Algebra Examples

$| - 3 p | = 15$

Step 1

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

$- 3 p = \pm 15$

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 $\pm$ to find the first solution.

$- 3 p = 15$

Step 2.2

Divide each term in $- 3 p = 15$ by $- 3$ and simplify.

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Step 2.2.1

Divide each term in $- 3 p = 15$ by $- 3$ .

$\frac{- 3 p}{- 3} = \frac{15}{- 3}$

Step 2.2.2

Simplify the left side.

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Step 2.2.2.1

Cancel the common factor of $- 3$ .

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Step 2.2.2.1.1

Cancel the common factor.

$\frac{- 3 p}{- 3} = \frac{15}{- 3}$

Step 2.2.2.1.2

Divide $p$ by $1$ .

$p = \frac{15}{- 3}$

Step 2.2.3

Simplify the right side.

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Step 2.2.3.1

Divide $15$ by $- 3$ .

$p = - 5$

Step 2.3

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

$- 3 p = - 15$

Step 2.4

Divide each term in $- 3 p = - 15$ by $- 3$ and simplify.

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Step 2.4.1

Divide each term in $- 3 p = - 15$ by $- 3$ .

$\frac{- 3 p}{- 3} = \frac{- 15}{- 3}$

Step 2.4.2

Simplify the left side.

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Step 2.4.2.1

Cancel the common factor of $- 3$ .

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Step 2.4.2.1.1

Cancel the common factor.

$\frac{- 3 p}{- 3} = \frac{- 15}{- 3}$

Step 2.4.2.1.2

Divide $p$ by $1$ .

$p = \frac{- 15}{- 3}$

Step 2.4.3

Simplify the right side.

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Step 2.4.3.1

Divide $- 15$ by $- 3$ .

$p = 5$

Step 2.5

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

$p = - 5, 5$