Linear Algebra Examples

$\frac{5}{3 y} + 1 \cdot (- \frac{3}{4 y}) + \frac{5}{2}$

Step 1

Set the denominator in $\frac{5}{3 y}$ equal to $0$ to find where the expression is undefined.

$3 y = 0$

Step 2

Divide each term in $3 y = 0$ by $3$ and simplify.

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Step 2.1

Divide each term in $3 y = 0$ by $3$ .

$\frac{3 y}{3} = \frac{0}{3}$

Step 2.2

Simplify the left side.

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Step 2.2.1

Cancel the common factor of $3$ .

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Step 2.2.1.1

Cancel the common factor.

$\frac{3 y}{3} = \frac{0}{3}$

Step 2.2.1.2

Divide $y$ by $1$ .

$y = \frac{0}{3}$

Step 2.3

Simplify the right side.

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Step 2.3.1

Divide $0$ by $3$ .

$y = 0$

Step 3

Set the denominator in $\frac{3}{4 y}$ equal to $0$ to find where the expression is undefined.

$4 y = 0$

Step 4

Divide each term in $4 y = 0$ by $4$ and simplify.

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Step 4.1

Divide each term in $4 y = 0$ by $4$ .

$\frac{4 y}{4} = \frac{0}{4}$

Step 4.2

Simplify the left side.

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Step 4.2.1

Cancel the common factor of $4$ .

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Step 4.2.1.1

Cancel the common factor.

$\frac{4 y}{4} = \frac{0}{4}$

Step 4.2.1.2

Divide $y$ by $1$ .

$y = \frac{0}{4}$

Step 4.3

Simplify the right side.

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Step 4.3.1

Divide $0$ by $4$ .

$y = 0$

Step 5

The domain is all values of $y$ that make the expression defined.

Interval Notation:

$(- \infty, 0) \cup (0, \infty)$

Set-Builder Notation:

${y | y \neq 0}$

Step 6