Absolute Value Calculator

How to Use the Absolute Value Calculator

1. Add Your Problem

   Use Calculator for keypad entry, or Math Input to type expressions, equations, inequalities, and functions. You can use vertical bars, such as \(|2x-5|=9\), when entering an equation. Use Canvas to draw, handwrite, or upload a photo of your problem.

2. Choose Your Output

   Step-by-step answers are selected by default. You can also choose Explain Like I’m 10, Create Practice Test, Create Study Guide, Create Flashcards, or Find My Mistake for your own work. Choose Create Graph when you enter a suitable absolute value function.

3. Choose a Language

   Use the language dropdown labeled “Default” if you want to change the response language.

4. Get the Result

   Select Solve to submit your problem and receive the chosen explanation or output.

How to Solve Absolute Value Problems

Absolute value measures distance from zero on a number line. Since distance cannot be negative, an absolute value result cannot be negative for real inputs.

The method changes depending on what you enter: an expression is evaluated, an equation is solved, an inequality becomes a range of values, and a function can be graphed.

Find the Absolute Value of an Expression

Simplify inside the vertical bars first when needed. Then ask: how far is that number from zero?

\[ |a|= \begin{cases} a, & a \ge0 \\ -a, & a < 0 \end{cases} \]

For example:

\[ |3-11|=|-8|=8 \]

The expression inside the bars equals \(-8\), which is eight units from zero.

Pay attention to where the bars begin and end. For instance,

\[ |-4|+|3|=4+3=7 \]

but

\[ |-4+3|=|-1|=1 \]

Absolute value can equal zero, as in \(|0|=0\). It cannot equal a negative real number.

Solve Absolute Value Equations

Use this method when an absolute value expression equals a number. First, isolate the bars so the equation looks like this:

\[ |A|=k \]

Then apply the right rule:

• If \(k > 0\), solve \(A=k\) and \(A=-k\).

• If \(k=0\), solve only \(A=0\).

• If \(k < 0\), there is no real solution.

For example:

\[ |2x-3|=7 \]

\[ 2x-3=7 \quad \text{or} \quad 2x-3=-7 \]

\[ x=5 \quad \text{or} \quad x=-2 \]

Those two cases represent values that place \(2x-3\) the same distance from zero. Absolute value equations are algebraic equations, so you can also solve for \(x\) step by step when you need help with the surrounding algebra.

This direct split works only after one absolute value expression is isolated. An equation with bars on both sides, such as \(|x-1|=|2x+3|\), may need piecewise cases or another method.

Solve Inequalities and Graph Functions

Absolute value inequalities describe distance from a number. When \(k>0\), values that are within a distance form an “and” statement:

\[ |A| < k \quad \Rightarrow \quad -k < A < k \]

\[ |A|\le k \quad \Rightarrow \quad -k\le A\le k \]

Values that are outside a distance form an “or” statement:

\[ |A|>k \quad \Rightarrow \quad A<-k \text{ or } A>k \]

\[ |A|\ge k \quad \Rightarrow \quad A\le-k \text{ or } A\ge k \]

For example:

\[ |x-4|\le3 \]

\[ -3\le x-4\le3 \]

\[ 1\le x\le7 \]

If the number on the right is zero or negative, pause before splitting. For example, \(|A|<-2\) has no real solution, while \(|A|\ge0\) is true for every real value of \(A\). Also, reverse an inequality sign whenever you divide or multiply by a negative number.

For graphing, a common form is:

\[ y=a|x-h|+k \]

The graph is V-shaped, with vertex \((h,k)\). A negative \(a\) reflects the V downward, and a larger \(|a|\) makes it narrower or steeper. The values of \(h\) and \(k\) shift the graph horizontally and vertically. You can see more Algebra tools for related graphing and equation practice.

A graph helps you see the function’s shape, but algebraic steps are still needed for exact equation or inequality solutions.

Understanding Absolute Value

Absolute value is distance, not a rule that simply “changes negatives to positives.” Both \(-6\) and \(6\) sit six units from zero, so:

\[ |-6|=6 \qquad \text{and} \qquad |6|=6 \]

The placement of a negative sign matters. In \(-|6|=-6\), the bars are evaluated first, giving \(6\), and the outside negative sign is applied afterward.

Distance also explains why an equation can have two, one, or no real solutions. A positive distance can be reached in two directions, zero distance reaches only zero, and a negative distance is impossible. The same distance idea creates the V shape of \(y=|x|\): on one side of zero, the graph uses \(x\); on the other, it uses \(-x\), keeping every output nonnegative.

Worked Example

Solve:

\[ |3x-5|=13 \]

1. \[ 3x-5=13 \quad \text{or} \quad 3x-5=-13 \]

   Since 13 is positive, the inside can be 13 or \(-13\).

2. \[ 3x=18 \quad \text{or} \quad 3x=-8 \]

   Add 5 to both sides in each equation.

3. \[ x=6 \quad \text{or} \quad x=-\frac{8}{3} \]

   Divide both sides by 3.

4. \[ |3(6)-5|=|13|=13 \]

   The first value checks correctly.

5. \[ \left|3\left(-\frac{8}{3}\right)-5\right|=|-13|=13 \]

   The second value also checks correctly.

The final answer is:

\[ \boxed{x=6 \quad \text{or} \quad x=-\frac{8}{3}} \]

If your equation changes into a quadratic after you remove the bars, you can compare methods for quadratics.

Common Mistakes

1. Treating \(|-9|\) as \(-9\)

   This happens when you focus on the negative sign instead of the distance. Since \(-9\) is nine units from zero, \(|-9|=9\).

2. Using \(\pm\) before isolating the absolute value

   Students sometimes split an equation while other terms are still outside the bars. First isolate \(|A|\), then split into \(A=k\) and \(A=-k\) only when \(k>0\).

3. Saying \(|A|=-k\) has two real solutions

   The \(\pm\) idea does not apply to a negative target. Because real-number distance cannot be negative, \(|A|=-4\) has no real solution.

4. Turning every absolute value inequality into “or”

   “Less than” means inside a distance, so both bounds must be true: an “and” statement. “Greater than” means outside the distance, so either outer region works: an “or” statement.

5. Forgetting to reverse an inequality after dividing by a negative number

   For example, after getting \(-2x>6\) while solving an absolute value inequality, dividing by \(-2\) gives \(x<-3\), not \(x>-3\). Dividing or multiplying by a negative reverses the inequality direction.

6. Assuming the absolute value of a product is different from the product of absolute values

   A common mistake is to think absolute value must be calculated only after multiplying. In fact, \(|ab|=|a||b|\). For example, \(|-3\times4|=|-12|=12\), which is the same as \(|-3||4|=3\times4=12\).

7. Confusing absolute value with the opposite of a number

   Absolute value and changing a number's sign are not the same operation. For example, \(|-7|=7), but (-(-7)=7\) for a different reason. The absolute value represents distance from zero, not simply a sign-change rule.

8. Ignoring parentheses when finding the absolute value of an expression

   When an expression contains multiple operations, applying absolute value to only one part can produce the wrong result. For example, \(|3-8|=5), while (3-|8|=-5\). The bars determine exactly which expression is being evaluated.

9. Assuming absolute value distributes over addition or subtraction

   The rule \(|a+b|=|a|+|b|\) is generally false. For example, \(|3+(-5)|=2\), whereas \(|3|+|-5|=8\). Absolute value can be distributed over multiplication, but not generally over addition or subtraction.

10. Forgetting that zero has an absolute value of zero

    Some users associate absolute value only with making negative numbers positive and overlook zero. Since zero is exactly zero units from zero, \(|0|=0\).