\left|\frac{x}{2}-\frac{1}{3}\right| &=\frac{1}{4} \\\left|\frac{1 / 6}{2}-\frac{1}{3}\right| &=\frac{1}{4} \\\left|\frac{1}{12}-\frac{1}{3}\right| &=\frac{1}{4} \end{aligned} \nonumber \]

Write equivalent fractions with a common denominator and subtract.

\[\begin{aligned}\left|\frac{1}{12}-\frac{4}{12}\right| &=\frac{1}{4} \\\left|-\frac{3}{12}\right| &=\frac{1}{4} \\\left|-\frac{1}{4}\right| &=\frac{1}{4} \end{aligned} \nonumber \]

Clearly, x = 1/6 checks. We’ll leave the check of the second solution to our readers.

Well, we’ve checked our solutions and they are correct, so it must be the case that

\[12\left|\frac{x}{2}-\frac{1}{3}\right|=\left|12\left(\frac{x}{2}-\frac{1}{3}\right)\right| \nonumber \]

But why? After all, absolute value bars, though they do act as grouping symbols, have a bit more restrictive meaning than ordinary grouping symbols such as parentheses, brackets, and braces.

We state the first property of absolute values.

If a and b are any real numbers, then

\[|a b|=|a||b| \nonumber \]

We can demonstrate the validity of this property by simply checking cases.

- If a and b are both positive real numbers, then so is ab and \(|a||b|=a b\). On the other hand, \(|a||b|=a b\). Thus, \(|ab| = |a||b|\).
- If a and b are both negative real numbers, then ab is positive and \(|ab| = ab\). On the other hand, \(|a||b| = (−a)(−b) = ab\). Thus, \(|ab| = |a||b|\).

We will leave the proof of the remaining two cases as exercises. We can use \(|a||b| = |ab|\) to demonstrate that

\[12\left|\frac{x}{2}-\frac{1}{3}\right|=|12|\left|\frac{x}{2}-\frac{1}{3}\right|=\left|12\left(\frac{x}{2}-\frac{1}{3}\right)\right| \nonumber \]

This validates the method of attack we used to solve equation (12) in Example \(\PageIndex{9}\).

On the other hand, it is not permissible to multiply by a negative number and simply slide the negative number inside the absolute value bars. For example,

\[-2|x-3|=|-2(x-3)| \nonumber \]

is clearly an error (well, it does work for x = 3). For any x except 3, the lefthand side of this result is a negative number, but the right-hand side is a positive number. They are clearly not equal.

In similar fashion, one can demonstrate a second useful property involving absolute value.

If a and b are any real numbers, then \[\left|\frac{a}{b}\right|=\frac{|a|}{|b|} \nonumber \] provided, of course, that \(b \neq 0\).

Again, this can be proved by checking four cases. For example, if a is a positive real number and b is a negative real number, then a/b is negative and \(|a/b| = −a/b\). On the other hand, \(|a|/|b| = a/(−b) = −a/b\).

We leave the proof of the remaining three cases as exercises.

This property is useful in certain situations. For example, should you desire to divide \(|2x − 4|\) by 2, you would proceed as follows.

\[\frac{|2 x-4|}{2}=\frac{|2 x-4|}{|2|}=\left|\frac{2 x-4}{2}\right|=|x-2| \nonumber \]

This technique is useful in several situations. For example, should you want to solve the equation \(|2x − 4| = 6\), you could divide both sides by 2 and apply the quotient property of absolute values.

## Distance Revisited

Recall that for any real number x, the absolute value of x is defined as the distance between the real number x and the origin on the real line. In this section, we will push this distance concept a bit further.

Suppose that you have two real numbers on the real line. For example, in the figure that follows, we’ve located 3 and −2 on the real line.

You can determine the distance between the two points by subtracting the number on the left from the number on the right. That is, the distance between the two points is d = 3 − (−2) = 5 units. If you subtract in the other direction, you get the negative of the distance, as in −2 − 3 = −5 units. Of course, distance is a nonnegative quantity, so this negative result cannot represent the distance between the two points. Consequently, to find the distance between two points on the real line, you must always subtract the number on the left from the number on the right.

However, if you take the absolute value of the difference, you’ll get the correct result regardless of the direction of subtraction.

\[d=|3-(-2)|=|5|=5 \quad \text { and } \quad d=|-2-3|=|-5|=5 \nonumber \]

This discussion leads to the following key idea.

## Property 16.

Suppose that a and b are two numbers on the real line

You can determine the distance d between a and b on the real line by taking the absolute value of their difference. That is,

\[d=|a-b| \nonumber \]

Of course, you could subtract in the other direction, obtaining \(d = |b − a|\). This is also correct.

Now that this geometry of distance has been introduced, it is useful to pronounce the symbolism |a−b| as “the distance between a and b” instead of saying “the absolute value of a minus b.”

## Example \(\PageIndex{10}\)

Solve the equation \[|x − 3| = 8 \nonumber \] for x.

Here’s the ideal situation to apply our new concept of distance. Instead of saying “the absolute value of x minus 3 is 8,” we pronounce the equation \(|x − 3| = 8\) as “the distance between x and 3 is 8.”

Draw a number line and locate the number 3 on the line.

Recall that the “distance between x and 3 is 8.” Having said this, mark two points on the real line that are 8 units away from 3.

Thus, the solutions of |x − 3| = 8 are x = −5 or x = 11

## Example \(\PageIndex{11}\)

Solve the equation \[|x + 5| = 2 \nonumber \] for x.

Rewrite the equation as a difference. \[|x − (−5)| = 2 \nonumber \] This is pronounced “the distance between x and −5 is 2.” Locate two points on the number line that are 2 units away from −5.

Hence, the solutions of \(|x + 5| = 2\) are x = −7 or x = −3.

For each of the equations in Exercises 1 - 4 , perform each of the following tasks.

- Set up a coordinate system on a sheet of graph paper. Label and scale each axis.
- Sketch the graph of each side of the equation without the aid of a calculator. Label each graph with its equation.
- Shade the solution of the equation on the x-axis (if any) as shown in Figure 5 (read "Expectations") in the narrative. That is, drop dashed lines from the points of intersection to the axis, then shade and label the solution set on the x-axis.

## Exercise \(\PageIndex{1}\)

|x| = −2

No solutions.

## Exercise \(\PageIndex{2}\)

Exercise \(\pageindex{3}\).

Solution: x = −3 or x = 3.

## Exercise \(\PageIndex{4}\)

For each of the equations in Exercises 5 - 8 , perform each of the following tasks.

- Load each side of the equation into the Y= menu of your calculator. Ad- just the viewing window so that all points of intersection of the two graphs are visible in the viewing window.
- Copy the image in your viewing screen onto your homework paper. Label each axis and scale each axis with xmin, xmax, ymin, and ymax. Label each graph with its equation.
- Use the intersect utility in the CALC menu to determine the points of intersection. Shade and label each solution as shown in Figure 5 (read "Expectations") in the narrative. That is, drop dashed lines from the points of intersection to the axis, then shade and label the solution set on the x-axis.

## Exercise \(\PageIndex{5}\)

|3−2x| = 5

Solutions: x = −1 or x = 4.

## Exercise \(\PageIndex{6}\)

Exercise \(\pageindex{7}\).

Solutions: x = −3 or x = 0.5.

## Exercise \(\PageIndex{8}\)

|5x−7| = 8

For each of the equations in Exercises 9 - 14 , provide a purely algebraic solution without the use of a calculator. Arrange your work as shown in Examples 6, 7, and 8 in the narrative, but do not use a calculator.

## Exercise \(\PageIndex{9}\)

\(x = −\frac{3}{4}\)

## Exercise \(\PageIndex{10}\)

|3x−11| = −5

## Exercise \(\PageIndex{11}\)

|2x+7| = 14

\(x = −\frac{21}{2}\) or \(x = \frac{7}{2}\)

## Exercise \(\PageIndex{12}\)

|7−4x| = 8

## Exercise \(\PageIndex{13}\)

|3−2x| = −1

## Exercise \(\PageIndex{14}\)

For each of the equations in Exercises 15 - 20 , perform each of the following tasks.

- Arrange each of the following parts on your homework paper in the same location. Do not do place the algebraic work on one page and the graphical work on another.
- Follow each of the directions given for Exercises 5 - 8 to find and record a solution with your graphing calculator.
- Provide a purely algebraic solution, showing all the steps of your work. Do these solutions compare favorably with those found using your graphing calculator in part (ii)? If not, look for a mistake in your work.

## Exercise \(\PageIndex{15}\)

|x−8| = 7

x = 1 or x = 15

## Exercise \(\PageIndex{16}\)

|2x−15| = 5

## Exercise \(\PageIndex{17}\)

|2x+11| = 6

x = −8.5 or x = −2.5

## Exercise \(\PageIndex{18}\)

|5x−21| = 7

## Exercise \(\PageIndex{19}\)

|x−12| = 6

x = 6 or x = 18

## Exercise \(\PageIndex{20}\)

Use a strictly algebraic technique to solve each of the equations in Exercises 21 - 28 . Do not use a calculator.

## Exercise \(\PageIndex{21}\)

|x+2|−3 = 4

x = −9 or x = 5

## Exercise \(\PageIndex{22}\)

Exercise \(\pageindex{23}\).

−2|3−2x| = −6

x = 0 or x = 3

## Exercise \(\PageIndex{24}\)

|4−x|+5 = 12

## Exercise \(\PageIndex{25}\)

3|x+2|−5 = |x+2|+7

x = −8 or x = 4

## Exercise \(\PageIndex{26}\)

4−3|4−x| = 2|4−x|−1

## Exercise \(\PageIndex{27}\)

\(|\frac{x}{3}−\frac{1}{4}| = \frac{1}{12}\)

\(x = \frac{1}{2}\) or x = 1

## Exercise \(\PageIndex{28}\)

\(|\frac{x}{4}−\frac{1}{2}| = \frac{2}{3}\)

Use the technique of distance on the number line demonstrated in Examples 16 and 17 to solve each of the equations in Exercises 29 - 32 . Provide number line sketches on your homework paper as shown in Examples 16 and 17 in the narrative.

## Exercise \(\PageIndex{29}\)

|x−5| = 8

x = −3 or x = 13

## Exercise \(\PageIndex{30}\)

|x−2| = 4

## Exercise \(\PageIndex{31}\)

x = −7 or x = −1

## Exercise \(\PageIndex{32}\)

Use the instructions provided in Exercises 5 - 8 to solve the equations in Exercises 33 - 34 .

## Exercise \(\PageIndex{33}\)

\(|x+2| = \frac{1}{3}x+5\)

## Exercise \(\PageIndex{34}\)

\(|x−3|=5−\frac{1}{2}x\)

In Exercises 35 - 36 , perform each of the following tasks.

- Set up a coordinate system on graph paper. Label and scale each axis.
- Without the use of a calculator, sketch the graphs of the left- and right-hand sides of the given equation. Label each graph with its equation.
- Drop dashed vertical lines from each point of intersection to the x-axis. Shade and label each solution on the x-axis (you will have to approximate).

## Exercise \(\PageIndex{35}\)

\(|x−2| = \frac{1}{3}x+2\)

## Exercise \(\PageIndex{36}\)

\(|x+4| = \frac{1}{3}x+4\)

## Exercise \(\PageIndex{37}\)

Given that a < 0 and b > 0, prove that |ab| = |a||b|.

If a is a negative real number and b is a positive real number, then ab is negative, so |ab| = −ab. On the other hand, a negative also means that |a| = −a, and b positive means that |b| = b, so that |a||b| = −a(b) = −ab. Comparing these results, we see that |ab| and |a||b| equal the same thing, and so they must be equal to one another.

## Exercise \(\PageIndex{38}\)

Given that a>0 and b<0, prove that |ab| = |a||b|.

## Exercise \(\PageIndex{39}\)

In the narrative, we proved that if a > 0 and b < 0, then \(|\frac{a}{b}| = \frac{|a|}{|b|}\). Prove the remaining three cases.

CaseI. (a, b > 0) If a and b are both positive real numbers, then \(\frac{a}{b}\) is positive and so \(|\frac{a}{b}| = \frac{a}{b}\). On the other hand, a positive also means that |a| = a, and b positive means that |b| = b, so that \(\frac{|a|}{|b|} = \frac{a}{b}\). Comparing these two results, we see that \(|\frac{a}{b}|\) and \(\frac{|a|}{|b|}\) equal the same thing, and so they must be equal to one another.

Case II. (a, b < 0) If a and b are both negative real numbers, then \(\frac{a}{b}\) is positive and so \(|\frac{a}{b}| = \frac{a}{b}\). On the other hand, a negative also means that |a| = −a, and b negative means that |b| = −b, so that \(\frac{|a|}{|b|} = \frac{−a}{(−b)} = \frac{a}{b}\). Comparing these two results, we see that \(|\frac{a}{b}|\) and \(\frac{|a|}{|b|}\) equal the same thing, and so they must be equal to one another.

Case III. (a < 0, b > 0) If a is a negative real number and b is a positive real number, then \(\frac{a}{b}\) is negative and so \(|\frac{a}{b}| =−(\frac{a}{b})\). On the other hand, a negative also means that |a| = −a, and b positive means that |b| = b, so that \(\frac{|a|}{|b|} = −\frac{a}{b} = −(\frac{a}{b})\). Comparing these two results, we see that \(|\frac{a}{b}|\) and \(\frac{|a|}{|b|}\) equal the same thing, and so they must be equal to one another.

- EXPLORE Random Article

## How to Solve Absolute Value Equations

Last Updated: January 31, 2023 References

This article was co-authored by wikiHow Staff . Our trained team of editors and researchers validate articles for accuracy and comprehensiveness. wikiHow's Content Management Team carefully monitors the work from our editorial staff to ensure that each article is backed by trusted research and meets our high quality standards. This article has been viewed 23,120 times.

## Setting up the Problem

- For example, |9| = 9; |-9| = -(-9) = 9.

## Calculating the Values

## Check Your Work

## Expert Q&A

- Remember that absolute value bars are distinct from parentheses and function differently. Thanks Helpful 0 Not Helpful 0
- Once you've solved for any variables, remember to simplify absolute values accordingly. Thanks Helpful 0 Not Helpful 0

## You Might Also Like

- ↑ http://tutorial.math.lamar.edu/Classes/Alg/SolveAbsValueEqns.aspx
- ↑ https://www.mathsisfun.com/numbers/absolute-value.html
- ↑ http://www.varsitytutors.com/high_school_math-help/solving-absolute-value-equations
- ↑ http://www.purplemath.com/modules/solveabs.htm
- ↑ https://www.khanacademy.org/math/algebra/absolute-value-equations-functions/absolute-value-equations/v/absolute-value-equations

## About this article

To solve absolute value equations, first isolate the absolute value terms by moving anything outside of the vertical bars to the other side of the equation. Next, solve for the positive value of the equation by isolating the variable. Since the absolute variable can represent 2 numbers, then solve for the negative value by putting a negative sign outside the vertical bars. Then, move the negative by dividing both sides by -1 and solve for the variable. If you want to learn how to check your answers for an absolute value equation, keep reading the article! Did this summary help you? Yes No

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## What is Absolute Value?

Practice questions, solving absolute value equations – methods & examples.

Solving equations containing an absolute value is as simple as working with regular linear equations . Before we can embark on solving absolute value equations, let’s take a review of what the word absolute value means.

In mathematics, the absolute value of a number refers to the distance of a number from zero, regardless of direction. The absolute value of a number x is generally represented as | x | = a, which implies that, x = + a and -a.

We say that the absolute value of a given number is the positive version of that number . For example, the absolute value of negative 5 is positive 5, and this can be written as: | − 5 | = 5.

Other examples of absolute values of numbers include: |− 9| = 9, |0| = 0, − |−12| = −12 etc. From these examples of absolute values, we simply define absolute value equations as equations containing expressions with absolute value functions.

## How to Solve Absolute Value Equations?

The following are the general steps for solving equations containing absolute value functions:

- Isolate the expression containing the absolute value function.
- Get rid of the absolute value notation by setting up the two equations so that in the first equation, the quantity inside absolute notation is positive. In the second equation, it is negative. You will remove the absolute notation and write the quantity with its suitable sign.
- Calculate the unknown value for the positive version of the equation.
- Solve for the negative version of the equation, in which you will first multiply the value on the other side of the equal sign by -1, and then solve.

In addition to the above steps, there are other important rules you should keep in mind when solving absolute value equations.

- The ∣x∣is always positive: ∣x∣ → +x.
- In | x| = a, if the a on the right is a positive number or zero, then there is a solution.

Solve the equation for x: |3 + x| − 5 = 4.

- Isolate the absolute value expression by applying the Law of equations. This means, we add 5 to both sides of the equation to obtain;

| 3 + x | − 5 + 5 = 4 + 5

| 3 + x |= 9

- Calculate for the positive version of the equation. Solve the equation by assuming the absolute value symbols.

| 3 + x | = 9 → 3 + x = 9

Subtract 3 from both sides of the equation.

3 – 3 + x = 9 -3

- Now calculate for the negative version of the equation by multiplying 9 by -1.

3 + x | = 9 → 3 + x = 9 × ( −1)

Also subtract 3 from both side to isolate x.

3 -3 + x = – 9 -3

Therefore 6 and -12 are the solutions.

Solve for all real values of x such that | 3x – 4 | – 2 = 3.

- Isolate the equation with absolute function by add 2 to both sides.

= | 3x – 4 | – 2 + 2 = 3 + 2

= | 3x – 4 |= 5

Assume the absolute signs and solve for the positive version of the equation.

| 3x – 4 |= 5→3x – 4 = 5

Add 4 to both sides of the equation.

3x – 4 + 4 = 5 + 4

Divide: 3x/3 =9/3

Now solve for the negative version by multiplying 5 by -1.

3x – 4 = 5→3x – 4 = -1(5)

3x – 4 = -5

3x – 4 + 4 = – 5 + 4

Divide by 3 on both sides.

Therefore, 3 and 1/3 are the solutions.

Solve for all real values of x: Solve | 2 x – 3 | – 4 = 3

Add 4 to both sides.

| 2 x – 3 | -4 = 3 →| 2 x – 3 | = 7

Assume the absolute symbols and solve for the positive version of x.

2 x – 3 = 7

2x – 3 + 3 = 7 + 3

Now solve for the negative version of x by multiplying 7 by -1

2 x – 3 = 7→2 x – 3 = -1(7)

Add 3 to both sides.

2x – 3 + 3 = – 7 + 3

x = – 2

Therefore, x = –2, 5

Solve for all real numbers of x: | x + 2 | = 7

Already the absolute value expression is isolated, therefore assume the absolute symbols and solve.

| x + 2 | = 7 → x + 2 = 7

Subtract 2 from both sides.

x + 2 – 2 = 7 -2

Multiply 7 by -1 to solve for the negative version of the equation.

x + 2 = -1(7) → x + 2 = -7

Subtract by 2 on both sides.

x + 2 – 2 = – 7 – 2

Therefore, x = -9, 5

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## Absolute Value Equation Calculator

What are absolute value equalities, how to use this absolute value equation calculator, how to solve absolute value equations by hand, how to graph absolute value equations, examples of absolute value equations.

Whenever you face absolute value equations, Omni's absolute value equation calculator is here to lend you a hand. With its help, you'll easily deal with all kinds of absolute value equalities , in particular equations where the absolute value is equal to 0. If you want to learn more about this topic, scroll down and learn:

- How to solve absolute value equations by hand ; and
- How to find an absolute value equation from a graph .

Once you're done here, take one step further in your mathematical journey and check out our absolute value inequalities calculator .

Let's briefly recall what the absolute value is, shall we? The absolute value of a real number x is the distance between this number and zero. We denote it by |x| . Algebraically:

- |x| = x if x is non-negative; and
- |x| = -x if x is negative.

For example:

- |0| = 0 ; and

In other words, the absolute value of a non-negative number is exactly this number, while for a negative number you have to throw away the minus sign .

💡 Did you know that absolute value is very popular in statistics? You can learn more via our tools:

- Mean absolute deviation calculator ; and
- Median absolute deviation calculator .

We can now move on and discuss what the absolute value equations are.

## Absolute value equalities

In general, any equation involving the absolute value of any expression is an absolute value equality. It can involve, e.g., polynomial expressions, roots, exponents, logarithms, etc.

In school, you're most likely to encounter an absolute value equation of the linear expression bx+c , that is, an equation of the form

a * |bx + c| + d = e ,

where a, b, c, d, e are real coefficients. And here's where our absolute value equation calculator enters the stage!

Omni's absolute value equation calculator can help you solve absolute value equalities of the form a * |bx + c| + d = e . Here's a brief instruction on how to solve them most efficiently:

- Enter the coefficients a, b, c, d, e of your absolute value equality. Remember that neither a nor b can be equal to zero - we don't want x to disappear from the equation!
- Your equation will appear at the bottom of the calculator - verify that

everything is all right. 3. Below the equation, you'll see the solution to your absolute value equation as well! By default, the calculator uses 4 decimal places to display the solution. Click the Advanced mode to adjust the precision . 4. Turn the Show steps? option to Yes to see some intermediate computations performed by our tool. 5. Turn the Show graph? option to Yes to see how to graph your absolute value equation .

If you want to solve by hand an absolute value equation of the form a * |bx + c| + d = e , follow these steps:

- Simplify your equation : transfer d to the right-hand side and divide both sides by a . You'll get |bx + c| = (e - d)/a .
- If (e - d)/a < 0 , then you've got an absolute value equation with no solutions ;
- If (e - d)/a = 0 , then you've got an absolute value equation with one solution , which is equal to -c/b ; and
- If (e - d)/a > 0 , then your equation has two solutions . Let's find them.
- Omitting the absolute value, we have bx + c = (e - d)/a or bx + c = -(e - d)/a . It now suffices to calculate x for each equation.
- Finally, x = (e - d)/(a * b) - c/b or x = (d - e)/(a * b) - c/b .

As you can see, it's not at all hard to solve absolute value equations by hand. Practice a bit with our absolute value equation calculator, and you'll quickly become an expert! As the next step, you may want to learn how to find an absolute value equation from a graph .

Graphing absolute value equations of the form a * |bx + c| + d = e is simple, you only need to remember a few basic rules:

- Start by plotting bx + c . It is a straight line that contains the points (0,c) and (-c/b,0) .
- Reflect the negative part of this line (i.e., where y<0 ) through the horizontal axis to get |bx + c| .
- Multiplying by a is equivalent to changing the slope. Additionally, if a<0 , you have to reflect your plot one more time through the horizontal axis.
- Adding d boils down to moving the plot upwards or downwards, depending on whether d is positive or negative. In any case, you now have the plot of a * |bx + c| + d .
- Draw the line y = e . The points where the two graphs intersect (if there are any) are the solutions of your equations.

Example 1. Solve |2x + 5| = x + 4

We resolve the absolute value on the left-hand side. Clearly 2x + 5 ≥ 0 is equivalent to x ≥ -2.5 Consequently, we obtain -2x - 5 = x + 4 if x < -2.5 and 2x + 5 = x + 4 if x ≥ -2.5

Simplifying, we get -3x = 9 if x < -2.5 and x = 1 if x ≥ -2.5

As a result, we have x = -3 or x = 1 You can use our absolute value solver to graph this solution.

Example 2. Solve |x| = -|x - 1| + 1

1.We resolve the absolute value on the right-hand side: |x| = -x + 1 + 1 if x ≥ 1 and |x| = x - 1 + 1 if x < 1

Simplify: |x| = -x + 2 if x ≥ 1 and |x| = x if x < 1

We will solve the equation |x| = -x + 2 if x ≥ 1 . Resolve the absolute value on the left-hand side:

- x = -x + 2 if x ≥ 1 and x ≥ 0
- -x = -x + 2 if x ≥ 1 and x < 0

Note, that in the second equation we have a contradiction in the conditions. So we ignore this equation and only solve the first one.

Solution of the first equation: 2x = 2 if x ≥ 1 . So x = 1 is indeed a solution.

Next, we will solve the equation |x| = x if x < 1 . Resolve the absolute value on the left-hand side:

- x = x if x < 1 and x ≥ 0
- -x = x if x < 1 and x < 0
- 0 = 0 if 0 ≤ x < 1
- 2x = 0 if x < 0
- All x ∈ [0 , 1) satisfy the equation in question
- x = 0 does not satisfy the condition x < 0 , so this is not a solution.

So all x in the interval [0,1) are solutions to the equation |x| = x if x < 1 .

Final solution: We found that x = 1 and x ∈ [0 , 1) are solutions. Taking all of this together, we see that every x ∈ [0 , 1] is a solution to our problem.

With the help of our absolute value solver, you can train yourself to solve this kind of problem, and become a master of absolute value equations!

## How many solutions can an absolute value equation have?

The number of solutions of an absolute value equation of the form a * |bx + c| + d = e depends on the sign of (e - d)/a . Namely, we have:

- Two solutions if (e -d)/a > 0 ;
- One solution if e = d ; and
- No solution if (e -d)/a < 0 .

## Can an absolute value be equal to a negative number?

No, absolute values cannot be equal to negative numbers. This follows from the very definition of the absolute value of a number, which is the distance between zero and this number, and distances, as we all know, are always non-negative.

## Can an absolute value be equal to 0?

Yes, but in a very special case. Namely, the only situation where the absolute value of a number is zero is when this number is zero: |x| = 0 if and only if x = 0 . A non-zero number will always have its absolute value equal to a positive number!

## Equation of a circle

Lowest common denominator, millionaire.

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## IMAGES

## VIDEO

## COMMENTS

Learn how to break down absolute value equations into two parts: positive and negative components. Follow the general approach and key points to solve different types of absolute value equations with examples and practice problems.

Learn how to solve absolute value equations with one, two, or no solutions, and how to graph absolute value functions with v-shaped curves. Watch a video lesson and see worked examples, tips, and questions from other learners.

Learn how to solve equations and inequalities involving absolute value, which represent the distance between two points on a number line. This section covers the basic properties of absolute value, how to isolate the absolute value expression, and how to apply the definition of absolute value to find the solutions. You will also see how to graph absolute value functions and inequalities using ...

How To Solve Absolute Value Equations The Organic Chemistry Tutor 7.41M subscribers Join Subscribe Subscribed 13K 707K views 5 years ago New Algebra Playlist This math video tutorial explains...

Learn how to solve absolute value equations in 3 easy steps with examples, a tutorial and an animated video. Find out why absolute value equations can have two solutions and how to check your answers.

Learn how to solve absolute value equations with one solution by identifying the expression in absolute value bars and making it equal to the right-hand side. See examples, graphs, and tips from Sal Khan and other users.

Solving the equation 8|x+7|+4 = -6|x+7|+6 which has two possible solutions. Created by Sal Khan. Questions Tips & Thanks Want to join the conversation? Sort by: Top Voted albertan123456789 8 years ago how does the absolute value thing separate the equation into 2 equations • ( 8 votes) Upvote Flag Hamda Khan 8 years ago

Mathematical Definition We can also give a strict mathematical/formula definition for absolute value. It is, |p| = {p if p ≥ 0 −p if p < 0 | p | = { p if p ≥ 0 − p if p < 0 This tells us to look at the sign of p p and if it's positive we just drop the absolute value bar.

To solve an absolute value equation, we first isolate the absolute value expression using the same procedures we used to solve linear equations. Once we isolate the absolute value expression we rewrite it as the two equivalent equations. How to Solve Absolute Value Equations. Example \(\PageIndex{4}\)

0:00 / 6:04 Solving Absolute Value Equations Explained! Mashup Math 171K subscribers Subscribe Subscribed 301 Share 49K views 3 years ago Algebra II & Trigonometry Practice Lessons On this...

14K 1M views 6 years ago This algebra video tutorial provides a basic introduction into absolute value equations. it explains how to solve absolute value equations the easy way. It...

Learn the general steps to solve absolute value equations with variables on one or both sides of the equation. See examples, practice problems and a solver tool.

When we attempt to solve the absolute-value equation | x | = 3, we are, in effect, setting two line equations equal to each other and finding where they cross. For instance: In the above, I've plotted the graph of y1 = | x | (being the blue line that looks like a "V") and y2 = 3 (being the green horizontal line).

Free absolute value equation calculator - solve absolute value equations with all the steps. Type in any equation to get the solution, steps and graph

For example, should you want to solve the equation \(|2x − 4| = 6\), you could divide both sides by 2 and apply the quotient property of absolute values. Distance Revisited Recall that for any real number x, the absolute value of x is defined as the distance between the real number x and the origin on the real line.

An absolute value equation is solved using the same rules as any other algebraic equation; however, this type of equation has two potential results, derived from a positive equation and a negative equation. Part 1 Setting up the Problem 1 Understand the mathematical definition of absolute value.

Solve an absolute value equation using the following steps: Get the absolve value expression by itself. Set up two equations and solve them separately. Absolute Value Equation Video Lesson Khan Academy Video: Absolute Value Equations Need more problem types? Try MathPapa Algebra Calculator Clear Absolute Value Equation Calculator » Show Keypad

Example 1. Solve the equation for x: |3 + x| − 5 = 4. Solution. Isolate the absolute value expression by applying the Law of equations. This means, we add 5 to both sides of the equation to obtain; | 3 + x | − 5 + 5 = 4 + 5. | 3 + x |= 9. Calculate for the positive version of the equation.

Learn how to solve absolute value equations with multiple steps. Absolute value of a number is the positive value of the number. For instance, the absolute v...

We can now move on and discuss what the absolute value equations are. Absolute value equalities. In general, any equation involving the absolute value of any expression is an absolute value equality. It can involve, e.g., polynomial expressions, roots, exponents, logarithms, etc. In school, you're most likely to encounter an absolute value ...

How: We are going to going to write the absolute value equations into a system with a constant function and an Absolute value functions and then solve graphically. Why: Any equation can be rewritten into a system of equations. This will allow us to solve any equation by graphing. HW #28 2.2 Day 1 Solve Absolute Value Equations Graphically

Learn how to solve absolute value equations in this free math video tutorial by Mario's Math Tutoring. We discuss how to rewrite an absolute value equation ...

http://www.greenemath.com/http://www.facebook.com/mathematicsbyjgreeneIn this lesson, we learn how to apply our rule for solving absolute value equations: |u...