Linear Algebra Examples

$\frac{2}{7} \cdot (14 q + \frac{7}{2}) - 3 q = 9$

Step 1

Simplify $\frac{2}{7} \cdot (14 q + \frac{7}{2}) - 3 q$ .

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Step 1.1

Simplify each term.

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Step 1.1.1

Apply the distributive property.

$\frac{2}{7} (14 q) + \frac{2}{7} \cdot \frac{7}{2} - 3 q = 9$

Step 1.1.2

Cancel the common factor of $7$ .

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Step 1.1.2.1

Factor $7$ out of $14 q$ .

$\frac{2}{7} (7 (2 q)) + \frac{2}{7} \cdot \frac{7}{2} - 3 q = 9$

Step 1.1.2.2

Cancel the common factor.

$\frac{2}{7} (7 (2 q)) + \frac{2}{7} \cdot \frac{7}{2} - 3 q = 9$

Step 1.1.2.3

Rewrite the expression.

$2 (2 q) + \frac{2}{7} \cdot \frac{7}{2} - 3 q = 9$

Step 1.1.3

Multiply $2$ by $2$ .

$4 q + \frac{2}{7} \cdot \frac{7}{2} - 3 q = 9$

Step 1.1.4

Cancel the common factor of $2$ .

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Step 1.1.4.1

Cancel the common factor.

$4 q + \frac{2}{7} \cdot \frac{7}{2} - 3 q = 9$

Step 1.1.4.2

Rewrite the expression.

$4 q + \frac{1}{7} \cdot 7 - 3 q = 9$

Step 1.1.5

Cancel the common factor of $7$ .

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Step 1.1.5.1

Cancel the common factor.

$4 q + \frac{1}{7} \cdot 7 - 3 q = 9$

Step 1.1.5.2

Rewrite the expression.

$4 q + 1 - 3 q = 9$

Step 1.2

Subtract $3 q$ from $4 q$ .

$q + 1 = 9$

Step 2

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

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Step 2.1

Subtract $1$ from both sides of the equation.

$q = 9 - 1$

Step 2.2

Subtract $1$ from $9$ .

$q = 8$

Step 3

The domain is the set of all valid $q$ values.

${q | q = 8}$

Step 4