Absolute Value Equation Calculator

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

Type of Equation:

One Absolute Value Two Absolute Values

a∙|b∙x + c| + d = e∙x + f

a =

b =

c =

d =

e =

f =

a∙|b∙x + c| + d = e∙|f∙x + g| + h

a =

b =

c =

d =

e =

f =

g =

h =

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 $\frac{21}{2}\nless -2$, this case is not possible.

SOLUTION

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