Precalculus Examples

| 2 x - 4 | = 10

$| 2 x - 4 | = 10$

Step 1

Remove the absolute value term. This creates a

\pm

$\pm$ on the right side of the equation because

| x | = \pm x

$| x | = \pm x$ .

2 x - 4 = \pm 10

$2 x - 4 = \pm 10$

Step 2

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

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Step 2.1

First, use the positive value of the

\pm

$\pm$ to find the first solution.

2 x - 4 = 10

$2 x - 4 = 10$

Step 2.2

Move all terms not containing

x

$x$ to the right side of the equation.

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Step 2.2.1

Add

4

$4$ to both sides of the equation.

2 x = 10 + 4

$2 x = 10 + 4$

Step 2.2.2

Add

10

$10$ and

4

$4$ .

2 x = 14

$2 x = 14$

2 x = 14

$2 x = 14$

Step 2.3

Divide each term in

2 x = 14

$2 x = 14$ by

2

$2$ and simplify.

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Step 2.3.1

Divide each term in

2 x = 14

$2 x = 14$ by

2

$2$ .

\frac{2 x}{2} = \frac{14}{2}

$\frac{2 x}{2} = \frac{14}{2}$

Step 2.3.2

Simplify the left side.

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Step 2.3.2.1

Cancel the common factor of

2

$2$ .

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Step 2.3.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{14}{2}$

Step 2.3.2.1.2

Divide

$x$ by

$1$ .

$x = \frac{14}{2}$

Step 2.3.3

Simplify the right side.

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Step 2.3.3.1

Divide

$14$ by

$2$ .

$x = 7$

Step 2.4

Next, use the negative value of the

$\pm$ to find the second solution.

$2 x - 4 = - 10$

Step 2.5

Move all terms not containing

$x$ to the right side of the equation.

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Step 2.5.1

Add

$4$ to both sides of the equation.

$2 x = - 10 + 4$

Step 2.5.2

Add

$- 10$ and

$4$ .

$2 x = - 6$

Step 2.6

Divide each term in

$2 x = - 6$ by

$2$ and simplify.

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Step 2.6.1

Divide each term in

$2 x = - 6$ by

$2$ .

$\frac{2 x}{2} = \frac{- 6}{2}$

Step 2.6.2

Simplify the left side.

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Step 2.6.2.1

Cancel the common factor of

$2$ .

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Step 2.6.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{- 6}{2}$

Step 2.6.2.1.2

Divide

$x$ by

$1$ .

$x = \frac{- 6}{2}$

Step 2.6.3

Simplify the right side.

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Step 2.6.3.1

Divide

$- 6$ by

$2$ .

$x = - 3$

Step 2.7

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

$x = 7, - 3$