The MathBlog Absolute Value Equation Calculator helps you quickly solve absolute value equations, providing detailed steps to enhance your understanding and learning. You can choose between two types of equations:

  • One absolute value equations:
    \( 3|2x+1|+3=2x+4 \)
    where a=3, b=2, c=1, d=3, e=2, and f=4.
  • Two absolute value equations:
    \( 3|x -1| -2 = 6| -2x + 4 |\)
    where a=3, b=1, c=-1, d=-2, e=6, f=-2, g=4, and h=0.

Choose your equation type below, enter the coefficients into the calculator, and click “Calculate” to see the results for \(x\).

Absolute Value Equation Calculator

a∙|b∙x + c| + d = e∙x + f

How to solve an absolute value equation by hand

ONE ABSOLUTE VALUE EQUATION

Solve \( 2|x+4|+3=9 \):

There are two cases:

CASE 1: \( (x+4) \ge 0 \)

\( 2|x+4|+3=9 \)
\( 2(x+4)+3=9 \)
\( 2x + 8 + 3 = 9 \)
\( 2x + 11 = 9 \)
\( 2x = 9 – 11 \)
\( 2x = -2 \)
\( x = -1 \)

TRUE: It verifies our condition.

CASE 2: \( (x+4) < 0 \)

\( 2|x+4|+3=9 \)
\( 2(-(x+4))+3=9 \)
\( 2(-x – 4) + 3 = 9 \)
\( -2x – 8 + 3 = 9 \)
\( -2x – 5 = 9 \)
\( -2x = 9 + 5 \)
\( -2x = 14 \)
\( x = -7 \)

TRUE: It verifies our condition.

SOLUTION

\( x = -1 \) or \( x = -7 \).

TWO ABSOLUTE VALUES EQUATION

Solve \( 4|x – 3| + 2 = |2x + 4| – 3 \):

There are multiple cases to consider based on the expressions inside the absolute values:

CASE 1: \( x – 3 \ge 0 \) and \( 2x + 4 \ge 0 \)

For \( x – 3 \ge 0 \Rightarrow x \ge 3 \)
For \( 2x + 4 \ge 0 \Rightarrow x \ge -2 \)
Combining these, we get \( x \ge 3 \)

\( 4(x – 3) + 2 = 2x + 4 – 3 \)
\( 4x – 12 + 2 = 2x + 1 \)
\( 4x – 10 = 2x + 1 \)
\( 2x – 10 = 1 \)
\( 2x = 11 \)
\( x = \frac{11}{2} \)

Since \( x = \frac{11}{2} \), it satisfies our condition.

CASE 2: \( x – 3 \ge 0 \) and \( 2x + 4 < 0 \)

For \( x – 3 \ge 0 \Rightarrow x \ge 3 \)
For \( 2x + 4 < 0 \Rightarrow x < -2 \)
There is no overlap, so this case is not possible.

CASE 3: \( x – 3 < 0 \) and \( 2x + 4 \ge 0 \)

For \( x – 3 < 0 \Rightarrow x < 3 \)
For \( 2x + 4 \ge 0 \Rightarrow x \ge -2 \)
Combining these, we get \( -2 \le x < 3 \)

\( 4(-(x – 3)) + 2 = 2x + 4 – 3 \)
\( 4(-x + 3) + 2 = 2x + 1 \)
\( -4x + 12 + 2 = 2x + 1 \)
\( -4x + 14 = 2x + 1 \)
\( -4x – 2x = 1 – 14 \)
\( -6x = -13 \)
\( x = \frac{13}{6} \)

Since \( -2 \le \frac{13}{6} < 3 \), it satisfies our condition.

CASE 4: \( x – 3 < 0 \) and \( 2x + 4 < 0 \)

For \( x – 3 < 0 \Rightarrow x < 3 \)
For \( 2x + 4 < 0 \Rightarrow x < -2 \)
Combining these, we get \( x < -2 \)

\( 4(-(x – 3)) + 2 = -(2x + 4) – 3 \)
\( 4(-x + 3) + 2 = -2x – 4 – 3 \)
\( -4x + 12 + 2 = -2x – 7 \)
\( -4x + 14 = -2x – 7 \)
\( -4x + 2x = -7 – 14 \)
\( -2x = -21 \)
\( x = \frac{21}{2} \)

Since \( \frac{21}{2} \not< -2 \), this case is not possible.

SOLUTION

\( x = \frac{11}{2} \) or \( x = \frac{13}{6} \).