Algebra Examples

$\frac{| 4 x - 12 |}{2} = 8$

Step 1

Multiply both sides by $2$ .

$\frac{| 4 x - 12 |}{2} \cdot 2 = 8 \cdot 2$

Step 2

Simplify.

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Step 2.1

Simplify the left side.

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Step 2.1.1

Simplify $\frac{| 4 x - 12 |}{2} \cdot 2$ .

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Step 2.1.1.1

Simplify the numerator.

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Step 2.1.1.1.1

Factor $4$ out of $4 x - 12$ .

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Step 2.1.1.1.1.1

Factor $4$ out of $4 x$ .

$\frac{| 4 (x) - 12 |}{2} \cdot 2 = 8 \cdot 2$

Step 2.1.1.1.1.2

Factor $4$ out of $- 12$ .

$\frac{| 4 x + 4 \cdot - 3 |}{2} \cdot 2 = 8 \cdot 2$

Step 2.1.1.1.1.3

Factor $4$ out of $4 x + 4 \cdot - 3$ .

$\frac{| 4 (x - 3) |}{2} \cdot 2 = 8 \cdot 2$

Step 2.1.1.1.2

Apply the distributive property.

$\frac{| 4 x + 4 \cdot - 3 |}{2} \cdot 2 = 8 \cdot 2$

Step 2.1.1.1.3

Multiply $4$ by $- 3$ .

$\frac{| 4 x - 12 |}{2} \cdot 2 = 8 \cdot 2$

Step 2.1.1.1.4

Factor $4$ out of $4 x - 12$ .

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Step 2.1.1.1.4.1

Factor $4$ out of $4 x$ .

$\frac{| 4 (x) - 12 |}{2} \cdot 2 = 8 \cdot 2$

Step 2.1.1.1.4.2

Factor $4$ out of $- 12$ .

$\frac{| 4 x + 4 \cdot - 3 |}{2} \cdot 2 = 8 \cdot 2$

Step 2.1.1.1.4.3

Factor $4$ out of $4 x + 4 \cdot - 3$ .

$\frac{| 4 (x - 3) |}{2} \cdot 2 = 8 \cdot 2$

Step 2.1.1.2

Remove non-negative terms from the absolute value.

$\frac{4 | x - 3 |}{2} \cdot 2 = 8 \cdot 2$

Step 2.1.1.3

Cancel the common factor of $2$ .

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Step 2.1.1.3.1

Cancel the common factor.

$\frac{4 | x - 3 |}{2} \cdot 2 = 8 \cdot 2$

Step 2.1.1.3.2

Rewrite the expression.

$4 | x - 3 | = 8 \cdot 2$

Step 2.2

Simplify the right side.

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Step 2.2.1

Multiply $8$ by $2$ .

$4 | x - 3 | = 16$

Step 3

Solve for $x$ .

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Step 3.1

Divide each term in $4 | x - 3 | = 16$ by $4$ and simplify.

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Step 3.1.1

Divide each term in $4 | x - 3 | = 16$ by $4$ .

$\frac{4 | x - 3 |}{4} = \frac{16}{4}$

Step 3.1.2

Simplify the left side.

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Step 3.1.2.1

Cancel the common factor of $4$ .

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Step 3.1.2.1.1

Cancel the common factor.

$\frac{4 | x - 3 |}{4} = \frac{16}{4}$

Step 3.1.2.1.2

Divide $| x - 3 |$ by $1$ .

$| x - 3 | = \frac{16}{4}$

Step 3.1.3

Simplify the right side.

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Step 3.1.3.1

Divide $16$ by $4$ .

$| x - 3 | = 4$

Step 3.2

Remove the absolute value term. This creates a $\pm$ on the right side of the equation because $| x | = \pm x$ .

$x - 3 = \pm 4$

Step 3.3

The complete solution is the result of both the positive and negative portions of the solution.

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Step 3.3.1

First, use the positive value of the $\pm$ to find the first solution.

$x - 3 = 4$

Step 3.3.2

Move all terms not containing $x$ to the right side of the equation.

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Step 3.3.2.1

Add $3$ to both sides of the equation.

$x = 4 + 3$

Step 3.3.2.2

Add $4$ and $3$ .

$x = 7$

Step 3.3.3

Next, use the negative value of the $\pm$ to find the second solution.

$x - 3 = - 4$

Step 3.3.4

Move all terms not containing $x$ to the right side of the equation.

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Step 3.3.4.1

Add $3$ to both sides of the equation.

$x = - 4 + 3$

Step 3.3.4.2

Add $- 4$ and $3$ .

$x = - 1$

Step 3.3.5

The complete solution is the result of both the positive and negative portions of the solution.

$x = 7, - 1$