Absolute Value Equations
Absolute value equations ask for values that make a distance or magnitude equal to a specified quantity. In brief: |x − 3| = 5 means x is five units from 3, so x = 8 or x = −2. The two solutions come from opposite directions on the number line, not from giving absolute value a negative output.
The main method is to isolate the absolute value, inspect the value on the other side and solve the appropriate cases. Some equations have two solutions, some have one and others have none. This guide explains those possibilities, handles linear expressions inside the bars and shows why checking the original equation matters.
What is the rule for solving |A| = c?
If c is positive, |A| = c means A = c or A = −c. The expression A may be a variable or a more complicated algebraic expression, but both equations must be considered. Their solutions are combined because satisfying either branch makes the absolute value equal to the required positive magnitude.
If c = 0, the only possibility is A = 0. The two apparent cases would be identical because positive zero and negative zero are the same number. If c is negative, there are no real solutions: an absolute value cannot equal a negative number.
These statements come directly from the distance definition. A positive distance from zero identifies two signed inputs, distance zero identifies one input and a negative distance identifies none. The number of solutions for the original variable can still depend on the expression A, so two branches do not always guarantee exactly two distinct variable values.
Why should you isolate the absolute value first?
An equation may include a coefficient or an added constant outside the bars. Undo those operations before splitting into cases, using the same operation on both sides. In 2|x − 1| + 3 = 11, subtract 3 to get 2|x − 1| = 8, then divide by 2 to obtain |x − 1| = 4.
Now solve x − 1 = 4 or x − 1 = −4, giving x = 5 or x = −3. Splitting before isolating can lead to negating only part of the expression or incorrectly changing an outside constant. Isolating makes the nonnegative quantity and its target value explicit.
A negative outside coefficient is allowed. For −3|x + 2| = −12, dividing by −3 gives |x + 2| = 4, which has ordinary positive-target cases. What matters is the value after the absolute value is isolated, not whether a negative number appears somewhere in the original equation.
How do you solve an equation with a linear expression inside?
Solve 3|2x − 1| − 4 = 11. Add 4 to both sides to get 3|2x − 1| = 15, then divide by 3 to obtain |2x − 1| = 5. The target is positive, so the inner expression must equal either 5 or −5.
The first case is 2x − 1 = 5, giving 2x = 6 and x = 3. The second case is 2x − 1 = −5, giving 2x = −4 and x = −2. The proposed solution set is therefore {−2, 3}, with the two cases connected by or rather than and.
Check both values in the original equation. At x = 3, the left side is 3|5| − 4 = 11. At x = −2, it is 3|−5| − 4 = 11. Both satisfy the equation, so both belong in the final solution set.
- Undo operations outside the absolute-value bars.
- Check whether the isolated target is positive, zero or negative.
- For a positive target, solve both signed inner equations.
- Substitute every candidate into the original equation.
- List the distinct valid solutions.
When does an absolute value equation have no solution?
The equation |x + 4| = −2 has no real solution because its left side is nonnegative and its right side is negative. Do not split it into x + 4 = −2 or x + 4 = 2; that procedure would create candidates for a different equation with a positive target.
For |3x − 6| = 0, solve only 3x − 6 = 0, giving x = 2. Checking confirms |0| = 0. This is one distinct solution, not two copies of the same value, even though a mechanical plus-or-minus notation might appear to generate two identical branches.
A contradiction may emerge only after isolating. In 4|x| + 7 = 3, subtracting 7 and dividing by 4 gives |x| = −1, so the original equation has no solution. Identifying that impossibility early saves unnecessary algebra and prevents invalid candidates from being presented as answers.
| Form | Condition | Equivalent requirement |
|---|---|---|
| |A| = c | c > 0 | A = c or A = −c |
| |A| = 0 | Zero target | A = 0 |
| |A| = c | c < 0 | No real solution |
| |A| = |B| | Real expressions | A = B or A = −B |
What if the other side contains a variable?
For |A(x)| = B(x), the right side must be nonnegative at every valid solution. You may form the cases A(x) = B(x) and A(x) = −B(x), but candidates must also satisfy B(x) ≥ 0 and the original equation. Ignoring this sign restriction can introduce values that solve a branch equation but not the absolute-value equation.
Consider |x − 2| = 2x. The right side requires x ≥ 0. The first case x − 2 = 2x gives x = −2, which violates that requirement. The second case x − 2 = −2x gives 3x = 2 and x = 2/3, which is nonnegative.
Substitution confirms the valid candidate: |2/3 − 2| = 4/3 and 2 × 2/3 = 4/3. The rejected candidate gives |−4| = 4 on the left and −4 on the right. Writing the sign requirement before solving makes the reason for rejection clear rather than leaving it as an unexplained checking step.
How do you solve equations with absolute values on both sides?
For real expressions A and B, |A| = |B| means A = B or A = −B. Two numbers have the same distance from zero exactly when they are equal or opposite. This rule avoids a four-case expansion when each side contains one absolute value and no additional outside terms remain.
Solve |2x − 1| = |x + 3|. The equal case is 2x − 1 = x + 3, giving x = 4. The opposite case is 2x − 1 = −(x + 3), giving 3x = −2 and x = −2/3. Both must be checked, and both satisfy the original equation.
An equation such as |x| + |x − 2| = 4 requires a different approach because one side contains a sum of absolute values. Split the number line at the points where the inner expressions change sign, then solve the appropriate linear expression on each interval. Do not replace a sum of absolute values by the absolute value of a sum.
How can a distance picture help solve an absolute value equation?
The equation |x − a| = r describes points at distance r from a. For |x + 1| = 4, the centre is −1 because x + 1 = x − (−1). Moving four units right and left from −1 gives the solutions 3 and −5, matching the two algebraic branches.
Graphically, solving |x − a| = r finds intersections of the V-shaped graph y = |x − a| with the horizontal line y = r. A positive horizontal level crosses twice, level zero touches the vertex and a negative level lies below the graph. This picture explains the basic solution counts.
Distance equations appear in target deviations and symmetric location problems. If a measurement must differ from 20 by exactly 0.4, the equation |m − 20| = 0.4 gives m = 19.6 or m = 20.4. An allowed range would instead require an inequality, because equality describes only the boundary values.
What mistakes should you avoid when solving absolute value equations?
Do not discard the second case when the isolated target is positive. Solving only x − 3 = 5 for |x − 3| = 5 misses the equally valid point on the opposite side of the centre. Conversely, do not create two cases when the target is negative, because the original equation is already impossible.
Negate the entire inner expression or target expression when forming an opposite case. Parentheses are essential in statements such as 2x − 1 = −(x + 3), which becomes 2x − 1 = −x − 3. Changing only the first term would alter the equation and produce an incorrect candidate.
Finally, retain domain restrictions if fractions or other restricted expressions occur inside the bars. Absolute value does not make a zero denominator valid. Check candidates in the original equation rather than only in a transformed branch, especially after squaring or when the other side contains a variable whose sign can change.