how to solve equations with absolute value and variables

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To solve an equation containing absolute value, isolate the absolute value on one side of the equation. Then set its contents equal to both the positive and negative value of the number on the other side of the equation and solve both equations.

Solve | x | + 2 = 5. 

Isolate the absolute value.

how to solve equations with absolute value and variables

Set the contents of the absolute value portion equal to +3 and –3.

how to solve equations with absolute value and variables

Answer: 3, –3 

Solve 3| x – 1| – 1 = 11. 

how to solve equations with absolute value and variables

Set the contents of the absolute value portion equal to +4 and –4.

Solving for x,

how to solve equations with absolute value and variables

Answer: 5, –3 

Solving inequalities containing absolute value and graphing

To solve an inequality containing absolute value , begin with the same steps as for solving equations with absolute value. When creating the comparisons to both the + and – of the other side of the inequality, reverse the direction of the inequality when comparing with the negative. 

Solve and graph the answer: | x – 1| > 2. 

Notice that the absolute value expression is already isolated.

| x – 1| > 2 

Compare the contents of the absolute value portion to both 2 and –2. Be sure to reverse the direction of the inequality when comparing it with –2.

Solve for x . 

how to solve equations with absolute value and variables

Graph the answer (see Figure 1). 

Figure 1. The graphic solution to | x – 1| > 2.

how to solve equations with absolute value and variables

Solve and graph the answer: 3| x | – 2 ≤ 1. 

how to solve equations with absolute value and variables

Compare the contents of the absolute value portion to both 1 and –1. Be sure to reverse the direction of the inequality when comparing it with –1.

how to solve equations with absolute value and variables

Graph the answer (see Figure 2). 

Figure 2. Graphing the solution to 3| x | – 2 ≤ 1.

how to solve equations with absolute value and variables

Solve and graph the answer: 2|1 – x | + 1 ≥ 3. 

how to solve equations with absolute value and variables

(Remember to switch the direction of the inequality when dividing by a negative)

how to solve equations with absolute value and variables

Graph the answer (see Figure 3). 

Figure 3. Graphing the solution 2|1 – x | + 1 ≥ 3.

how to solve equations with absolute value and variables

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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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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.

x

Setting up the Problem

Step 1 Understand the mathematical definition of absolute value.

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

Step 2 Understand what an absolute value represents.

Calculating the Values

Step 1 Set up the equation for the positive value.

Check Your Work

Step 1 Check the result of your positive equation.

Expert Q&A

Video . by using this service, some information may be shared with youtube..

  • 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

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  • ↑ 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

wikiHow Staff

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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1.2: Absolute Value Equations

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  • Darlene Diaz
  • Santiago Canyon College via ASCCC Open Educational Resources Initiative

When solving equations with absolute value, the solution may result in more than one possible answer because, recall, absolute value is just distance from zero. Since the integer \(−4\) has distance \(4\) units from zero, and 4 has distance 4 units from zero, then there are two integers that have distance \(4\) from zero, \(−4,\: 4\). We extend this concept to algebraic absolute value equations. This is illustrated in the following example.

Example \(\PageIndex{1}\)

Solve for \(x\): \(|x|=7\)

\[\begin{array}{rl}|x|=7&\text{Expression in the absolute value can be positive or negative} \\ x=7\text{ or }x=-7&\text{Solution}\end{array}\nonumber\]

Let’s think about the solution set. The equation is asking for all numbers in which the distance from zero is \(7\). Well, there are two integers that have a distance \(7\) from zero, \(−7\) and \(7\). Hence, the solution set \(\{−7, 7\}\).

The first set of rules for working with negative numbers came from \(7^{\text{th}}\) century India. However, in 1758, more than a thousand years later, British mathematician Francis Maseres claimed that negatives “Darken the very whole doctrines of the equations and make dark of the things which are in their nature excessively obvious and simple.”

Definition: Absolute Value

Absolute value for linear equations in one variable is given by \[\text{If }|x|=a,\text{ then }x=a\text{ or }x=-a\nonumber\] where \(a\) is a real number.

When we have an equation with absolute value, it is important to first isolate the absolute value, then remove the absolute value by applying the definition.

Example \(\PageIndex{2}\)

Solve for \(x\): \(5+|x|=8\)

\[\begin{array}{rl}5+|x|=8&\text{Isolate the absolute value by subtracting }5\text{ from each side} \\ |x|=3&\text{Rewrite as two linear equations} \\ x=3\text{ or }x=-3&\text{Solution}\end{array}\nonumber\]

Thus, the solution set is \(\{-3,3\}\).

Example \(\PageIndex{3}\)

Solve for \(x\): \(-4|x|=-20\)

\[\begin{array}{rl}-4|x|=-20&\text{Isolate the absolute value} \\ \frac{-4|x|}{-4}=\frac{-20}{-4}&\text{Divide each side by }-4 \\ |x|=5&\text{Rewrite as two linear equations} \\ x=5\text{ or }x=-5&\text{Solution}\end{array}\nonumber\]

Thus, the solution set is \(\{-5,5\}\).

Never combine the inside of the absolute value with factors or terms from outside the absolute value. We always have to isolate the absolute value first, then apply the definition to obtain two equations without the absolute value.

Example \(\PageIndex{4}\)

Solve for \(y\): \(5|y| − 4 = 26\)

\[\begin{array}{rl}5|y|-4=26&\text{Isolate the absolute value term by adding }4\text{ to each side} \\ 5|y|=30&\text{Divide each side by }5 \\ \\ |y|=6&\text{Rewrite as two linear equations} \\ y=6\quad\text{or}\quad y=-6&\text{Solution}\end{array}\nonumber\]

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

Absolute Value Equations with Different Solutions

Often, we will have linear arguments inside the absolute value which changes the solution. Previously, all solution sets have been opposite integers, but in these cases, the solution sets contain different sized numbers.

Example \(\PageIndex{5}\)

Solve for \(t\): \(|2t − 1| = 7\)

\[\begin{array}{rl} |2t-1|=7&\text{The absolute value term is isolated. Rewrite as two linear equations.} \\ 2t-1=7\quad\text{or}\quad 2t-1=-7&\text{Solve each equation.}\end{array}\nonumber\]

Notice we have two equations to solve where each equation results in a different solution. In any case, we solve as usual.

\[\begin{array}{lll} 2t-1=7&& 2t-1=-7 \\ 2t=8&\text{or}& 2t=-6 \\ t=4&&t=-3\end{array}\nonumber\]

Thus, the solution set is \(\{-3,4\}\).

Multiple-Step Absolute Value Equations

Example \(\pageindex{6}\).

Solve for \(x\): \(2 − 4|2x + 3| = −18\)

To isolate the absolute value, we first apply the addition rule for equations. Then apply the multiplication rule for equations.

\[\begin{array}{rl}2-4|2x+3|=-18&\text{Isolate the absolute value term by subtracting }2\text{ from each side} \\ -4|2x+3|=-20&\text{Divide each side by }-4 \\ |2x+3|=5&\text{Rewrite as two linear equations.} \\ 2x+3=5\quad\text{or}\quad 2x+3=-5\end{array}\nonumber\]

Solve each equation.

\[\begin{array}{lll}2x+3=5&&2x+3=-5 \\ 2x=2&\text{or}& 2x=-8 \\ x=1&&x=-4\end{array}\nonumber\]

We now have obtained two solutions, \(x = 1\) and \(x = −4\). Thus, the solution set is \(\{−4, 1\}\).

Equations with Two Absolute Values

In this case, we have an absolute value on each side of the equals sign. However, even though there are two absolute values, we apply the same process. Recall, methods never change, only problems.

Example \(\PageIndex{7}\)

Solve for \(m\): \(|2m-7|=|4m+6|\)

In order to apply the definition, we rewrite this equation as two linear equations, but with the left side as its positive and negative value: \[\begin{array}{rl}|2m-7|=|4m+6|&\text{Rewrite as two linear equations} \\ 2m-7=\color{blue}{4m+6}\color{black}{}\quad\text{or}\quad 2m-7=\color{blue}{-(4m+6)}\color{black}{}\end{array}\nonumber\]

Now, we can solve as usual. Be sure to distribute the negative for the equation on the right.

\[\begin{array}{lll} &&2m-7=\color{blue}{-}\color{black}{}(4m+6) \\ 2m-7=4m+6&&2m-7=\color{blue}{-}\color{black}{}4m\color{blue}{-}\color{black}{}6 \\ -13=2m&&6m-7=-6 \\ -\frac{13}{2}=m&\text{or}&6m=1 \\ &&m=\frac{1}{6}\end{array}\nonumber\]

Thus gives the solutions, \(m=-\frac{13}{2}\) or \(m=\frac{1}{6}\). Thus, the solution set is \(\left\{-\frac{13}{2},\frac{1}{6}\right\}\).

In Example \(\PageIndex{7}\) , because there are absolute value expressions on both sides of the equation, we could have easily applied the definition to the left side and obtained \[\color{blue}{2m-7}\color{black}{}=4m-6\quad\text{or}\quad\color{blue}{-(2m-7)}\color{black}{}=4m-6\nonumber\]

Then solved each linear equation as usual and obtained the same results.

Special Cases

As we are solving absolute value equations, it is important to be aware of special cases. Remember, the result after evaluating absolute value must always be non-negative.

Example \(\PageIndex{8}\)

Solve for \(x\): \(7 + |2x − 5| = 4\)

\[\begin{array}{rl}7+|2x-5|=4&\text{Isolate the absolute value term by subtracting }7\text{ from each side} \\ |2x-5|=-3&X\text{ False}\end{array}\nonumber\]

Careful! Observe the absolute value of \(2x − 5\) is a negative number. This is impossible with absolute value because the result after evaluating absolute value must always be non-negative . Thus, we say this equation has no solution .

Absolute Value Equations Homework

Exercise \(\pageindex{1}\).

\(|x| = 8\)

Exercise \(\PageIndex{2}\)

\(|b| = 1\)

Exercise \(\PageIndex{3}\)

\(|5 + 8a| = 53\)

Exercise \(\PageIndex{4}\)

\(|3k + 8| = 2\)

Exercise \(\PageIndex{5}\)

\(|9 + 7x| = 30\)

Exercise \(\PageIndex{6}\)

\(|8 + 6m| = 50\)

Exercise \(\PageIndex{7}\)

\(|6 − 2x| = 24\)

Exercise \(\PageIndex{8}\)

\(−7| − 3 − 3r| = −21\)

Exercise \(\PageIndex{9}\)

\(7| − 7x − 3| = 21\)

Exercise \(\PageIndex{10}\)

\(\frac{|−4b − 10|}{8} = 3\)

Exercise \(\PageIndex{11}\)

\(8|x + 7| − 3 = 5\)

Exercise \(\PageIndex{12}\)

\(5|3 + 7m| + 1 = 51\)

Exercise \(\PageIndex{13}\)

\(3 + 5|8 − 2x| = 63\)

Exercise \(\PageIndex{14}\)

\(|6b − 2| + 10 = 44\)

Exercise \(\PageIndex{15}\)

\(−7 + 8| − 7x − 3| = 73\)

Exercise \(\PageIndex{16}\)

\(|5x + 3| = |2x − 1|\)

Exercise \(\PageIndex{17}\)

\(|3x − 4| = |2x + 3|\)

Exercise \(\PageIndex{18}\)

\(\left|\frac{4x-2}{5}\right|=\left|\frac{6x+3}{2}\right|\)

Exercise \(\PageIndex{19}\)

\(|n| = 7\)

Exercise \(\PageIndex{20}\)

\(|x| = 2\)

Exercise \(\PageIndex{21}\)

\(|9n + 8| = 46\)

Exercise \(\PageIndex{22}\)

\(|3 − x| = 6\)

Exercise \(\PageIndex{23}\)

\(|5n + 7| = 23\)

Exercise \(\PageIndex{24}\)

\(|9p + 6| = 3\)

Exercise \(\PageIndex{25}\)

\(|3n − 2| = 7\)

Exercise \(\PageIndex{26}\)

\(|2 + 2b| + 1 = 3\)

Exercise \(\PageIndex{27}\)

\(\frac{|-4-3n|}{4}=2\)

Exercise \(\PageIndex{28}\)

\(8|5p + 8| − 5 = 11\)

Exercise \(\PageIndex{29}\)

\(3 − |6n + 7| = −40\)

Exercise \(\PageIndex{30}\)

\(4|r + 7| + 3 = 59\)

Exercise \(\PageIndex{31}\)

\(5 + 8| − 10n − 2| = 101\)

Exercise \(\PageIndex{32}\)

\(7|10v − 2| − 9 = 5\)

Exercise \(\PageIndex{33}\)

\(8|3 − 3n| − 5 = 91\)

Exercise \(\PageIndex{34}\)

\(|2 + 3x| = |4 − 2x|\)

Exercise \(\PageIndex{35}\)

\(\left|\frac{2x-5}{3}\right|=\left|\frac{3x+4}{2}\right|\)

Exercise \(\PageIndex{36}\)

\(\frac{|-n+6|}{6}=0\)

Calcworkshop

How to Solve Absolute Value Equations? 9+ Examples!

// Last Updated: January 20, 2020 - Watch Video //

Absolute Value… makes everything positive, right? But, why?

Jenn (B.S., M.Ed.) of Calcworkshop® teaching abs value equations

Jenn, Founder Calcworkshop ® , 15+ Years Experience (Licensed & Certified Teacher)

Well, Absolute Value means a distance away from the origin (zero), and a distance is always positive!

Alright, that makes sense. But what about Absolute Value Equations and Inequalities ?

So glad you asked!

Okay, so there is a big difference between taking the Absolute Value of a number versus taking the Absolute Value of a variable.

When you take the Absolute Value of a number, it just turns that number positive, like |3| = 3 and |-3|=3.

Easy, right?

But when you take the Absolute Value of a variable, you must account for both positive and negative direction.

For example, if we had |x|=5 that means we are either a distance of 5 units to the right of zero or we are a distance of 5 units to the left of zero, so we need to provide for both possibilities.

Steps for solving absolute value equations and inequalities

Absolute Value Equation

In this lesson we’re going to remind ourself of the most important, and often overlooked, step in solving Absolute Value Equations and Inequalities…

…getting the Absolute Value by itself before you solve!

But don’t worry, the process is straightforward, and simple to follow.

And, no matter if your solving an equation or a compound (conjunction) inequality, as Paul’s Online Notes accurately states, with less then (i.e., less “and”) or greater (i.e., great “or”) symbols, the steps are all the same!

Together we’re going to walk through 6 examples of how to solve Absolute Value Equations and 4 examples of how to solve Absolute Value Inequalities, all while expressing our answers in Interval Notation .

How to Solve Absolute Value Equations – Video

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how to solve equations with absolute value and variables

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How to Solve Absolute Value Equations

how to solve equations with absolute value and variables

When Do You Flip the Inequality Sign?

Absolute value equations can be a little intimidating at first, but if you keep at it you'll soon be solving them easily. When you're trying to solve absolute value equations, it helps to keep the meaning of absolute value in mind.

The ​ absolute value ​ of a number ​ x ​, written | ​ x ​ |, is its distance from zero on a number line. For instance, −3 is 3 units away from zero, so the absolute value of −3 is 3. We write it like this: | −3 | = 3.

Another way to think about it is that ​ absolute value ​ is the positive "version" of a number. So the absolute value of −3 is 3, while the absolute value of 9, which is already positive, is 9.

Algebraically, we can write a ​ formula for absolute value ​ that looks like this:

Take an example where ​ x ​ = 3. Since 3 ≥ 0, the absolute value of 3 is 3 (in absolute value notation, that's: | 3 | = 3).

Now what if ​ x ​ = −3? It's less than zero, so | −3 | = −( −3). The opposite, or "negative," of −3 is 3, so | −3 | = 3.

Now for some absolute value equations. The general steps for solving an absolute value equation are:

Isolate the absolute value expression.

Solve the positive "version" of the equation.

Solve the negative "version" of the equation by multiplying the quantity on the other side of the equals sign by −1.

Take a look at the problem below for a concrete example of the steps.

Example: Solve the equation for ​ x ​:

You'll need to get | 3 + ​ x ​ | by itself on the left side of the equals sign. To do this, add 5 to both sides:

Solve for ​ x ​ as if the absolute value sign weren't there!

That's easy: Just subtract 3 from both sides.

So one solution to the equation is that ​ x ​ = 6.

Start again at | 3 + ​ x ​ | = 9. The algebra in the previous step showed that ​ x ​ could be 6. But since this is an absolute value equation, there's another possibility to consider. In the equation above, the absolute value of "something" (3 + ​ x ​) equals 9. Sure, the absolute value of positive 9 equals 9, but there's another option here too! The absolute value of −9 also equals 9. So the unknown "something" could also equal −9.

In other words:

The quick way to arrive at this second version is to multiply the quantity on the other side of the equals from the absolute value expression (9, in this case) by −1, then solve the equation from there.

Subtract 3 from both sides to get:

So the two solutions are: ​ x ​ = 6 or ​ x ​ = −12.

And there you have it! These kinds of equations take practice, so don't worry if you're struggling at first. Keep at it and it will get easier!

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SOLVING EQUATIONS WITH ABSOLUTE VALUES ON BOTH SIDES

The picture shown below explains how to solve the equations in which we have absolute value sign on both sides.

how to solve equations with absolute value and variables

Example 1 :

Solve for x :

|x - 3| = |3x + 2|

Based on the idea given above, we have

x - 3 = (3x + 2) 

x - 3 = 3x + 2

x - 3 = -(3x + 2) 

x - 3 = - 3x - 2

Justify and evaluation :

Substitute x = -5/2 and x = 1/4 in the given absolute value equation. 

|-5/2 - 3| = |3(-5/2) + 2|

|-11/2| = |-15/2 + 2|

|-11/2| = |-11/2|

11/2 = 11/2

|1/4 - 3| = |3(1/4) + 2|

|-11/4| = |3/4 + 2|

|11/4| = |11/4|

11/4 = 11/4

Substituting x = -5/2 and x = 1/4 into the original equation results in true  statements.

Both the answers x = -5/2 and x = 1/4 are correct and acceptable.

Problem 2 :

|x - 7| = |2x - 2|

x - 7 = 2x - 2

x - 7 = - (2x - 2) 

x - 7 = -2x + 2

Substitute x = -5 and x = 3 in the given absolute value equation.

|-5 - 7| = |2(-5) - 2|

|-12| = |-10 - 2|

|-12| = |-12|

|3 - 7| = |2(3) - 2|

|-4| = |6 - 2|

Substituting x = -5 and x = 3 into the original equation results in true  statements.

Both the answers x = -5 and x = 3 are correct and acceptable.

Problem 3 :

Solve for z :

|2z + 5| = |2z - 1|

2z + 5 = 2z - 1

The above statement is false. 

No solution here.

2z + 5 = - (2z - 1) 

2z + 5 = -2z + 1

Substitute z = -1 in the given absolute value equation.

|2(-1) + 5| = |2(-1) - 1|

|-2 + 5| = |-2 - 1|

Substituting z = -1 into the original equation results in true  statement. 

So, the answer z  = -1 is correct and acceptable.

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NEW CHAPTER 5 Variables, Expressions and Equations

5.4 introduction to equations, learning objectives.

  • Explain what an equation in one variable represents.
  • Determine if a given value for a variable is a solution of an equation.
  • Classify an equation as conditional, a contradiction or an identity.
  • solution : a value that can be substituted for a variable to make an equation true.
  • unknown : a variable in an equation that needs to be solved for.
  • equation : a mathematical statement that asserts the equivalence of two expressions.
  • conditional equation : an equation that has a solution
  • contradiction : an equation that has no solution
  • identity : an equation that is always true

What is an Equation?

An equation is a mathematical statement that asserts the equivalence of two expressions. For example, the assertion that “two plus five equals seven” is represented by the equation [latex]2 + 5 = 7[/latex].

In most cases, an equation contains one or more variables. For example, the equation [latex]x + 3 = 5[/latex], read “[latex]x[/latex] plus three equals five”, asserts that the expression [latex]x+3[/latex] is equal to the value [latex]5[/latex].

It is possible for equations to have more than one variable. For example, [latex]x + y + 7 = 13[/latex] is an equation in two variables, while [latex]5x^2+y^2+9z^2=36[/latex] is an equation in three variables.

Solving Equations

When an equation contains a variable such as [latex]x[/latex], this variable is considered an  unknown  value. In many cases, we can find the values for [latex]x[/latex] that make the equation true. These values are called  solutions  of the equation.

For example, consider the equation we were talking about above: [latex]x + 3 =5[/latex]. You have probably already guessed that the only possible value of [latex]x[/latex] that makes the equation true is 2, because [latex]2 + 3 = 5[/latex]. We use an equals sign to show that we know the value of a given variable. In this case, [latex]x=2[/latex] is the only solution of the equation[latex]x + 3 =5[/latex].

The values of the variables that make an equation true are called the  solutions  of the equation. In turn, solving an equation means determining what values for the variables make the equation a true statement.

The equation above was fairly straightforward; it was easy for us to identify the solution as [latex]x = 2[/latex]. However, it becomes useful to have a process for finding solutions for unknowns as problems become more complex.

Verifying Solutions

If a number is found as a solution of an equation, then substituting that number back into the equation in place of the variable should make the equation true. Thus, we can easily check whether a number is a genuine solution to a given equation.

For example, let’s examine whether [latex]x=3[/latex] is a solution to the equation  [latex]2x + 31 = 37[/latex].

Substituting 3 for [latex]x[/latex], we have:

[latex]2x + 31 = 37 \\ 2\color{blue}{(3)} + 31 = 37 \\ 6 + 31 = 37 \\ 37 = 37[/latex]

This equality is a true statement. Therefore, we can conclude that [latex]x = 3[/latex] is, in fact, a solution of the equation [latex]2x+31=37[/latex].

Determine whether or not [latex]x=-2[/latex] is a solution of the following equations:

1. [latex]3x+7=1[/latex]

2. [latex]-3x^2-x+10=0[/latex]

3. [latex]\sqrt{x^2}=x[/latex]

Replace [latex]x[/latex] i each equation with [latex]-2[/latex] and check if the equation is true.

1. [latex]3x+7=1 \\ 3\color{blue}{(-2)}+7=1 \\ -6+7=1 \\ 1=1[/latex] TRUE [latex]x=-2[/latex] is a solution.

2. [latex]-3x^2-x+10=0 \\ -3(\color{blue}{(-2)}^2-\color{blue}{(-2)}+10=0 \\ -3\cdot 4 + 2 + 10 = 0 \\ -12 + 2 + 10 = 0 \\ 0 = 0[/latex] TRUE [latex]x=-2[/latex] is a solution.

3. [latex]\sqrt{x^2}=x \\ \sqrt{\color{blue}{(-2)}^2}=\color{blue}{(-2)} \\ \sqrt{4}=-2 \\ 2=-2[/latex] FALSE [latex]x=-2[/latex] is NOT a solution.

Determine whether or not [latex]x=3[/latex] is a solution of the following equations:

1. [latex]-2x+5=-1[/latex]

2. [latex]-2x^2+4x+30=0[/latex]

3. [latex]\sqrt{4x^2}=2x[/latex]

  • [latex]x=3[/latex] is a solution of [latex]-2x+5=-1[/latex]
  • [latex]x=3[/latex] is NOT a solution of [latex]-2x^2+4x-6=0[/latex]
  • [latex]x=3[/latex] is a solution of [latex]\sqrt{4x^2}=2x[/latex]

Determine whether the pair of values [latex]x=1 \text{ and }y=-2[/latex] is a solution of the equation.

1. [latex]4x+y=2[/latex]

2. [latex]x^2 + y^2=-3[/latex]

Replace [latex]x[/latex] with [/altex]1[/latex] and [latex]y[/latex] with [latex]-2[/latex].

1. [latex]4x+y=2 \\ 4\color{blue}{(1)}+\color{blue}{(-2)}=2 \\ 4 + (-2) = 2 \\ 2 = 2[/latex]  TRUE. [latex]x=1,\,y=-2[/latex] is a solution of the equation.

2.[latex]x^2 + y^2 = -3 \\\color{blue}{(1)}^2 +\color{blue}{(-2)}^2 = -3 \\ 1 + 4 = 3 \\ 5 = 3 [/latex]  FALSE. [latex]x=1,\,y=-2[/latex] is NOT a solution of the equation.

Determine whether the pair of values [latex]x=2 \text{ and }y=-3[/latex] is a solution of the equation.

1. [latex]x-y=-1[/latex]

2. [latex]x^2 - y^2=-5[/latex]

  • [latex]x=2,\,y=-3[/latex] is a NOT solution of the equation.
  • [latex]x=2,\,y=-3[/latex] is a solution of the equation.

Classes of Equations

Equations can be broadly classified into three categories:

  • Conditional equations

Contradictions

Let’s take a closer look at equations in each of these categories.

Conditional Equations

Case 1: exactly one solution.

The one-variable equation [latex]x+2=5[/latex] has only one solution [latex]x=3[/latex] because the number 3 is the only value that satisfies the equation (makes the left side equal to the right side). The truth of the equation depends upon the value that is put in for [latex]x[/latex]. In other words, the truth of the equation is conditional on the value of [latex]x[/latex].

Case 2: Infinitely Many Solutions (with a condition on the values)

Consider the equation [latex]x+y=2[/latex]. There are infinite number of solution pairs (x and y) for the equation.

For example,

[latex]1+1=2[/latex] [latex]2+0=2[/latex] [latex]4+(–2)=2[/latex] [latex]\frac{1}{2}+\frac{3}{2}=2[/latex]

Any pair of two real numbers with a sum of 2 is a solution of this equation. Even though there are infinitely many solutions, the equation is CONDITIONAL because only those pairs that sum to 2 are solutions of the equation. All other pairs whose sum is not 2 cannot be solutions. A conditional equation means the solution of the equation is constrained to specific value(s).

No Solution:

Consider the equation [latex]x^2=–4[/latex]. There is no solution for this equation because the square of any real number is never negative.

[latex](+)^2 = +[/latex] [latex](–)^2= +[/latex]

Therefore, the equation [latex]x^2=–4[/latex] is a contradiction  as it has no real solution.

Similarly, the equation [latex]|\,x\, |=–4[/latex] is also a contradiction because the absolute value of any number is never negative.

Infinitely Many Solutions (with no condition on the values):

An identity is an equation where any value can be a solution of the equation.

For example, the equation [latex]x=x[/latex] is an identity because the left side of the equation is ALWAYS equal to the right side of the equation, regardless of the value of [latex]x[/latex]. The left side and the right side are identical. Any value for the variable [latex]x[/latex] will be a solution because the left side is always equal to the right side no matter what value we plug in for the variable [latex]x[/latex].

Classify each equation as conditional, a contradiction, or an identity, and identify how many solutions there are.

1. [latex]5x-7=3[/latex]

2. [latex]\sqrt{x-6}=-10[/latex]

3. [latex]x^4+5x^2+12=0[/latex]

4. [latex]4(3x-5)=12x-20[/latex]

1. [latex]5x-7=3[/latex] is a conditional equation because only [latex]x=2[/latex] is a solution. One solution.

2. [latex]\sqrt{x-6}=-10[/latex] is a  contradiction because the square root of any number cannot be negative. No solutions.

3. [latex]x^4+5x^2+12=0[/latex] is a  contradiction because each term is positive and the sum of the terms cannot equal zero. No solutions.

4. [latex]4(3x-5)=12x-20[/latex] is an  identity because this is an example of the distributive property that is true for all values of [latex]x[/latex]. Infinite solutions.

1. [latex]-2x + 2 = 0[/latex]

2. [latex]-\sqrt{x} = -8[/latex]

3. [latex]\large | \normalsize -3x \large | \normalsize = -6[/latex]

4. [latex]2x - y= 4[/latex]

  • Conditional. One solution. [latex]\left ( x = 1 \right )[/latex]
  • Conditional. One solution. [latex]\left ( x = 64 \right )[/latex]
  • Contradiction. No solutions. (absolute values cannot be negative)
  • Conditional. Infinite solutions.

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Solving equations

Here you will learn about solving equations, including linear and quadratic algebraic equations, and how to solve them.

Students will first learn about solving equations in grade 8 as a part of expressions and equations, and again in high school as a part of reasoning with equations and inequalities.

What is solving an equation?

Solving equations is a step-by-step process to find the value of the variable. A variable is the unknown part of an equation, either on the left or right side of the equals sign. Sometimes, you need to solve multi-step equations which contain algebraic expressions.

To do this, you must use the order of operations, which is a systematic approach to equation solving. When you use the order of operations, you first solve any part of an equation located within parentheses. An equation is a mathematical expression that contains an equals sign.

For example,

\begin{aligned}y+6&=11\\\\ 3(x-3)&=12\\\\ \cfrac{2x+2}{4}&=\cfrac{x-3}{3}\\\\ 2x^{2}+3&x-2=0\end{aligned}

There are two sides to an equation, with the left side being equal to the right side. Equations will often involve algebra and contain unknowns, or variables, which you often represent with letters such as x or y.

You can solve simple equations and more complicated equations to work out the value of these unknowns. They could involve fractions, decimals or integers.

What is solving an equation?

Common Core State Standards

How does this relate to 8 th grade and high school math?

  • Grade 8 – Expressions and Equations (8.EE.C.7) Solve linear equations in one variable.
  • High school – Reasoning with Equations and Inequalities (HSA.REI.B.3) Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters.

[FREE] Math Equations Check for Understanding Quiz (Grade 6 to 8)

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Use this quiz to check your grade 6 to 8 students’ understanding of math equations. 10+ questions with answers covering a range of 6th, 7th and 8th grade math equations topics to identify areas of strength and support!

How to solve equations

In order to solve equations, you need to work out the value of the unknown variable by adding, subtracting, multiplying or dividing both sides of the equation by the same value.

  • Combine like terms .
  • Simplify the equation by using the opposite operation to both sides.
  • Isolate the variable on one side of the equation.

Solving equations examples

Example 1: solve equations involving like terms.

Solve for x.

Combine like terms.

Combine the q terms on the left side of the equation. To do this, subtract 4q from both sides.

The goal is to simplify the equation by combining like terms. Subtracting 4q from both sides helps achieve this.

After you combine like terms, you are left with q=9-4q.

2 Simplify the equation by using the opposite operation on both sides.

Add 4q to both sides to isolate q to one side of the equation.

The objective is to have all the q terms on one side. Adding 4q to both sides accomplishes this.

After you move the variable to one side of the equation, you are left with 5q=9.

3 Isolate the variable on one side of the equation.

Divide both sides of the equation by 5 to solve for q.

Dividing by 5 allows you to isolate q to one side of the equation in order to find the solution. After dividing both sides of the equation by 5, you are left with q=1 \cfrac{4}{5} \, .

Example 2: solve equations with variables on both sides

Combine the v terms on the same side of the equation. To do this, add 8v to both sides.

7v+8v=8-8v+8v

After combining like terms, you are left with the equation 15v=8.

Simplify the equation by using the opposite operation on both sides and isolate the variable to one side.

Divide both sides of the equation by 15 to solve for v. This step will isolate v to one side of the equation and allow you to solve.

15v \div 15=8 \div 15

The final solution to the equation 7v=8-8v is \cfrac{8}{15} \, .

Example 3: solve equations with the distributive property

Combine like terms by using the distributive property.

The 3 outside the parentheses needs to be multiplied by both terms inside the parentheses. This is called the distributive property.

\begin{aligned}& 3 \times c=3 c \\\\ & 3 \times(-5)=-15 \\\\ &3 c-15-4=2\end{aligned}

Once the 3 is distributed on the left side, rewrite the equation and combine like terms. In this case, the like terms are the constants on the left, –15 and –4. Subtract –4 from –15 to get –19.

Simplify the equation by using the opposite operation on both sides.

The goal is to isolate the variable, c, on one side of the equation. By adding 19 to both sides, you move the constant term to the other side.

\begin{aligned}& 3 c-19+19=2+19 \\\\ & 3 c=21\end{aligned}

Isolate the variable to one side of the equation.

To solve for c, you want to get c by itself.

Dividing both sides by 3 accomplishes this.

On the left side, \cfrac{3c}{3} simplifies to c, and on the right, \cfrac{21}{3} simplifies to 7.

The final solution is c=7.

As an additional step, you can plug 7 back into the original equation to check your work.

Example 4: solve linear equations

Combine like terms by simplifying.

Using steps to solve, you know that the goal is to isolate x to one side of the equation. In order to do this, you must begin by subtracting from both sides of the equation.

\begin{aligned} & 2x+5=15 \\\\ & 2x+5-5=15-5 \\\\ & 2x=10 \end{aligned}

Continue to simplify the equation by using the opposite operation on both sides.

Continuing with steps to solve, you must divide both sides of the equation by 2 to isolate x to one side.

\begin{aligned} & 2x \div 2=10 \div 2 \\\\ & x= 5 \end{aligned}

Isolate the variable to one side of the equation and check your work.

Plugging in 5 for x in the original equation and making sure both sides are equal is an easy way to check your work. If the equation is not equal, you must check your steps.

\begin{aligned}& 2(5)+5=15 \\\\ & 10+5=15 \\\\ & 15=15\end{aligned}

Example 5: solve equations by factoring

Solve the following equation by factoring.

Combine like terms by factoring the equation by grouping.

Multiply the coefficient of the quadratic term by the constant term.

2 x (-20) = -40

Look for two numbers that multiply to give you –40 and add up to the coefficient of 3. In this case, the numbers are 8 and –5 because 8 x -5=–40, and 8+–5=3.

Split the middle term using those two numbers, 8 and –5. Rewrite the middle term using the numbers 8 and –5.

2x^2+8x-5x-20=0

Group the terms in pairs and factor out the common factors.

2x^2+8x-5x-20=2x(x + 4)-5(x+4)=0

Now, you’ve factored the equation and are left with the following simpler equations 2x-5 and x+4.

This step relies on understanding the zero product property, which states that if two numbers multiply to give zero, then at least one of those numbers must equal zero.

Let’s relate this back to the factored equation (2x-5)(x+4)=0

Because of this property, either (2x-5)=0 or (x+4)=0

Isolate the variable for each equation and solve.

When solving these simpler equations, remember that you must apply each step to both sides of the equation to maintain balance.

\begin{aligned}& 2 x-5=0 \\\\ & 2 x-5+5=0+5 \\\\ & 2 x=5 \\\\ & 2 x \div 2=5 \div 2 \\\\ & x=\cfrac{5}{2} \end{aligned}

\begin{aligned}& x+4=0 \\\\ & x+4-4=0-4 \\\\ & x=-4\end{aligned}

The solution to this equation is x=\cfrac{5}{2} and x=-4.

Example 6: solve quadratic equations

Solve the following quadratic equation.

Combine like terms by factoring the quadratic equation when terms are isolated to one side.

To factorize a quadratic expression like this, you need to find two numbers that multiply to give -5 (the constant term) and add to give +2 (the coefficient of the x term).

The two numbers that satisfy this are -1 and +5.

So you can split the middle term 2x into -1x+5x: x^2-1x+5x-5-1x+5x

Now you can take out common factors x(x-1)+5(x-1).

And since you have a common factor of (x-1), you can simplify to (x+5)(x-1).

The numbers -1 and 5 allow you to split the middle term into two terms that give you common factors, allowing you to simplify into the form (x+5)(x-1).

Let’s relate this back to the factored equation (x+5)(x-1)=0.

Because of this property, either (x+5)=0 or (x-1)=0.

Now, you can solve the simple equations resulting from the zero product property.

\begin{aligned}& x+5=0 \\\\ & x+5-5=0-5 \\\\ & x=-5 \\\\\\ & x-1=0 \\\\ & x-1+1=0+1 \\\\ & x=1\end{aligned}

The solutions to this quadratic equation are x=1 and x=-5.

Teaching tips for solving equations

  • Use physical manipulatives like balance scales as a visual aid. Show how you need to keep both sides of the equation balanced, like a scale. Add or subtract the same thing from both sides to keep it balanced when solving. Use this method to practice various types of equations.
  • Emphasize the importance of undoing steps to isolate the variable. If you are solving for x and 3 is added to x, subtracting 3 undoes that step and isolates the variable x.
  • Relate equations to real-world, relevant examples for students. For example, word problems about tickets for sports games, cell phone plans, pizza parties, etc. can make the concepts click better.
  • Allow time for peer teaching and collaborative problem solving. Having students explain concepts to each other, work through examples on whiteboards, etc. reinforces the process and allows peers to ask clarifying questions. This type of scaffolding would be beneficial for all students, especially English-Language Learners. Provide supervision and feedback during the peer interactions.

Easy mistakes to make

  • Forgetting to distribute or combine like terms One common mistake is neglecting to distribute a number across parentheses or combine like terms before isolating the variable. This error can lead to an incorrect simplified form of the equation.
  • Misapplying the distributive property Incorrectly distributing a number across terms inside parentheses can result in errors. Students may forget to multiply each term within the parentheses by the distributing number, leading to an inaccurate equation.
  • Failing to perform the same operation on both sides It’s crucial to perform the same operation on both sides of the equation to maintain balance. Forgetting this can result in an imbalanced equation and incorrect solutions.
  • Making calculation errors Simple arithmetic mistakes, such as addition, subtraction, multiplication, or division errors, can occur during the solution process. Checking calculations is essential to avoid errors that may propagate through the steps.
  • Ignoring fractions or misapplying operations When fractions are involved, students may forget to multiply or divide by the common denominator to eliminate them. Misapplying operations on fractions can lead to incorrect solutions or complications in the final answer.

Related math equations lessons

  • Math equations
  • Rearranging equations
  • How to find the equation of a line
  • Solve equations with fractions
  • Linear equations
  • Writing linear equations
  • Substitution
  • Identity math
  • One step equation

Practice solving equations questions

1. Solve 4x-2=14.

GCSE Quiz False

Add 2 to both sides.

Divide both sides by 4.

2. Solve 3x-8=x+6.

Add 8 to both sides.

Subtract x from both sides.

Divide both sides by 2.

3. Solve 3(x+3)=2(x-2).

Expanding the parentheses.

Subtract 9 from both sides.

Subtract 2x from both sides.

4. Solve \cfrac{2 x+2}{3}=\cfrac{x-3}{2}.

Multiply by 6 (the lowest common denominator) and simplify.

Expand the parentheses.

Subtract 4 from both sides.

Subtract 3x from both sides.

5. Solve \cfrac{3 x^{2}}{2}=24.

Multiply both sides by 2.

Divide both sides by 3.

Square root both sides.

6. Solve by factoring:

Use factoring to find simpler equations.

Set each set of parentheses equal to zero and solve.

x=3 or x=10

Solving equations FAQs

The first step in solving a simple linear equation is to simplify both sides by combining like terms. This involves adding or subtracting terms to isolate the variable on one side of the equation.

Performing the same operation on both sides of the equation maintains the equality. This ensures that any change made to one side is also made to the other, keeping the equation balanced and preserving the solutions.

To handle variables on both sides of the equation, start by combining like terms on each side. Then, move all terms involving the variable to one side by adding or subtracting, and simplify to isolate the variable. Finally, perform any necessary operations to solve for the variable.

To deal with fractions in an equation, aim to eliminate them by multiplying both sides of the equation by the least common denominator. This helps simplify the equation and make it easier to isolate the variable. Afterward, proceed with the regular steps of solving the equation.

The next lessons are

  • Inequalities
  • Types of graph
  • Coordinate plane

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  1. How To Solve Absolute Value Equations

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  2. How To Solve Absolute Value Equations, Basic Introduction, Algebra

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  3. Solving Absolute Value Equations (solutions, examples, videos

    how to solve equations with absolute value and variables

  4. PPT

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  5. Solving Absolute Value Equations: Complete Guide

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  6. Solving Linear Absolute Value Equations with variables on both sides

    how to solve equations with absolute value and variables

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  6. How to Solve Absolute Value Equations [Algebra basics]

COMMENTS

  1. 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. 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. Questions Tips & Thanks

  3. 2.6: Solving Absolute Value Equations and Inequalities

    Step 2: Set the argument of the absolute value equal to ± p. Here the argument is 5x − 1 and p = 6. 5x − 1 = − 6 or 5x − 1 = 6. Step 3: Solve each of the resulting linear equations. 5x − 1 = − 6 or 5x − 1 = 6 5x = − 5 5x = 7 x = − 1 x = 7 5. Step 4: Verify the solutions in the original equation. Check x = − 1.

  4. Algebra

    There are two ways to define absolute value. There is a geometric definition and a mathematical definition. We will look at both. Geometric Definition In this definition we are going to think of |p| | p | as the distance of p p from the origin on a number line. Also, we will always use a positive value for distance.

  5. How To Solve Absolute Value Equations

    This math video tutorial explains how to solve absolute value equations with variables on both sides. It contains plenty of examples and practice problems. ...more ...more Extraneous Solutions...

  6. Worked example: absolute value equation with two solutions

    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

  7. Solving Absolute value equations with a variable on both sides

    Learn how to solve absolute value equations with extraneous solutions. Absolute value of a number is the positive value of the number. For instance, the abso...

  8. How To Solve Absolute Value Equations, Basic Introduction, Algebra

    The Organic Chemistry Tutor 7.4M subscribers Join Subscribe Subscribed 14K 1M views 6 years ago This algebra video tutorial provides a basic introduction into absolute value equations. it...

  9. Solving Equations Containing Absolute Value

    To solve an equation containing absolute value, isolate the absolute value on one side of the equation. Then set its contents equal to both the positive and negative value of the number on the other side of the equation and solve both equations. Solve | x | + 2 = 5. Isolate the absolute value.

  10. 6.3

    How To: Given an absolute value equation, solve it. Isolate the absolute value expression on one side of the equal sign. If c > 0 c > 0, write and solve two equations: ax+b = c a x + b = c and ax+b =−c a x + b = − c. In the next video, we show examples of solving a simple absolute value equation.

  11. How to solve absolute value equations

    Steps 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

  12. 2.8: Linear Inequalities and Absolute Value Inequalities

    To solve absolute value inequalities, just as with absolute value equations, we write two inequalities and then solve them independently. Definition: Absolute Value Inequalities For an algebraic expression \(X\) and \(k>0,\) an absolute value inequality is an inequality of the form

  13. How to Solve Absolute Value Equations: 10 Steps (with Pictures)

    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.

  14. Solving Absolute Value Equations

    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.

  15. 1.2: Absolute Value Equations

    1.2: Absolute Value Equations. When solving equations with absolute value, the solution may result in more than one possible answer because, recall, absolute value is just distance from zero. Since the integer has distance units from zero, and 4 has distance 4 units from zero, then there are two integers that have distance from zero, .

  16. How to Solve Absolute Value Equations? (9+ Examples!)

    But when you take the Absolute Value of a variable, you must account for both positive and negative direction. For example, if we had |x|=5 that means we are either a distance of 5 units to the right of zero or we are a distance of 5 units to the left of zero, so we need to provide for both possibilities. Absolute Value Equation.

  17. Solving Absolute-Value Equations

    A linear absolute value equation is an equation that takes the form | ax + b | = c . Taking the equation at face value, you don't know if you should change what

  18. 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.

  19. Algebra

    Here is a set of practice problems to accompany the Absolute Value Equations section of the Solving Equations and Inequalities chapter of the notes for Paul Dawkins Algebra course at Lamar University.

  20. How to Solve Absolute Value Equations

    The general steps for solving an absolute value equation are: Isolate the absolute value expression. Solve the positive "version" of the equation. Solve the negative "version" of the equation by multiplying the quantity on the other side of the equals sign by −1. Take a look at the problem below for a concrete example of the steps.

  21. Easiest Way To Solve an Absolute Value Equation and Determine ...

    Learn how to solve absolute value equations with extraneous solutions. Absolute value of a number is the positive value of the number. For instance, the abso...

  22. SOLVING EQUATIONS WITH ABSOLUTE VALUES ON BOTH SIDES

    SOLVING EQUATIONS WITH ABSOLUTE VALUES ON BOTH SIDES. The picture shown below explains how to solve the equations in which we have absolute value sign on both sides. Example 1 : Solve for x : |x - 3| = |3x + 2|. Solution : Based on the idea given above, we have. x - 3 = (3x + 2) x - 3 = 3x + 2.

  23. 5.4 Introduction to Equations

    It is possible for equations to have more than one variable. For example, [latex]x + y + 7 = 13[/latex] is an equation in two variables, while [latex]5x^2+y^2+9z^2=36[/latex] is an equation in three variables. Solving Equations. When an equation contains a variable such as [latex]x[/latex], this variable is considered an unknown value. In many ...

  24. Video: Absolute Value

    Video: Absolute Value Expression | Evaluation, Simplification & Examples Video: Solving Absolute Value Functions & Equations | Rules & Examples

  25. Solving Equations

    What is solving an equation? Solving equations is a step-by-step process to find the value of the variable. A variable is the unknown part of an equation, either on the left or right side of the equals sign. Sometimes, you need to solve multi-step equations which contain algebraic expressions.

  26. Absolute Value Equation Calculator

    Calculate it! Example: 3|2x+1|+4=25 Example (Click to try) 3|2x+1|+4=25 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. Absolute Value Equation Video Lesson Khan Academy Video: Absolute Value Equations