The Absolute Value Equation Calculator provided by Hesapstan solves structured equations of the form p|ax+b|+q=r step by step. It does not solve equations with multiple absolute value groups, nonlinear inside expressions such as |x²−1|, complex numbers, graphs, or symbolic parameter cases.
An absolute value equation is first isolated as |ax+b| = c
This calculator works with p|ax+b|+q=r. The first step is to isolate the absolute value expression: |ax+b| = (r−q)/p. The value on the right is c, and it determines whether the equation can have real solutions.
Absolute value measures the distance of a number from zero on the number line. |3| = 3 because 3 is 3 units from zero; |−3| = 3 because −3 is also 3 units from zero. This distance interpretation explains the two-case method: if |ax+b| = c, the expression ax+b must sit exactly c units from zero, which can happen in two directions.
Equations such as |x+1| + |x−2| = 5 or |x²−1| = 3 are outside this runtime. The supported structure is p|ax+b|+q=r only.
A negative isolated right side means no real solution
Absolute value cannot be negative. After isolation, if |ax+b| = c has c < 0, the solution set is empty. For example, |x| = −1 has no real solution.
If |ax+b| = 0, the inside expression must be exactly 0. The two cases collapse to the same value, so there is one solution.
A positive c creates two linear cases
When |ax+b| = c and c > 0, the inside expression may equal c or −c. The calculator therefore solves ax+b=c and ax+b=−c separately, simplifies the x-values as exact fractions when needed, and verifies each answer by substitution.
For 2|3x−1|+4=10, isolation gives |3x−1|=3. Case 1: 3x−1=3 gives x=4/3. Case 2: 3x−1=−3 gives x=−2/3. Both values are checked in the original equation.
Worked example step by step: 2|3x−1|+4=10
The following steps show how the calculator solves this equation and verifies each answer.
- Isolation: 2|3x−1| = 10−4 = 6, so |3x−1| = 3
- Since c = 3 > 0, two cases apply.
- Case 1: 3x−1 = 3 → 3x = 4 → x = 4/3
- Case 2: 3x−1 = −3 → 3x = −2 → x = −2/3
- Verify Case 1: 2|3(4/3)−1|+4 = 2|4−1|+4 = 2×3+4 = 10 ✓
- Verify Case 2: 2|3(−2/3)−1|+4 = 2|−2−1|+4 = 2×3+4 = 10 ✓
- Solution set: { 4/3, −2/3 }
For |2x+5|+8=3, isolation gives |2x+5| = 3−8 = −5. Because absolute value can never be negative, this equation has no real solution and the solution set is empty.
Verification helps catch missing or invalid roots
A common student error is solving only ax+b=c or missing the fact that the isolated right side is negative. This runtime substitutes each candidate value back into p|ax+b|+q=r and shows whether it verifies.
The coefficients p and a cannot be zero. Decimal comma and decimal dot are accepted, but symbolic parameters, complex numbers, and graphing are not supported.
Frequently Asked Questions
What type of absolute value equations does this solve?
It solves equations with one absolute value group in the form p|ax+b|+q=r. The coefficients a and p must be nonzero real numbers.
What is the two-case method?
If |ax+b|=c and c is positive, the inside expression can be c or −c. So the calculator solves ax+b=c and ax+b=−c as separate linear cases.
What if the isolated right side is negative?
There is no real solution, because an absolute value cannot equal a negative number.
Why is there one solution when c = 0?
If |ax+b|=0, then ax+b must be exactly 0. The positive and negative cases are the same, so only one x-value remains.
Can I solve |x+1| = |x−2| here?
No. That equation has two absolute value groups. This calculator supports only the single-group form p|ax+b|+q=r.
Are answers shown as decimals or fractions?
The runtime shows exact rational fractions when the denominator is not 1, so values such as −8/3 are not rounded into decimals.
What does verification mean?
Each candidate solution is substituted back into the original equation so you can see whether it satisfies the equation.
Does this calculator graph the equation?
No. It is an algebraic step solver for the supported structure; it does not provide graphing, complex solutions, or symbolic parameter solving.