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 $$x = \frac{13}{6}$$.