Algebra Examples

$\frac{y^{2}}{y - 3} \cdot \frac{y^{2} - y - 6}{y^{2} + 1 y}$

Step 1

Set the denominator in $\frac{y^{2}}{y - 3}$ equal to $0$ to find where the expression is undefined.

$y - 3 = 0$

Step 2

Add $3$ to both sides of the equation.

$y = 3$

Step 3

Set the denominator in $\frac{y^{2} - y - 6}{y^{2} + 1 y}$ equal to $0$ to find where the expression is undefined.

$y^{2} + 1 y = 0$

Step 4

Solve for $y$ .

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Step 4.1

Multiply $y$ by $1$ .

$y^{2} + y = 0$

Step 4.2

Factor $y$ out of $y^{2} + y$ .

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Step 4.2.1

Factor $y$ out of $y^{2}$ .

$y \cdot y + y = 0$

Step 4.2.2

Raise $y$ to the power of $1$ .

$y \cdot y + y = 0$

Step 4.2.3

Factor $y$ out of $y^{1}$ .

$y \cdot y + y \cdot 1 = 0$

Step 4.2.4

Factor $y$ out of $y \cdot y + y \cdot 1$ .

$y (y + 1) = 0$

Step 4.3

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$y = 0$

$y + 1 = 0$

Step 4.4

Set $y$ equal to $0$ .

$y = 0$

Step 4.5

Set $y + 1$ equal to $0$ and solve for $y$ .

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Step 4.5.1

Set $y + 1$ equal to $0$ .

$y + 1 = 0$

Step 4.5.2

Subtract $1$ from both sides of the equation.

$y = - 1$

Step 4.6

The final solution is all the values that make $y (y + 1) = 0$ true.

$y = 0, - 1$

Step 5

The equation is undefined where the denominator equals $0$ , the argument of a square root is less than $0$ , or the argument of a logarithm is less than or equal to $0$ .

$y = - 1, y = 0, y = 3$

Step 6