Linear Algebra Examples

$c + \frac{c - 3}{2} = 15$

Step 1

Simplify $c + \frac{c - 3}{2}$ .

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Step 1.1

To write $c$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$c \cdot \frac{2}{2} + \frac{c - 3}{2} = 15$

Step 1.2

Simplify terms.

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Step 1.2.1

Combine $c$ and $\frac{2}{2}$ .

$\frac{c \cdot 2}{2} + \frac{c - 3}{2} = 15$

Step 1.2.2

Combine the numerators over the common denominator.

$\frac{c \cdot 2 + c - 3}{2} = 15$

Step 1.3

Simplify the numerator.

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Step 1.3.1

Add $c \cdot 2$ and $c$ .

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Step 1.3.1.1

Reorder $c$ and $2$ .

$\frac{2 \cdot c + c - 3}{2} = 15$

Step 1.3.1.2

Add $2 \cdot c$ and $c$ .

$\frac{3 \cdot c - 3}{2} = 15$

Step 1.3.2

Factor $3$ out of $3 \cdot c - 3$ .

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Step 1.3.2.1

Factor $3$ out of $3 \cdot c$ .

$\frac{3 (c) - 3}{2} = 15$

Step 1.3.2.2

Factor $3$ out of $- 3$ .

$\frac{3 (c) + 3 (- 1)}{2} = 15$

Step 1.3.2.3

Factor $3$ out of $3 (c) + 3 (- 1)$ .

$\frac{3 (c - 1)}{2} = 15$

Step 2

Multiply both sides of the equation by $\frac{2}{3}$ .

$\frac{2}{3} \cdot \frac{3 (c - 1)}{2} = \frac{2}{3} \cdot 15$

Step 3

Simplify both sides of the equation.

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Step 3.1

Simplify the left side.

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Step 3.1.1

Simplify $\frac{2}{3} \cdot \frac{3 (c - 1)}{2}$ .

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Step 3.1.1.1

Cancel the common factor of $2$ .

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Step 3.1.1.1.1

Cancel the common factor.

$\frac{2}{3} \cdot \frac{3 (c - 1)}{2} = \frac{2}{3} \cdot 15$

Step 3.1.1.1.2

Rewrite the expression.

$\frac{1}{3} (3 (c - 1)) = \frac{2}{3} \cdot 15$

Step 3.1.1.2

Cancel the common factor of $3$ .

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Step 3.1.1.2.1

Cancel the common factor.

$\frac{1}{3} (3 (c - 1)) = \frac{2}{3} \cdot 15$

Step 3.1.1.2.2

Rewrite the expression.

$c - 1 = \frac{2}{3} \cdot 15$

Step 3.2

Simplify the right side.

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Step 3.2.1

Simplify $\frac{2}{3} \cdot 15$ .

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Step 3.2.1.1

Cancel the common factor of $3$ .

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Step 3.2.1.1.1

Factor $3$ out of $15$ .

$c - 1 = \frac{2}{3} \cdot (3 (5))$

Step 3.2.1.1.2

Cancel the common factor.

$c - 1 = \frac{2}{3} \cdot (3 \cdot 5)$

Step 3.2.1.1.3

Rewrite the expression.

$c - 1 = 2 \cdot 5$

Step 3.2.1.2

Multiply $2$ by $5$ .

$c - 1 = 10$

Step 4

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

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Step 4.1

Add $1$ to both sides of the equation.

$c = 10 + 1$

Step 4.2

Add $10$ and $1$ .

$c = 11$

Step 5

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

${c | c = 11}$

Step 6