Absolute Value Calculator

What Is Absolute Value?

Absolute value describes magnitude without direction. For a real number, it is the number’s distance from zero on a number line. The notation uses vertical bars: \(\lvert x\rvert\) is read as “the absolute value of \(x\).” Because distance cannot be negative, every real-number absolute value is nonnegative. That one idea explains the basic facts \(\lvert 7\rvert=7\), \(\lvert-7\rvert=7\), and \(\lvert0\rvert=0\).

A signed number communicates both size and direction. The number \(8\) is eight units to the right of zero; \(-8\) is eight units to the left. Their positions are different, but each is eight units from zero. Absolute value preserves the shared distance and ignores the direction. It is therefore not quite accurate to say that absolute value “removes a negative sign.” That shortcut happens to work for a single negative number, but it does not explain expressions, equations, inequalities, or functions. The distance interpretation works in all of those settings.

The formal definition is piecewise. If the input is already nonnegative, its absolute value stays the same. If the input is negative, absolute value takes its opposite:

The second line deserves attention. When \(x=-9\), the expression \(-x\) means \(-(-9)\), which equals \(9\). Here the leading minus sign means “the opposite of,” not “make the answer negative.” This distinction is the foundation for later algebra. It also explains why zero is special: \(\lvert0\rvert=0\), and zero is the only real number with absolute value zero.

Absolute Value as Distance on a Number Line

A number line gives the most reliable mental model for absolute value. Put zero in the center. Positive numbers lie to its right, and negative numbers lie to its left. To find \(\lvert x\rvert\), count the units from \(x\) to zero without caring which direction you move. The point at \(-5\) is five units left of zero, so \(\lvert-5\rvert=5\). The point at \(5\) is five units right of zero, so \(\lvert5\rvert=5\).

This immediately reveals a useful symmetry: every nonzero number and its opposite have the same absolute value. Thus \(\lvert a\rvert=\lvert-a\rvert\). A number line also helps prevent sign errors in subtraction. For example, the distance from \(-3\) to \(4\) is seven units. Algebra records that distance as \(\lvert4-(-3)\rvert=\lvert7\rvert=7\), or equally as \(\lvert-3-4\rvert=\lvert-7\rvert=7\). The expressions inside the bars differ by a sign, but the distances are identical.

Thinking in distances changes the way you read algebraic expressions. The expression \(\lvert x-6\rvert\) means the distance between \(x\) and \(6\), not merely “the positive version of \(x-6\).” If \(\lvert x-6\rvert=2\), then \(x\) is two units from \(6\). There are two possible positions: \(4\) and \(8\). The two answers are not a mysterious rule; they are the two sides of the center point \(6\).

Similarly, \(\lvert x+3\rvert\) can be rewritten as \(\lvert x-(-3)\rvert\). It measures distance from \(-3\). This rewrite is often helpful because the center becomes visible. For a function such as \(y=\lvert x+3\rvert\), the V-shaped graph has its vertex at \((-3,0)\), not at \((3,0)\). A careful number-line interpretation keeps the sign attached to the center instead of reversing it by accident.

When working with negative integers, it is useful to be comfortable with direction before applying absolute value. The verified negative numbers learning resources provide additional practice with signs, ordering, and operations that support this topic.

How to Evaluate an Absolute Value Expression

For a single number, absolute value is quick: identify whether the number is negative, zero, or positive. A positive input remains unchanged, a negative input becomes its opposite, and zero remains zero. The calculator at the top is useful for checking several values quickly, but the same three-case reasoning applies every time.

The last case is where many mistakes begin. Absolute value bars group the entire expression inside them, much like parentheses do. In \(\lvert3-10\rvert\), first compute \(3-10=-7\). Only then evaluate \(\lvert-7\rvert=7\). Do not take the absolute values of \(3\) and \(10\) separately; \(\lvert3-10\rvert\) is not the same expression as \(\lvert3\rvert-\lvert10\rvert\).

Fractions and decimals follow exactly the same rule. For instance, \(\left\lvert-\frac{11}{8}\right\rvert=\frac{11}{8}\), \(\lvert-0.045\rvert=0.045\), and \(\left\lvert\frac{2}{3}\right\rvert=\frac{2}{3}\). If a fraction contains a negative sign in either the numerator or denominator, the value is negative before the bars are applied: \(\left\lvert\frac{-5}{12}\right\rvert=\frac{5}{12}\) and \(\left\lvert\frac{5}{-12}\right\rvert=\frac{5}{12}\). A negative numerator and negative denominator make a positive fraction, so \(\left\lvert\frac{-5}{-12}\right\rvert=\frac{5}{12}\).

For mixed or multi-step expressions, preserve the usual order of operations inside the bars. In \(\lvert2(4-7)+1\rvert\), first calculate the parentheses: \(4-7=-3\). Multiply: \(2(-3)=-6\). Add \(1\): \(-6+1=-5\). Finally, take the absolute value: \(\lvert-5\rvert=5\). The vertical bars apply after the inside has been reduced to a single value.

Exponents require the same care. Compare \(\lvert-4\rvert^2\) with \(\lvert(-4)^2\rvert\). Both happen to equal \(16\), but their steps differ. In the first expression, calculate the absolute value first, then square: \(4^2=16\). In the second, square inside the bars first: \((-4)^2=16\), then take the absolute value. By contrast, \(-\lvert-4\rvert^2=-16\), because the negative sign is outside the complete squared absolute-value expression.

The Difference Between \(-x\), \(\lvert x\rvert\), and \(-\lvert x\rvert\)

These three expressions look similar but communicate different ideas. The expression \(-x\) is the opposite of \(x\). The expression \(\lvert x\rvert\) is the nonnegative distance of \(x\) from zero. The expression \(-\lvert x\rvert\) is the negative of that distance, so it is always zero or negative. Keeping them separate avoids a large share of algebra errors.

For negative inputs, \(-x\) and \(\lvert x\rvert\) give the same numerical result. That coincidence can hide the difference between them. For a positive input such as \(x=6\), however, \(-x=-6\) while \(\lvert x\rvert=6\). They are not interchangeable formulas. In fact, the piecewise definition says that \(\lvert x\rvert=-x\) only when \(x<0\); when \(x\geq0\), \(\lvert x\rvert=x\).

This distinction matters when simplifying. If \(x\) is known to be negative, then \(\lvert x\rvert=-x\). If \(x\) is known to be positive, then \(\lvert x\rvert=x\). If the sign of \(x\) is unknown, you usually cannot remove the bars without more information or a piecewise case. For example, \(\lvert x\rvert=x\) is not true for all real \(x\), and \(\lvert x\rvert=-x\) is not true for all real \(x\).

Essential Absolute Value Properties

Absolute value has a small set of dependable properties. They help simplify expressions, verify answers, and make sense of algebraic steps. Each property comes from the distance interpretation, although some are easiest to remember after seeing examples.

The first fact says absolute value is never negative. The second says only zero has zero distance from zero. The third says opposite points are equally distant from zero. These rules are useful when checking equations. If a calculation leads to \(\lvert x-2\rvert=-5\), stop immediately: no real number can make the left side negative, so there is no real solution.

Products and quotients distribute cleanly across absolute value. For example, \(\lvert(-3)(5)\rvert=\lvert-15\rvert=15\), while \(\lvert-3\rvert\lvert5\rvert=3\cdot5=15\). Likewise, \(\left\lvert\frac{-18}{6}\right\rvert=3\), and \(\frac{\lvert-18\rvert}{\lvert6\rvert}=\frac{18}{6}=3\). These properties are particularly useful for rational expressions and for solving equations where a product is inside the bars.

Addition and subtraction are different. In general, \(\lvert a+b\rvert\) is not equal to \(\lvert a\rvert+\lvert b\rvert\). For example, \(\lvert4+(-4)\rvert=\lvert0\rvert=0\), but \(\lvert4\rvert+\lvert-4\rvert=8\). There is, however, an inequality that is always true:

This is called the triangle inequality. In plain language, going directly from one point to another is never farther than taking a route that makes an extra stop. At an introductory level, the most important lesson is simply not to split absolute-value bars across addition or subtraction unless a rule specifically justifies the step.

Another helpful fact is \(\lvert x^2\rvert=x^2\) for every real \(x\), because a square is already nonnegative. More generally, \(\lvert x\rvert^2=x^2\). This is why the distance formula uses squared differences: squaring removes sign before the final square root returns a nonnegative distance.

Absolute Value with Fractions, Decimals, and Radicals

The definition of absolute value does not depend on the type of real number. An integer, a terminating decimal, a repeating decimal, a fraction, or a radical all have positions on the number line. Their absolute values are their nonnegative distances from zero.

With fractions, determine the sign first. A fraction is negative if exactly one of its numerator or denominator is negative. Therefore:

Do not confuse the absolute value of a fraction with changing only one part of it. The rule \(\left\lvert\frac{a}{b}\right\rvert=\frac{\lvert a\rvert}{\lvert b\rvert}\) is valid because both resulting absolute values are nonnegative. This form can be useful when simplifying a rational expression, provided the denominator is not zero.

Decimals are no different. The negative decimal \(-0.006\) lies \(0.006\) units from zero, so \(\lvert-0.006\rvert=0.006\). Trailing zeros do not change the value: \(\lvert-4.20\rvert=4.20=4.2\). For a measurement problem, keeping the shown decimal places may still matter because it reflects the precision of the measurement even though the numerical value is unchanged.

Radicals often need simplification before interpretation. Since the principal square root is nonnegative, \(\lvert\sqrt{13}\rvert=\sqrt{13}\). But \(\lvert-\sqrt{13}\rvert=\sqrt{13}\). Similarly, \(\lvert3-\sqrt{10}\rvert\) requires checking the sign of \(3-\sqrt{10}\). Because \(\sqrt{10}\) is a little more than \(3\), the inside is negative, so \(\lvert3-\sqrt{10}\rvert=\sqrt{10}-3\). A decimal estimate can help you identify the sign, but an exact comparison is preferable when possible.

Students working on fraction operations can reinforce these skills with the verified basic fractions resources. The key connection is that signs belong to the whole value, while absolute value reports the magnitude of that value.

Absolute Difference: Distance Between Two Numbers

The absolute difference between two real numbers is their distance apart on a number line. For numbers \(a\) and \(b\), the formula is:

The order does not matter because reversing the subtraction changes only the sign inside the bars. For example, the distance between \(-4\) and \(9\) is \(\lvert-4-9\rvert=\lvert-13\rvert=13\). Starting with the other number gives \(\lvert9-(-4)\rvert=\lvert13\rvert=13\). In either direction, the two points are thirteen units apart.

Absolute difference is often more informative than a signed difference when the question asks “how much” rather than “in which direction.” If a target score is \(90\) and a score is \(83\), the shortfall has size \(\lvert83-90\rvert=7\). If a forecast is \(28^\circ\) and the actual temperature is \(31^\circ\), the error size is \(\lvert31-28\rvert=3^\circ\). The sign of the ordinary difference tells whether the actual result is above or below the reference; absolute difference tells the size of the gap.

It is important not to confuse absolute difference with simply subtracting the smaller-looking number. First write a valid subtraction, including parentheses around negative values if needed, then take the absolute value. For instance, the distance from \(-8.5\) to \(-2\) is \(\lvert-8.5-(-2)\rvert=\lvert-6.5\rvert=6.5\). The negative signs are part of the values; the parentheses keep the subtraction readable.

On a coordinate plane, absolute differences describe horizontal and vertical changes. The horizontal separation of \((x_1,y_1)\) and \((x_2,y_2)\) is \(\lvert x_2-x_1\rvert\), and the vertical separation is \(\lvert y_2-y_1\rvert\). When the points are not aligned horizontally or vertically, those differences become parts of the Pythagorean-theorem distance formula. For that next step, the verified 2D distance calculator is a useful companion.

Understanding Absolute Value Equations

An absolute value equation asks for inputs that are a specified distance from a center. The basic form is \(\lvert u\rvert=c\). Before solving, inspect \(c\). If \(c<0\), there is no real solution because absolute value cannot be negative. If \(c=0\), there is one solution: \(u=0\). If \(c>0\), there are usually two solutions because a point can lie \(c\) units to either side of zero.

For example, \(\lvert x\rvert=5\) has solutions \(x=5\) and \(x=-5\). Both values are five units from zero. The equation \(\lvert x-4\rvert=5\) has solutions \(x-4=5\) or \(x-4=-5\). Solving those two linear equations gives \(x=9\) or \(x=-1\). In distance language, the solutions are five units from the center \(4\).

The first algebraic task is normally to isolate the absolute-value expression. Consider \(2\lvert3x-1\rvert-4=10\). Add \(4\) to both sides to get \(2\lvert3x-1\rvert=14\), then divide by \(2\): \(\lvert3x-1\rvert=7\). Now split into two cases:

The first branch gives \(3x=8\), so \(x=\frac83\). The second gives \(3x=-6\), so \(x=-2\). Checking both values in the original equation is a good habit, especially when the absolute value was not isolated at the start.

Some equations have one solution, none, or infinitely many solutions because the expression inside the bars is constant. For example, \(\lvert0x+3\rvert=3\) is true for every real \(x\), while \(\lvert0x+3\rvert=5\) has no solution. Likewise, \(\lvert2x-6\rvert=0\) has exactly one solution because the only way an absolute value is zero is for its inside to be zero: \(2x-6=0\), so \(x=3\).

When an equation has more than one absolute-value expression, do not automatically split every bar independently. An equation such as \(\lvert x-1\rvert=\lvert x+3\rvert\) says that \(x\) is equally distant from \(1\) and \(-3\). The answer is their midpoint, \(-1\). Squaring both sides can be appropriate in that particular case, but the distance picture often gives the result faster and explains why it is unique.

How Absolute Value Inequalities Describe Intervals

Absolute value inequalities are also distance statements. The symbol tells you whether points must lie close to a center or far from it. For a positive number \(c\), the expression \(\lvert u\rvertc\) means the distance is greater than \(c\), so \(u\) lies outside that middle region.

The language gives a memory aid: “less than” produces an and statement because the solution stays between two endpoints; “greater than” produces an or statement because the solution lies in either outside region. The equality bar in \(\leq\) or \(\geq\) determines whether endpoints are included.

For example, solve \(\lvert x+2\rvert<5\). Rewrite the inside as a distance from \(-2\). The rule gives \(-5

Now solve \(\lvert2x-1\rvert\geq7\). Use two branches: \(2x-1\leq-7\) or \(2x-1\geq7\). The first gives \(2x\leq-6\), so \(x\leq-3\). The second gives \(2x\geq8\), so \(x\geq4\). The solution is \(x\leq-3\) or \(x\geq4\). These are the values whose distance from the center \(\frac12\) is at least \(\frac72\).

Always isolate the bars before applying the patterns. For \(3\lvert x-4\rvert+2<14\), subtract \(2\) and divide by \(3\): \(\lvert x-4\rvert<4\). Then \(-4

The sign of the number on the right is critical. Since \(\lvert u\rvert\geq0\), the inequality \(\lvert u\rvert<-2\) has no solution. The inequality \(\lvert u\rvert>-2\) is true for every real number. For non-strict forms, \(\lvert u\rvert\leq-2\) has no solution, while \(\lvert u\rvert\geq-2\) is true for every real number. These cases are easy to decide before doing any algebra.

The Absolute Value Function and Its Graph

The parent absolute value function is \(f(x)=\lvert x\rvert\). Its graph is a V with vertex at \((0,0)\). The right branch follows \(y=x\) because inputs on the right are nonnegative. The left branch follows \(y=-x\) because negative inputs are replaced by their opposites. The graph is continuous at the vertex, but it has a sharp corner there because the slope changes from \(-1\) on the left to \(1\) on the right.

The domain is all real numbers: any real \(x\) can be placed inside the bars. The range is \(y\geq0\): outputs cannot be negative. Points such as \((-3,3)\), \((-2,2)\), \((-1,1)\), \((0,0)\), \((1,1)\), \((2,2)\), and \((3,3)\) make the symmetry clear.

Most graphing problems use the transformed form \(y=a\lvert x-h\rvert+k\). The vertex is \((h,k)\), and the vertical line \(x=h\) is the axis of symmetry. The number \(a\) controls opening and steepness. A positive \(a\) opens the V upward; a negative \(a\) reflects it downward. If \(\lvert a\rvert>1\), the graph is narrower because each arm changes more quickly. If \(0<\lvert a\rvert<1\), it is wider.

Take \(y=-2\lvert x-3\rvert+4\). The vertex is \((3,4)\). Because \(a=-2\), the graph opens downward and is steeper than the parent graph. The axis of symmetry is \(x=3\). Starting at the vertex, move one unit right or left; the output falls by \(2\), producing points \((2,2)\) and \((4,2)\). Moving two units from the vertex gives \((1,0)\) and \((5,0)\), which are the x-intercepts.

The expression inside the bars is a common source of sign mistakes. In \(y=\lvert x+5\rvert\), write \(x+5\) as \(x-(-5)\). The vertex is at \((-5,0)\), so the graph shifts left five units. In \(y=\lvert x-5\rvert\), the vertex is at \((5,0)\), so it shifts right five units. The visible sign in the formula is opposite the horizontal shift because the standard form is \(x-h\).

Finding Intercepts, Domain, and Range

For an absolute value function in vertex form, the domain is normally all real numbers. The range depends on whether the graph opens up or down. If \(a>0\), the vertex gives the minimum output, so \(y\geq k\). If \(a<0\), the vertex gives the maximum output, so \(y\leq k\). If \(a=0\), the expression is no longer a V-shaped absolute value graph; it is the constant function \(y=k\).

To find y-intercepts, substitute \(x=0\). For \(y=3\lvert x-2\rvert-1\), the y-intercept is \(3\lvert-2\rvert-1=6-1=5\), so the point is \((0,5)\). To find x-intercepts, set \(y=0\):

The two possible x-values are \(x-2=\frac13\) and \(x-2=-\frac13\), so \(x=\frac73\) or \(x=\frac53\). A V that opens upward may have two x-intercepts, one x-intercept when its vertex touches the x-axis, or no x-intercepts when the whole graph stays above the axis. A downward V follows the corresponding reversed pattern.

Use the vertex first whenever you are asked to sketch or analyze the graph. It gives the center, the opening, and the range in one step. Then use the coefficient \(a\) to make a small symmetric table of points. This is generally more efficient and less error-prone than plotting many unrelated inputs.

Why Absolute Value Matters Outside a Worksheet

Absolute value is practical whenever the size of a difference matters more than the direction of that difference. A signed difference is still useful when you need to know whether a value is high or low, above or below, ahead or behind. Absolute value becomes useful when both directions should count equally.

In measurement, absolute error compares a measured value with an accepted or target value:

If a rod should be \(50.0\text{ cm}\) long but measures \(49.7\text{ cm}\), the absolute error is \(\lvert49.7-50.0\rvert=0.3\text{ cm}\). The signed error is \(-0.3\text{ cm}\), which shows the rod is short; the absolute error reports only the size of the discrepancy. Both quantities can be useful, but they answer different questions.

Tolerances are naturally written with absolute value. Suppose a machine part must stay within \(0.02\) millimeters of a target diameter \(12.00\) millimeters. The condition is \(\lvert d-12.00\rvert\leq0.02\). Expanding it gives \(11.98\leq d\leq12.02\). The compact absolute-value form describes “within a fixed distance of a target” in one line.

In finance, an account might end a day above or below its planned balance. The signed variation tells direction, while \(\lvert\text{actual}-\text{planned}\rvert\) gives the amount of variation. In sports, the margin between two scores is the absolute difference. In navigation, the difference between elevations relative to sea level has a size even when one elevation is below sea level. In programming, an abs function is used in the same way to compare distances, enforce tolerances, and calculate unsigned differences.

Statistics uses absolute deviations to describe how far observations are from a center. If a data value is \(x\) and a reference value is \(m\), then \(\lvert x-m\rvert\) measures the distance between them. The mean absolute deviation averages those distances. This prevents positive and negative deviations from cancelling one another, which would happen if ordinary signed deviations were added without first accounting for direction.

Absolute Value in Algebra, Geometry, and Data

Absolute value connects several major topics in mathematics. In algebra, it describes solutions a fixed distance from a center. In coordinate geometry, it expresses horizontal or vertical separation. In data analysis, it measures the size of a residual or error. The same notation works because each application is fundamentally about nonnegative distance.

For example, the condition \(\lvert x-100\rvert\leq8\) says that \(x\) must be within \(8\) of \(100\). The solution \(92\leq x\leq108\) could describe an acceptable test-score band, a manufacturing tolerance, or a safe operating range. The condition \(\lvert x-100\rvert\geq8\) describes the opposite: values at least \(8\) units away from the target.

In geometry, a vertical line through \(x=4\) can be viewed as the set of points whose horizontal coordinate has zero distance from \(4\): \(\lvert x-4\rvert=0\). Parallel vertical boundary lines \(x=2\) and \(x=6\) can be summarized by \(\lvert x-4\rvert=2\). This relationship between a center and equal-distance boundaries is one reason absolute-value graphs and inequalities are so useful for describing regions.

In data work, absolute residuals avoid cancellation. Imagine predicted values that are sometimes \(4\) units too high and sometimes \(4\) units too low. The signed errors can add to zero even though every prediction is off. Taking absolute values first reports the actual average size of the errors. This is why “absolute” often means “magnitude only” in scientific and analytical settings.

A solid algebra foundation helps connect these applications to symbolic rules. The verified algebra overview and linear-equations resources are useful next steps when you want more practice solving the linear branches that arise after an absolute-value equation is split.

A Reliable Problem-Solving Strategy

Most absolute value problems become manageable when you identify what the bars represent before doing calculations. Use the following process for a number, expression, equation, or inequality:

1. Read the bars as a group. Everything between \(\lvert\) and \(\rvert\) belongs to one expression.
2. Simplify inside first. Use the normal order of operations before evaluating the absolute value.
3. Ask for the type of problem. A direct evaluation produces one nonnegative number. An equation may produce zero, one, two, or infinitely many solutions. An inequality describes an interval or intervals.
4. Use distance language. Interpret \(\lvert x-a\rvert\) as distance from \(a\). This often shows the answer pattern before any algebra.
5. Check sign restrictions. An absolute value cannot equal a negative number, and right-side negative values in inequalities frequently settle the problem immediately.
6. Verify in the original expression. Substitute possible solutions back, especially after several algebra steps.

A calculator is best used as a check, not as a replacement for the structure. For direct values, enter the number exactly as it appears after simplification. For fractional inputs, use a slash, such as -13/6. For distances, type both values with their signs. The result should support your own number-line reasoning: a distance cannot come out negative.

Common Absolute Value Mistakes to Avoid

Mistake 1: Removing the bars before simplifying. In \(\lvert2-9\rvert\), the answer is not \(2-9\). Simplify to \(-7\), then take the absolute value: \(7\). The bars apply to the result of the whole expression inside them.

Mistake 2: Splitting bars across subtraction. The statement \(\lvert a-b\rvert=\lvert a\rvert-\lvert b\rvert\) is not generally true. A quick counterexample is \(\lvert3-8\rvert=5\), while \(\lvert3\rvert-\lvert8\rvert=-5\). Products and quotients have distribution rules; sums and differences do not.

Mistake 3: Giving only one answer to an equation. If \(\lvert x-2\rvert=6\), both \(x=8\) and \(x=-4\) work. Unless the right side is zero or the structure creates a special case, an absolute-value equation with a positive right side has two branches.

Mistake 4: Forgetting that a negative right side can be impossible. The equation \(\lvert4x+1\rvert=-2\) has no real solution. No amount of algebra can make a real-number distance negative.

Mistake 5: Using “and” for every inequality. The inequality |x| < 5 gives −5 < x < 5, an and statement. The inequality |x| > 5 gives x < −5 or x > 5, an or statement. A number line makes the difference visible: inside versus outside.

Mistake 6: Reversing a graph shift. In \(y=\lvert x+4\rvert\), the vertex is \((-4,0)\), not \((4,0)\). Convert \(x+4\) to \(x-(-4)\) to identify \(h=-4\) in \(y=a\lvert x-h\rvert+k\).

Mistake 7: Dropping parentheses around negative numbers. Write \(\lvert-6-(-2)\rvert\), not an ambiguous chain of signs. Parentheses preserve the subtraction and make the calculation easier to check.

Worked Absolute Value Examples

Example 1: A negative integer

Evaluate \(\lvert-14\rvert\).

The point \(-14\) is fourteen units from zero. The result is positive because it is a distance.

Example 2: An expression inside the bars

Evaluate \(\lvert6-15\rvert\).

Subtract first. The expression inside the bars is \(-9\), whose distance from zero is \(9\).

Example 3: A fraction

Evaluate \(\left\lvert-\frac{17}{5}\right\rvert\).

The negative fraction lies \(\frac{17}{5}\) units from zero. Its absolute value is positive.

Example 4: Distance between two negative numbers

Find the distance between \(-11\) and \(-3\).

The points are eight units apart. Notice that subtracting a negative number creates addition: \(-11-(-3)=-8\).

Example 5: An equation with two solutions

Solve \(\lvert x+1\rvert=4\).

The expression \(\lvert x+1\rvert\) measures distance from \(-1\). The two solutions, \(3\) and \(-5\), are each four units from \(-1\).

Example 6: An equation with no real solution

Solve \(\lvert2x-7\rvert=-1\).

There is no real solution. The left side is always at least zero, so it cannot equal \(-1\). Recognizing this immediately saves unnecessary algebra.

Example 7: A less-than inequality

Solve \(\lvert3x-6\rvert<12\).

The solution is the interval \((-2,6)\). Every value in that interval makes \(3x-6\) less than twelve units from zero.

Example 8: A greater-than-or-equal-to inequality

Solve \(\lvert x-5\rvert\geq3\).

The expression says that \(x\) is at least three units from \(5\). Values between \(2\) and \(8\) are too close; the endpoints are included because the sign is \(\geq\).

Example 9: A transformed graph

Describe \(y=\frac12\lvert x+2\rvert-3\).

Write the inside as \(x-(-2)\). The vertex is \((-2,-3)\). The graph opens upward because \(a=\frac12>0\), and it is wider than \(y=\lvert x\rvert\) because \(\left\lvert\frac12\right\rvert<1\). The axis of symmetry is \(x=-2\), the domain is all real numbers, and the range is \(y\geq-3\).

Example 10: A tolerance statement

A temperature sensor must read within \(0.5^\circ\) of \(22^\circ\). Write and solve an absolute value inequality for its reading \(T\).

The absolute-value statement is compact, while the final compound inequality gives the permitted interval.