## Solving Absolute Value Equations

Solving absolute value equations is as easy as working with regular linear equations. The only additional key step that you need to remember is to separate the original absolute value equation into two parts: positive and negative ( ± ) components.

Below is the general approach on how to break them down into two equations:

In addition, we also need to keep in mind the following key points regarding the setup above:

## Key Points to Remember when Solving Absolute Value Equations

Key Point #1 : The sign of $\left| x \right|$ must be positive. For emphasis, $\left| x \right| \to + \left| x \right|$.

Key Point #2 : The $x$ inside the absolute value symbol, $\left| {\,\,\,\,\,} \right|$, could be any expressions.

Key Point #3 : The $a$ on the right side of the equation must be either a positive number or zero  to have a solution.

Key Point #4 : If the $a$ on the right side is a negative number , then it has no solution.

Given the considerations above, there are instances where the right-hand side contains a variable. However, it could also be on the left, provided the opposite side has the absolute value expression.

To solve such cases, we follow the standard procedure, but it’s imperative to validate each solution by plugging it back into the original absolute value equation. If the substitution yields a true statement, the answer we found is included in the solution set. If not, it’s excluded.

See examples #8 and #9 to see the demonstrations of this concept in the practice problems below:

Absolute Value Expressions Practice Problems with Answers

Finally, there is another case that usually shows up when dealing with absolute value equations. That is, the equation contains absolute value expressions on both sides. Here’s the procedure how to work it out.

Suppose we have the absolute value equation

$\left| {ax + b} \right| = \left| {cx + d} \right|$

Then, the two expressions must either be equal to each other or be negatives of each other.

$ax + b = cx + d$

$ax + b = – \left( {cx + d} \right)$

See examples #10 and #11 to see how it works in the practice problems found below:

## Examples of How to Solve Absolute Value Equations

Example 1: Solve the absolute value equation $\left| x \right| =\, – 5$ .

The absolute value of any number is either positive or zero. But this equation suggests that there is a number whose absolute value is negative. Can you think of any numbers that can make the equation true? Well, there is none.

Since there’s no value of $x$ that can satisfy the equation, we say that it has no solution .

In fact, the following absolute value equations don’t have solutions as well.

Example 2: Solve the absolute value equation $– \left| x \right| =\, – 5$ .

Don’t be quick to conclude that this equation has no solution. Although the right side of the equation is negative, the absolute value expression itself must be positive. But it is not, right?

What we need is to eliminate first the negative sign of the absolute value symbol before we can proceed.

Observe that the given equation has a coefficient of −1 . Divide both sides of the equation by this value to get rid of the negative sign.

Since the absolute value expression and the number are both positive, we can now apply the procedure to break it down into two equations.

Therefore, the solution to the problem becomes

Therefore, the solution set is $\{- 5,5\}$.

You may verify our answers by substituting them back to the original equation. I’ll leave it to you.

Example 3: Solve the absolute value equation $\left| {x – 5} \right| = 3$ .

This problem is getting interesting since the expression inside the absolute value symbol is no longer just a single variable. Don’t worry; the set-up remains the same. Just be careful when you break up the given absolute value equation into two simpler linear equations, then proceed how you usually solve equations.

Therefore, the solution set is $\{2,8\}$.

You may check the answers back to the original equation.

Example 4: Solve the absolute value equation $\left| { – 2x + 7} \right| = 25$ .

You may think that this problem is complex because of the $–2$ next to the variable $x$. However, that shouldn’t intimidate you because the key idea remains the same. We have the absolute value symbol isolated on one side and a positive number on the other. Solving this is just like another day in the park!

Break it up into the $+$ and $-$ components, then solve each equation.

Therefore, the solution set is $\{- 9,16\}$.

Example 5 : Solve the absolute value equation $\left| { – 6x + 3} \right| – 7 = 20$.

This one is  not ready  just yet to be separated into two components. Why? It is because the absolute value symbol is not by itself on one side of the equation. If you look at it, there is a $-7$ on the left side that must be eliminated first. Once we get rid of that, then we should be okay to proceed as usual.

Eliminate the $-7$ on the left side by adding both sides by $\color{blue}7$.

Now, we have an absolute value equation that can be broken down into two pieces.

Therefore, the solution set is $\{-4,5\}$.

Example 6 : Solve the absolute value equation $– 7\left| {9\, – 2x} \right| + 9 =\, – 12$.

The absolute value expression is not isolated yet. Eliminate the $+9$ first and then the $-7$ which is currently multiplying the absolute value expression.

Now, let’s split them into two cases, and solve each equation.

Therefore, the solution set is $\{3,6\}$.

Example 7 : Solve the absolute value equation $\left| {{x^2} + 2x – 4} \right| = 4$.

This is an interesting problem because we have a quadratic expression inside the absolute value symbol. I hope you don’t get distracted by how it looks! If you’re faced with a situation in which you’re not sure how to proceed, stick to the basics and things that you already know.

We don’t care about the “stuff” inside the absolute value symbol. As long as it is isolated, and the other side is a positive number, we can definitely apply the rule to split the equation into two cases.

In fact, the only difference between this problem from what you’ve been doing so far is that you will be solving quadratic equations instead of linear equations.

We can verify that our four answers or solutions are $x = – \,4$, $-2$, $0$, and $2$, by graphing the two functions and looking at their points of intersections.

Therefore, the solution set is $\{-4,-2,0,2\}$.

Take a quiz:

Absolute Value Equations Quiz

You might also like these tutorials:

• Absolute Value Equations Practice Problems with Answers
• Tough Absolute Value Equations
• Graphing Absolute Value Functions
• Solving Absolute Value Inequalities

## ABSOLUTE VALUE EQUATIONS WITH TWO ABSOLUTE VALUES

When we solve absolute value functions with two absolute signs, we follow the rules given below.

More simply,

u = v and -u = v

Solve the following absolute value equations.

Problem 1 :

|x + 4| = |2x - 7|

|x + 4| = |2x - 7|

 x + 4 = 2x - 7x - 2x = -7 - 4-x = - 11x = 11 x + 4 = - (2x - 7)x + 4 = -2x + 7x + 2x = 7 - 43x = 3x = 1

So, the solution is x = 1 or x = 11.

Problem 2 :

|3x + 5| = |x - 6|

 3x + 5 = x - 63x - x = -6 - 52x = -11x = -11/2 3x + 5 = -(x - 6)3x + 5 = -x + 63x + x = 6 - 54x = 1x = 1/4

So, the solution is x = -11/2 or x = 1/4.

Problem 3 :

|x - 9| = |x + 6|

 x - 9 = x + 6x - x = 6 - 90 = -3it is not true. |x - 9| = |x + 6|x - 9 = - (x + 6)x - 9 = -x - 6x + x = -6 + 92x = 3x = 3/2

So, the solution is 3/2.

|x + 4| = |x - 3|

 x + 4 = x - 3x - x = -3 - 40 = -7it is not true. |x + 4| = |x - 3|x + 4 = -(x - 3)x + 4 = -x + 3x + x = 3 - 42x = -1x = -1/2

So, the solution is -1/2.

Problem 5 :

|5t + 7| = |4t + 3|

 5t + 7 = 4t + 35t - 4t = 3 - 7t = -4 |5t + 7| = |4t + 3|5t + 7 = - (4t + 3)5t + 7 = -4t - 35t + 4t = -3 - 79t = -10t = -10/9

o, the solution is t = -4 or t = -10/9.

Problem 6 :

|3a - 1| = |2a + 4|

 3a - 1 = 2a + 43a - 2a = 4 + 1a = 5 |3a - 1| = |2a + 4|3a - 1 = - (2a + 4)3a - 1 = -2a - 43a + 2a = -4 + 15a = -3a = -3/5

So, the solution is a = 5 or a = -3/5.

Problem 7 :

|n - 3| = |3 - n|

True for all real values of n. So, infinitely many solution.

Problem 8 :

|y - 2| = |2 - y|

True for all real values of y. So, infinitely many solution.

Problem 9 :

|7 - a| = |a + 5|

 7 - a = a + 5-a - a = 5 - 7-2a = -2a = 1 |7 - a| = |a + 5|7 - a = - (a + 5)7 - a = -a - 5-a + a = -5 - 70 = -12

So, the solution is a = 1.

Problem 10 :

|6 - t| = |t + 7|

 6 - t = t + 7-t - t = 7 - 6-2t = 1t = -1/2 |6 - t| = |t + 7|6 - t = - (t + 7)6 - t = -t - 7-t + t = -7 - 60 = -13

So, the solution is t = -1/2.

Problem 11 :

|1/2x - 5| = |1/4x + 3|

 |1/2x - 5| = |1/4x + 3|1/2x - 5 = 1/4x + 31/2x - 1/4x = 3 + 5(2x - x)/4 = 8x/4 = 8x = 32 |1/2x - 5| = |1/4x + 3|1/2x - 5 = - (1/4x + 3)1/2x - 5 = -1/4x - 31/2x + 1/4x = -3 + 5(2x + x)/4 = 23x/4 = 23x = 8x = 8/3

So, the solution is x = 32 or x = 8/3.

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## 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 Absolute Value Equations – Methods & Examples

## 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

| 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)

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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## What is the best way to solve an equation involving multiple absolute values?

An absolute value expression such as $|ax-b|$ can be rewritten in two cases as $|ax-b|=\begin{cases} ax-b & \text{ if } x\ge \frac{b}{a} \\ b-ax & \text{ if } x< \frac{b}{a} \end{cases}$, so an equation with $n$ separate absolute value expressions can be split up into $2^n$ cases, but is there a better way?

For example, with $|2x-5|+|x-1|+|4x+3|=13$, is there a better way to handle all the possible combinations of $x\ge\frac{5}{2}$ versus $x<\frac{5}{2}$, $x\ge 1$ versus $x< 1$, and $x\ge-\frac{3}{4}$ versus $x<-\frac{3}{4}$?

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• absolute-value

• 6 $\begingroup$ For someone who has to solve a lot of equations of this sort, I would always recommend that the first thing to do is to make a plot of the functions of interest, if only to reckon which intervals one should be looking at. $\endgroup$ –  J. M. ain't a mathematician Commented Aug 23, 2010 at 22:41
• 1 $\begingroup$ @J. Mangaldan: The plot is also a good way to verify the answers. There are something like 24 signed-number operations involved in my algebraic solution below, so even with 99% accuracy per operation, that's only a 78.6% chance of correct calculations throughout. $\endgroup$ –  Isaac Commented Aug 23, 2010 at 23:32

For an equation with $n$ absolute values, the $n$ places where each absolute value splits into 2 cases divide the number line into $n+1$ regions where, within each region, each absolute value can be replaced by either the expression inside the absolute value or its opposite. Each resulting equation can then be solved, restricting solutions to the corresponding region on the number line.

In the given example,

$\small{\begin{matrix} \leftarrow & -\frac{3}{4} & \text{---} & 1 & \text{---} & \frac{5}{2} & \rightarrow \\ \begin{matrix}-(2x-5)-(x-1)\\-(4x+3)=13\end{matrix} & \begin{matrix}|\\|\end{matrix} & \begin{matrix}-(2x-5)-(x-1)\\+(4x+3)=13\end{matrix} & \begin{matrix}|\\|\end{matrix} & \begin{matrix}-(2x-5)+(x-1)\\+(4x+3)=13\end{matrix} & \begin{matrix}|\\|\end{matrix} & \begin{matrix}(2x-5)+(x-1)\\+(4x+3)=13\end{matrix} \\ -7x+3=13 & | & x+9=13 & | & 3x+7=13 & | & 7x-3=13 \\ x=-\frac{10}{7} & | & x=4\notin[-\frac{3}{4},1] & | & x=2 & | & x=\frac{16}{7}\notin[\frac{5}{2},\infty) \end{matrix}}$

So, the solutions are $x=-\frac{10}{7}$ and $x=2$ (the values that were solutions to an equation for a particular region and were within that region).

• $\begingroup$ This table is too wide to display correctly for me :/ $\endgroup$ –  Larry Wang Commented Aug 24, 2010 at 0:22
• $\begingroup$ @Kaestur Hakarl: Sorry--with the \small{} wrapper, it just barely fits for me, but I couldn't think of any way to make it smaller and still legible. I'm open to suggestions. $\endgroup$ –  Isaac Commented Aug 24, 2010 at 0:33
• $\begingroup$ See my answer: what you propose is a special case of the CAD algorithm. $\endgroup$ –  Bill Dubuque Commented Aug 24, 2010 at 0:49
• $\begingroup$ How do you determine wich expression belongs in which region? $\endgroup$ –  freethinker36 Commented Nov 15, 2019 at 3:14
• $\begingroup$ @freethinker36 In the original question, $ax-b$ versus $b-ax$ depending on which side of $\frac{b}{a}$ $x$ is on. $\endgroup$ –  Isaac Commented Nov 18, 2019 at 20:54

This is merely a very special case of the powerful CAD (cylindrical algebraic decomposition) algorithm for quantifier elimination in real-closed fields, e.g. see Jirstrand's paper [1] for a nice introduction.

[1] M. Jirstrand. Cylindrical algebraic decomposition - an introduction . 1995 Technical report S-58183, Automatic Control group, Department of Electrical Engineering Linkoping University, Linkoping, Sweden. Freely available here or here .

• 1 $\begingroup$ I knew somebody would invoke CAD at some point; this is in fact what Mathematica uses internally for such sets, FYI. On the other hand, I can't imagine making a student do CAD manually! $\endgroup$ –  J. M. ain't a mathematician Commented Aug 24, 2010 at 2:06
• 3 $\begingroup$ In fact I implemented the CAD-related substrate in Macsyma - long before Mathematica existed - which is why I'm familiar with such. The ideas behind CAD are quite simple and certainly can be done manually for small problems like this. $\endgroup$ –  Bill Dubuque Commented Aug 24, 2010 at 2:55
• $\begingroup$ ...so that's why your name was familiar! :D Macsyma really was a gem. Making a student algorithmically go through the steps of CAD for this still seems like "cruel and unusual punishment" to me though. $\endgroup$ –  J. M. ain't a mathematician Commented Aug 24, 2010 at 3:01
• 3 $\begingroup$ I think perhaps you misunderstand my comment. When I say that one can employ CAD manually I don't mean mechanically . One can employ insight to use only the simple parts of CAD needed for this problem - yielding something similar to Isaac's answer. That's hardly punishing at all. Rather, it's instructive. $\endgroup$ –  Bill Dubuque Commented Aug 24, 2010 at 3:11
• 1 $\begingroup$ When you put it that way, then I certainly agree. $\endgroup$ –  J. M. ain't a mathematician Commented Aug 24, 2010 at 3:27

Not the answer you're looking for browse other questions tagged algebra-precalculus roots absolute-value faq ..

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• Isolate the absolute value on one side of the equation.
• Is the number on the other side of the equation negative? If you answered yes, then the equation has no solution. If you answered no, then go on to step 3.
• Write two equations without absolute values. The first equation will set the quantity inside the bars equal to the number on the other side of the equal sign; the second equation will set the quantity inside the bars equal to the opposite of the number on the other side.
• Solve the two equations.
• Write two equations without absolute values.  The first equation will set the quantity inside the bars on the left side equal to the quantity inside the bars on the right side.  The second equation will set the quantity inside the bars on the left side equal to the opposite of the quantity inside the bars on the right side.
So the only solution to this problem is x = 1/2
Since 3 is included in the set of real numbers, we will just say that the solution to this equation is All Real Numbers

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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).

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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• a special character: @$#!%*?& ## Absolute Value Equation Calculator What do you want to calculate. • Solve for Variable • Practice Mode • Step-By-Step ## Example (Click to try) About absolute value equations. • Get the absolve value expression by itself. • Set up two equations and solve them separately. ## Number Line • \left|3x+1\right|=4 • \left|a-6\right|=\left|6-a\right| • 1.5\left|3p\right|=4.5 • 4\left|2x-8\right|=9 • \left|3x+4\right|=-3 • 5-\left|2(x+3)\right|=0 absolute-equation-calculator • High School Math Solutions – Absolute Value Equation Calculator Solving absolute value equations is somewhat tricky; it requires understanding of the absolute value property.... Please add a message. Message received. Thanks for the feedback. #### IMAGES 1. How To Solve Absolute Value Equations 2. PPT 3. Solving Equations with Two Absolute Values 4. 24.2 Solving Absolute Value Equations Containing Two Absolute Value Expressions (1.6) 5. How To Solve Absolute Value Equations, Basic Introduction, Algebra 6. Solving an Absolute Value Equation with Two Absolute Value Expressions (Example) #### VIDEO 1. HOW to solve ABSOLUTE value equations 2. Evaluating Absolute Value Expressions 3. 2|4x + 3|=12, Let’s solve the ABSOLUTE VAULE equation step-by-step… 4. Equations with Two Absolute Values (ex2) 5. Solving Double Absolute Value Equations (Introduction) 6. Introduction to solving an absolute value equation #### COMMENTS 1. Solving Absolute Value Equations Solving Absolute Value Equations. Solving absolute value equations is as easy as working with regular linear equations. The only additional key step that you need to remember is to separate the original absolute value equation into two parts: positive and negative (±) components.Below is the general approach on how to break them down into two equations: 2. Solving Equations with Two Absolute Value Expressions From Thinkwell's College AlgebraChapter 2 Equations and Inequalities, Subchapter 2.4 Other Types of Equations 3. Worked example: absolute value equation with two solutions Lesson 2: Solving absolute value equations. Intro to absolute value equations and graphs. Worked example: absolute value equation with two solutions. ... So you could almost treat this expression-- the absolute value of x plus 7, you can just treat it as a variable, and then once you solve for that, it becomes a simpler absolute value problem ... 4. Solving an Absolute Value Equation with Two Absolute Value Expressions This video walks through an example of solving an absolute value equation that contains two absolute value expressions.For more math help and resources, visi... 5. Solving Absolute-Value Equations: A Special Case I'll solve to find that interval: x − 3 > 0. x > 3. The argument of this absolute value will be negative before the breakpoint (at x= 3) and positive after. (It's equal to zero at the breakpoint.) The second absolute-value expression, in the right-hand side of the equation, is positive for: 3 x + 2 > 0. 3 x > −2. 6. 2.6: Solving Absolute Value Equations and Inequalities 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 ... 7. Intro to absolute value equations and graphs To solve absolute value equations, find x values that make the expression inside the absolute value positive or negative the constant. To graph absolute value functions, plot two lines for the positive and negative cases that meet at the expression's zero. The graph is v-shaped. Created by Sal Khan and CK-12 Foundation. 8. Absolute Value Equations with Two Absolute Values When we solve absolute value functions with two absolute signs, we follow the rules given below. More simply, |u| = |v|. u = v and -u = v. Solve the following absolute value equations. So, the solution is x = 1 or x = 11. So, the solution is x = -11/2 or x = 1/4. it is not true. 9. Absolute value equations, functions, & inequalities This topic covers: - Solving absolute value equations - Graphing absolute value functions - Solving absolute value inequalities. ... Rational expressions, equations, & functions. Unit 14. Trigonometric functions. Unit 15. Algebraic modeling. Unit 16. Complex numbers. Unit 17. Conic sections. Unit 18. Series & induction. 10. How to solve absolute value equations 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. 11. Solving Absolute Value Equations Solution. 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. x = 5. 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. 12. Solving Harder Absolute-Value Equations When one absolute-value expression is inside another one, this is called a "nested" expression. The solution process works the same as what we've seen before; the difference will be the number of times we have to split things into cases. Solve | | 2x + 5 | − 3 | = 8. This equation has one absolute-value expression nested inside another one. 13. What is the best way to solve an equation involving multiple absolute Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site 14. Solving Absolute Value Equations Containing TWO Absolute ... Thanks to all of you who support me on Patreon. You da real mvps!$1 per month helps!! :) https://www.patreon.com/patrickjmt !! Solving Absolute Value Equ...

15. Absolute Value Equations

Solve the two equations. Follow these steps to solve an absolute value equality which contains two absolute values (one on each side of the equation): Write two equations without absolute values. The first equation will set the quantity inside the bars on the left side equal to the quantity inside the bars on the right side.

16. Solving Simpler Absolute-Value Equations

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 ...

17. Algebra Examples

Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor. ... Absolute Value Expressions and Equations. Solve for x. Step 1. Rewrite the equation as . Step 2. Move all terms not containing to the right side of the equation. Tap for ...

18. Absolute Value Equation Calculator

About absolute value equations. Solve an absolute value equation using the following steps: Get the absolve value expression by itself. Set up two equations and solve them separately.

19. Solving Absolute Value Equations

To solve an equation like this, with a variable inside absolute value bars, we must separate the two possible cases and solve for each. The expression inside the absolute value bars might be positive. In that case, it equals the absolute value: x = 3. Or the expression inside the absolute value bars might be negative.

20. Absolute Value Equation Calculator

Free absolute value equation calculator - solve absolute value equations with all the steps. Type in any equation to get the solution, steps and graph ... System of Equations System of Inequalities Basic Operations Algebraic Properties Partial Fractions Polynomials Rational Expressions Sequences Power Sums Interval Notation Pi ...

21. The study of new fixed-point iteration schemes for solving absolute

Absolute value equations (AVEs) play a crucial role in solving various complexities across scientific computing, engineering, management science, and operations research. This article presents two fixed-point iteration schemes designed for such AVEs. Furthermore, we show some convergence conditions under specific circumstances. Lastly, the theoretical findings are validated using numerical ...

22. Worked example: absolute value equations with one solution

To solve absolute value problems with one solution, identify expression in absolute value bars. Recall absolute value is zero only if expression is zero. Make expression equal right-hand side. Use algebra to find x values that satisfy equation. Check solutions by plugging in. Graph solutions on number line. Mark points.