Basic Math Examples

$| \frac{1}{5} d - 15 | = 6$

Step 1

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

$\frac{1}{5} d - 15 = \pm 6$

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.

$\frac{1}{5} d - 15 = 6$

Step 2.2

Combine $\frac{1}{5}$ and $d$ .

$\frac{d}{5} - 15 = 6$

Step 2.3

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

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Step 2.3.1

Add $15$ to both sides of the equation.

$\frac{d}{5} = 6 + 15$

Step 2.3.2

Add $6$ and $15$ .

$\frac{d}{5} = 21$

Step 2.4

Multiply both sides of the equation by $5$ .

$5 \frac{d}{5} = 5 \cdot 21$

Step 2.5

Simplify both sides of the equation.

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Step 2.5.1

Simplify the left side.

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Step 2.5.1.1

Cancel the common factor of $5$ .

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Step 2.5.1.1.1

Cancel the common factor.

$5 \frac{d}{5} = 5 \cdot 21$

Step 2.5.1.1.2

Rewrite the expression.

$d = 5 \cdot 21$

Step 2.5.2

Simplify the right side.

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Step 2.5.2.1

Multiply $5$ by $21$ .

$d = 105$

Step 2.6

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

$\frac{1}{5} d - 15 = - 6$

Step 2.7

Combine $\frac{1}{5}$ and $d$ .

$\frac{d}{5} - 15 = - 6$

Step 2.8

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

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Step 2.8.1

Add $15$ to both sides of the equation.

$\frac{d}{5} = - 6 + 15$

Step 2.8.2

Add $- 6$ and $15$ .

$\frac{d}{5} = 9$

Step 2.9

Multiply both sides of the equation by $5$ .

$5 \frac{d}{5} = 5 \cdot 9$

Step 2.10

Simplify both sides of the equation.

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Step 2.10.1

Simplify the left side.

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Step 2.10.1.1

Cancel the common factor of $5$ .

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Step 2.10.1.1.1

Cancel the common factor.

$5 \frac{d}{5} = 5 \cdot 9$

Step 2.10.1.1.2

Rewrite the expression.

$d = 5 \cdot 9$

Step 2.10.2

Simplify the right side.

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Step 2.10.2.1

Multiply $5$ by $9$ .

$d = 45$

Step 2.11

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

$d = 105, 45$